locally most powerful test for the random coefficient
TRANSCRIPT
Research ArticleLocally Most Powerful Test for the RandomCoefficient Autoregressive Model
Li Bi1 Feilong Lu1 Kai Yang2 and Dehui Wang 1
1School of Mathematics Jilin University Changchun 130012 China2School of Mathematics and Statistics Changchun University of Technology Changchun 130012 China
Correspondence should be addressed to Dehui Wang wangdhjlueducn
Received 3 April 2019 Accepted 17 June 2019 Published 27 June 2019
Academic Editor Jixiang Yang
Copyright copy 2019 Li Bi et al This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
In this article we study the problem of testing the constancy of the coefficient in a class of stationary first-order random coefficientautoregressive (RCAR(1)) model We construct a new test statistic based on the locally most powerful-type (LMP) test Under thenull hypothesis we derive the limiting distribution of the proposed test statistic In the simulation we compare the power betweenLMP test and empirical likelihood (EL) test and find that the accuracy of using LMP is 67 288 and 261 higher than that ofEL test under normal studentrsquos 119905 and symmetric contamination errors respectively A real life data is given to illustrate the practicaleffectiveness of our test
1 Introduction
In time series analysis autoregressive and linear processes arewidely used due to their mathematical tractability In fact theautoregressive model has two prominently advantages theirestimation procedures are well established and the existenceof stationary solutions are easily derived Therefore thismodel has been investigated in the field of signal detectionand classification psychometry and biomedical engineeringSee for example Hoque [1] Ogawa et al [2] Subasi et al[3] andMaleki et al [4]The first-order autoregressive model(AR(1)) is defined as119883119905 = 120572119883119905minus1 + 120576119905 119905 ge 1 (1)
where 120576119905 is an independent and identically distributed(iid) random error sequence with probability density func-tion 119891120576 E(120576119905) = 0 Var(120576119905) = 1205902120576 and 120576119905 independent of1198830 for all 119905 However it has been found that a variety ofdata sets cannot be modelled precisely by assuming linearityFinancial data for instance present heteroscedasticity andbiological data suffer from random perturbations To addressthis problem Conlisk [5] considered a random coefficientautoregressive model defined as follows119883119905 = 120572119905119883119905minus1 + 120576119905 119905 ge 1 (2)
where 120572119905 is a sequence of iid randomvariables with E(120572119905 ) =120572 and Var(120572119905) = 1205902120572 120576119905 is an iid random error sequencewith mean zero and variance 1205902120576 The random variable 1198830 isassumed to be independent of 120572119905 which is independent of120576119905 Note that when 1205902120572 = 0 the RCAR(1) model reduces tothe standard AR(1) model
The RCAR(1) models have been widely applied in math-ematical literature due to its strong application value inpractice For instance Nicholls and Quinn [6ndash8] derivedthe least square (LS) estimates of the model parameterand showed that the LS estimates are strongly consistentand obey the central limit theorem Brandt [9] restated thenecessary and sufficient conditions for the existence of a strictstationarity and ergodicity solution in the RCAR(1) modelWang and Ghosh [10] considered the Basyesian estimationmethod to estimate RCAR model parameter and exploredthe frequentist properties of the Bayes estimator Wang etal [11] obtained the asymptotic properties of the maximumlikelihood estimators of these parameters in RCAR(1) modelunder the unit root assumption
More recently the problem of testing traditional AR(1)model against RCAR(1) model is thus of essential issue inthis settings This detection problem was firstly investigatedby Nicholls and Quinn [8] where a Gaussian Lagrange
HindawiMathematical Problems in EngineeringVolume 2019 Article ID 6593821 11 pageshttpsdoiorg10115520196593821
2 Mathematical Problems in Engineering
multiplier test is derived for the problem However thislikelihood ratio approach has several weaknesses as the truevalue of the parameter under the null hypothesis lies on theboundary of the parameter space the asymptotics will not beeasily established Ramanathan and Rajarshi [12] considereda non-Gaussian test method to handle the change problemfor parameters in RCAR(1) case Their signed rank testhowever require a symmetry assumption on the innovationdensity which is highly unnatural in this context Still forthe RCAR(1) model Lee [13] proposed the cumulative sumtest for parameter change in RCAR(1) model But it detectsa change only in a function of the parameter rather thanthe parameter themselves Recently Moreno and Romo [14]studied robust unit root test with autoregressive errors Theunit root test shows some size distortions and requires manyestimates in the construction of test
To deal with these drawbacks we construct a locallymost powerful-type (LMP) test for testing the constancy ofparameters in the RCAR(1) model The LMP test has beeninvestigated by several authors Rohatgi et al [15] Chikk-agoudar and Biradar [16] and Manik et al [17] consideredthe LMP test for parameter constancy The test has meritin that its calculation process is less cumbersome and itproduces stable sizes especially when parameter nearby thetrue value In contrast the traditional tests mentioned aboveshow some size distortions and require many estimates in theconstruction of tests see Ramanathan and Rajarshi [12] Ina general way our test can conventionally discard correlationeffects and enhance the performance of the test In this paperwe illustrate how our proposed method can be implementedfor finite samples under normal studentrsquos t and symmetriccontamination errors (see Huber [18] and Hettmansperger[19]) Through numerical simulation studies we can see thatour test has a stronger power in terms of maintaining theempirical level and power than the empirical likelihood (EL)test suggested by Zhao et al [20]
The outline of this paper is organized as follows InSection 2 we introduce our test statistic and derive its limitingdistribution under the null hypothesis Numerical simula-tions to evaluate the empirical size and power of our testtechnique are discussed in Section 3 Section 4 presents a reallife data example to illustrate the superior of ourmethodsWeprovide brief concluding remarks in Conclusions Appendixprovides the proofs of the main results
2 Methodology and Main Results
In this section we will construct a test statistic to test whether120572119905 is a constant To achieve this task we set up the null andalternative hypotheses 1198670 1205902120572 = 0
v119904 1198671 1205902120572 gt 0 (3)
In what follows we give our main results Suppose that thetime series data 1199091 1199092 119909119899 are generated from (2) Let120572119905 be an iid sequence of random variables with commonprobability distribution 119865120572 120576119905 is an iid sequence of randomvariableswith density function119891120576 In this article we regard1198830
as a givennumber alternatively if1198830 is a randomvariable weshall consider only inferences conditionally on 1198830 are fixedNote that 1198830 120572119905 and 120576119905 are independent
Next we assume the following conditions to establish theasymptotic properties of the test statistic
(C1) 1205722 + 1205902120572 lt 1(C2)The distribution 119865120572 of 120572119905 is such that 119864|120572119905|3 lt infin(C3) The third and mixed derivatives of log119891120576 with
respect to 120572 and 1205902120576 are uniformly bounded on (120572 1205902120576 )(C4) Differentiation thrice with respect to (120572 1205902120576 ) of 119891120576
under the integration is boundedIt is easy to derive that119883119905 is aMarkov chain on 0 1 2 sdot sdot sdot
with the following transition probabilities
119891119883119905|119883119905minus1 (119909119905 | 119909119905minus1) = int119891119883119905|119883119905minus1 (119909119905 | 119909119905minus1 120572119905) 119889119865120572= int119891120576 (119909119905 minus 120572119905119909119905minus1) 119889119865120572 (4)
The Markovity follows from (2) and the fact that 120572119905 is aniid sequence The conditional density 119891119883119905|119883119905minus1(119909119905 | 119909119905minus1)is obtained by noting that conditional on 120572119905 Conditionalon (119883119905minus1 120572119905) and 120576119905 is an iid sequence with the density119891120576 hence the probability density function of 119883119905 is given by119891120576(119909119905 minus 120572119905119909119905minus1) So we get the result in the above equation
Therefore we can write down the likelihood function 1198711198671for RCAR(1) model
1198711198671 (1199091 119909119905) = 1198911198830 (1199090) 119899prod119905=1
int119891120576 (119909119905 minus 120572119905119909119905minus1) 119889119865120572 (5)
Furthermore employing this spirit of LMP approach intro-duced by Manik et al [17] we can obtain our test statistic
119876119899 (120573) = 119899sum119905=1
12 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) = 119899sum119905=1119877119905 (120573) (6)
where
119877119905 (120573) = 12 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) (7)
And the detailed procedures can be found in Appendix ALet120573 = (120572 1205902120576 )119879 denote vector of all unknown parameters
of the RCAR(1) process Hereafter we use the notation = ( 1205902120576 ) to represent the maximum likelihood estimator(MLE) of 120573 In addition the actual test statistic can beobtained by replacing 120573 with their MLE It can be easilyderived that 119877119905(120573)F119877119905 is a zero mean martingale
Before we state our main results the following assump-tions will be made
(A1) sup119905119864|119877119905(120573)|2+120578 lt infin for some 120578 gt 0Then about the asymptotic distribution of119876119899() we have
the following theorem
Theorem 1 Let 119909119905 be a sequence of strictly stationaryergodic and F119905-measurable solutions to equation (2) under
Mathematical Problems in Engineering 3
Table 1 Empirical sizes of LMP and EL tests at nominal level 005 for model N1
119873 (120590119873 120572)(05 03) (05 05) (05 07) (05 09) (1 03) (1 05) (1 07) (1 09)100 MP 0095 0105 0109 0101 0116 0103 0089 0098
EL 0008 0004 0002 0003 0003 0002 0003 0002200 LMP 0089 0082 0085 0086 0085 0082 0083 0087
EL 0006 0007 0004 0005 0004 0004 0003 0003300 LMP 0074 0077 0082 0069 0076 0071 0076 0074
EL 0006 0009 0007 0006 0005 0007 0008 0009400 LMP 0063 0070 0075 0068 0067 0063 0070 0072
EL 0007 0009 0006 0004 0007 0008 0005 0004500 LMP 0066 0060 0077 0060 0065 0062 0060 0055
EL 0009 0005 0006 0005 0006 0009 0014 0008600 LMP 0062 0057 0067 0053 0065 0060 0059 0068
EL 0009 0005 0011 0007 0009 0007 0008 0007700 LMP 0064 0059 0068 0056 0057 0062 0058 0063
EL 0005 0010 0008 0007 0007 0008 0005 0003800 LMP 0060 0061 0060 0056 0058 0055 0055 0063
EL 0009 0007 0007 0010 0011 0012 0008 0009900 LMP 0055 0062 0054 0048 0053 0053 0062 0062
EL 0015 0015 0008 0010 0011 0007 0010 00091000 LMP 0063 0051 0057 0052 0058 0062 0057 0054
EL 0019 0009 0012 0009 0011 0012 0007 00095000 LMP 0051 0051 0051 0053 0050 0048 0049 0053
EL 0020 0022 0018 0020 0016 0021 0024 0022
conditions of (C1) minus (C4) and above (A1) Then under 1198670we have asymptotic normality
(radic119899)minus1119876119899 () 119889997888rarr 119873(0 2) (8)
here 2 = 1205902 minus119882119879Σminus1119882(The expressions for 1205902 and Σminus1 are derived in
Appendix B)
3 Simulation
In this section we carry out some simulation studies tocompare performances of the locally most powerful-typetest and the empirical likelihood test in terms of empiricalsize and power The empirical size and power for the twotests in Tables 1ndash6 are based on 1000 repetitions with thehelp of R software Within each study we set the initialvalues 1198830 equiv 1 and employ the significance level at 005Throughout this simulation we used notation LMP for locallymost powerful-type test by our algorithm EL for empiricallikelihood method
31 Empirical Size To calculate empirical sizes we pay ourattention to three kinds of model (1) defined as follows
(N1) 120572119905 = 120572 120576119905 sim 119873(0 1205902119873)(N2) 120572119905 = 120572 120576119905 sim 119905(119898)
(N3) 120572119905 = 120572 120576119905 sim F120576(119909)The distribution function of 120585-contamination is
F120576 (119909) = 120585Φ( 1199091205901) + (1 minus 120585)Φ( 1199091205902) (9)
where 120585 is a fixed constant satisfying 0 lt 120585 lt 1 andΦ(119909) is thedistribution function of standard normal random variable120590119894 gt 0 119894 = 1 2
Let us now describe how the simulated results areobtained First of all we use the model N1 to generate datawhen the 120590119873 = 05 1 and 120572 = 03 05 07 09 Secondly wegenerate data from model N2 with the degrees of freedom119898 = 8 10 and 120572 = 03 05 07 09 For model N1 and N2we take the sample size 119873 = 100 200 300 400 500 600700 800 900 1000 and 5000 Thirdly we simulate samplesfrommodel N3 with the (120585 1205901 1205902 ) = (08 1 3) (120585 1205901 1205902 ) =(09 1 3) and 120572 = 03 05 07 09 For model N3 we set thesample size119873 = 100 300 500 1000 and 5000 The empiricallevels of threemodels are presented inTables 1ndash3 respectivelyAs seen from Tables 1ndash3 the LMP test give similar resultsunder normal studentrsquos t and symmetric contaminationerrors In addition the LMP test produces sizes closer to thenominal significance level 005 quite satisfactorily especiallyfor larger sample sizes However the EL method has lowerlevels for each sample size so one may make wrong decisionsin tests based on them Looking at three models results wecan conclude that our method has a greater effect than the ELtest in terms of the empirical level
4 Mathematical Problems in Engineering
Table 2 Empirical sizes of LMP and EL tests at nominal level 005 for model N2
119873 (119898 120572)(8 03) (8 05) (8 07) (8 09) (10 03) (10 05) (10 07) (10 09)100 MP 0060 0065 0066 0061 0065 0068 0075 0067
EL 0003 0003 0002 0003 0004 0003 0004 0001200 LMP 0059 0061 0065 0059 0066 0063 0068 0061
EL 0002 0004 0002 0002 0004 0004 0002 0004300 LMP 0058 0057 0059 0055 0063 0058 0063 0055
EL 0005 0004 0003 0009 0006 0006 0004 0002400 LMP 0058 0055 0058 0055 0062 0054 0061 0058
EL 0004 0003 0005 0003 0004 0003 0005 0006500 LMP 0056 0055 0056 0052 0057 0059 0055 0054
EL 0006 0004 0005 0005 0007 0005 0006 0007600 LMP 0060 0057 0058 0060 0062 0059 0056 0053
EL 0003 0005 0005 0005 0008 0007 0009 0006700 LMP 0055 0056 0058 0057 0060 0060 0056 0058
EL 0006 0005 0005 0008 0006 0006 0004 0008800 LMP 0050 0054 0054 0052 0056 0054 0054 0054
EL 0005 0007 0008 0006 0006 0008 0008 0008900 LMP 0051 0052 0058 0056 0056 0060 0049 0051
EL 0005 0006 0006 0009 0006 0008 0010 00051000 LMP 0052 0056 0057 0055 0050 0053 0050 0054
EL 0007 0008 0005 0006 0012 0009 0009 00075000 LMP 0052 0050 0050 0050 0051 0049 0054 0052
EL 0015 0014 0012 0016 0017 0014 0012 0013
Table 3 Empirical sizes of LMP and EL tests at nominal level 005 for model N3
119873 (1205901 1205902 120572)(1 3 01) (1 3 02) (1 3 03) (1 3 04) (1 3 05) (1 3 06) (1 3 07) (1 3 08) (1 3 09)120585 = 08100 LMP 0064 0068 0065 0072 0053 0078 0071 0053 0047
EL 0000 0000 0001 0000 0000 0001 0000 0000 0001300 LMP 0045 0066 0053 0038 0054 0081 0051 0043 0040
EL 0000 0001 0000 0001 0001 0002 0000 0001 0000500 LMP 0058 0057 0057 0049 0050 0060 0043 0049 0044
EL 0001 0000 0001 0000 0000 0001 0000 0000 00011000 LMP 0050 0059 0052 0051 0045 0054 0036 0043 0036
EL 0002 0001 0001 0001 0001 0001 0003 0001 00015000 LMP 0053 0050 0054 0050 0048 0050 0047 0050 0046
EL 0003 0004 0003 0007 0007 0006 0008 0006 0011120585 = 09100 LMP 0078 0081 0089 0069 0089 0095 0083 0091 0086
EL 0000 0002 0001 0000 0000 0000 0000 0000 0001300 LMP 0080 0070 0074 0070 0070 0056 0063 0057 0055
EL 0000 0000 0001 0000 0000 0001 0002 0000 0001500 LMP 0072 0081 0068 0067 0058 0059 0060 0046 0053
EL 0001 0000 0001 0000 0000 0001 0001 0000 00041000 LMP 0064 0061 0067 0067 0055 0055 0052 0051 0045
EL 0000 0000 0000 0001 0002 0003 0003 0000 00015000 LMP 0055 0053 0050 0050 0055 0046 0050 0047 0050
EL 0004 0002 0003 0002 0002 0006 0003 0003 0009
Mathematical Problems in Engineering 5
Table 4 Empirical powers of LMP and EL tests at nominal level 005 for model A1
119873 (120590119873 119886 119887)(05 05 05) (05 05 1) (05 1 05) (05 1 1) (1 05 05) (1 05 1) (1 1 05) (1 1 1)100 LMP 0165 0102 0171 0110 0171 0107 0174 0090
EL 0055 0024 0040 0023 0050 0022 0041 0025200 LMP 0318 0143 0294 0147 0320 0140 0311 0160
EL 0191 0070 0156 0073 0202 0049 0155 0067300 LMP 0513 0209 0494 0253 0511 0175 0449 0241
EL 0405 0144 0332 0173 0397 0124 0294 0168400 LMP 0674 0232 0561 0330 0663 0264 0590 0338
EL 0578 0176 0456 0272 0559 0189 0450 0260500 LMP 0791 0377 0720 0427 0770 0374 0725 0438
EL 0722 0298 0629 0366 0705 0303 0625 0374600 LMP 0842 0415 0792 0521 0832 0424 0790 0508
EL 0798 0352 0717 0460 0793 0365 0712 0458700 LMP 0912 0483 0862 0598 0919 0478 0855 0598
EL 0882 0429 0809 0541 0885 0423 0794 0536800 LMP 0964 0560 0915 0645 0947 0568 0918 0643
EL 0938 0512 0877 0604 0928 0516 0872 0599900 LMP 0967 0633 0936 0727 0971 0623 0955 0703
EL 0952 0586 0910 0696 0959 0586 0933 06871000 LMP 0988 0712 0963 0799 0978 0702 0962 0776
EL 0981 0678 0942 0774 0969 0676 0938 07495000 LMP 1000 1000 1000 1000 1000 1000 1000 1000
EL 1000 1000 1000 1000 1000 1000 1000 1000
Table 5 Empirical powers of LMP and EL tests at nominal level 005 for model A2119873 (119898 119886 119887)(6 05 05) (6 05 1) (6 1 05) (6 1 1) (10 05 05) (10 05 1) (10 1 05) (10 1 1)100 LMP 0289 0110 0220 0149 0241 0096 0183 0131
EL 0045 0016 0027 0019 0047 0018 0027 0029200 LMP 0596 0252 0519 0310 0525 0199 0452 0272
EL 0161 0052 0118 0055 0186 0057 0146 0063300 LMP 0790 0400 0722 0475 0762 0351 0686 0429
EL 0336 0111 0276 0124 0373 0124 0293 0152400 LMP 0899 0538 0861 0644 0862 0462 0815 0564
EL 0486 0147 0399 0196 0520 0173 0440 0223500 LMP 0964 0624 0933 0757 0957 0596 0906 0699
EL 0622 0218 0529 0285 0695 0262 0590 0345600 LMP 0979 0701 0951 0819 0965 0657 0947 0754
EL 0708 0237 0606 0351 0767 0301 0650 0395700 LMP 0995 0791 0974 0862 0983 0727 0978 0824
EL 0814 0351 0669 0404 0839 0389 0773 0483800 LMP 0996 0837 0990 0927 0994 0809 0984 0882
EL 0845 0404 0767 0512 0895 0488 0839 0585900 LMP 0999 0874 0995 0934 0996 0831 0993 0924
EL 0879 0430 0801 0548 0929 0514 0877 06221000 LMP 1000 0936 1000 0971 1000 0889 0996 0951
EL 0913 0515 0887 0626 0946 0601 0930 07065000 LMP 1000 1000 1000 1000 1000 1000 1000 1000
EL 0992 0967 0994 0985 1000 1000 1000 0999
6 Mathematical Problems in Engineering
Table 6 Empirical powers of LMP and EL tests at nominal level 005 for model A3 with (1205901 1205902) = (1 3)119873 (119886 119887)(05 05) (05 1) (05 15) (1 05) (1 1) (1 15) (15 05) (15 1) (15 15)120585 = 08100 LMP 0141 0054 0034 0120 0073 0038 0084 0082 0046
EL 0039 0009 0006 0026 0012 0008 0018 0012 0009300 LMP 0775 0388 0183 0628 0493 0299 0475 0448 0331
EL 0325 0102 0030 0231 0097 0052 0143 0114 0061500 LMP 0959 0675 0381 0880 0791 0594 0754 0729 0583
EL 0656 0236 0070 0500 0268 0143 0347 0222 01311000 LMP 1000 0967 0753 0996 0992 0934 0982 0982 0939
EL 0945 0599 0255 0891 0673 0399 0720 0609 04025000 LMP 1000 1000 1000 1000 1000 1000 1000 1000 1000
EL 1000 1000 0981 1000 1000 0997 0999 1000 0997120585 = 09100 LMP 0118 0043 0035 0071 0041 0025 0049 0052 0037
EL 0042 0013 0015 0032 0021 0014 0020 0013 0011300 LMP 0673 0278 0101 0545 0372 0190 0388 0313 0220
EL 0301 0077 0021 0199 0101 0043 0128 0093 0055500 LMP 0914 0558 0277 0868 0681 0423 0700 0622 0467
EL 0571 0183 0054 0486 0239 0128 0353 0212 01131000 LMP 1000 0929 0627 0995 0965 0849 0970 0954 0865
EL 0927 0503 0204 0852 0637 0375 0702 0548 03685000 LMP 1000 1000 1000 1000 1000 1000 1000 1000 1000
EL 1000 1000 0963 1000 1000 0996 1000 1000 0994
32 Empirical Power In order to investigate the empiricalpowers we consider the alternatives under three versions ofmodel (2)
(A1) 120572119905 sim 119861119890119905119886(119886 119887) 120576119905 sim 119873(0 1205902119873)(A2) 120572119905 sim 119861119890119905119886(119886 119887) 120576119905 sim 119905(119898)(A3)120572119905 sim 119861119890119905119886(119886 119887) 120576119905 sim F120576(119909)Thedistribution function
of 120585-contamination is
F120576 (119909) = 120585Φ( 1199091205901) + (1 minus 120585)Φ( 1199091205902) (10)
where 120585 is a fixed constant satisfying 0 lt 120585 lt 1 andΦ(119909) is thedistribution function of standard normal random variable120590119894 gt 0 119894 = 1 2
To calculate the empirical powers of the two tests inthe first we generate samples from the model A1 in whichthe parameters 119886 = 05 1 119887 = 05 1 and 120590119873 = 05 1In the second setting we simulate samples from model A2with the degrees of freedom 119898 = 6 10 the parameters119886 = 05 1 and 119887 = 05 1 For model A1 and A2 we takethe sample size 119873 = 100 200 300 400 500 600 700800 900 1000 and 5000 In the last case we generate datafrom the model A3 in which the (120585 1205901 1205902 ) = (08 1 3)(120585 1205901 1205902 ) = (09 1 3) and the parameters 119886 = 05 1 15119887 = 05 1 15 For model A3 we employ the sample size119873 = 100 300 500 1000 and 5000 The empirical power ofthe above three models are summed up in Tables 4ndash6 Inthree cases the power of two test statistics increase with thesample size while our test produces relatively better powers
than the EL test Furthermore both tests powers are closeto 1 at the sample size 119873 = 5000 Overall from thesesimulations we conclude that the accuracy of using LMP is67 288 261 higher than that of EL test under normalstudentrsquos 119905 and symmetric contamination errors respectivelyAs anticipated our finding shows that the LMP test is afunctional tool to detect a parameter change for RCAR(1)model Therefore we recommend the LMP for practicaluse because its computation is not difficult and the overallperformance is better than that EL under normal studentrsquos tand symmetric contamination errors
4 A Real Life Data Analysis
In this section we illustrate how the LMP method can beapplied to a practical application This data consist of 78monthly number of annulus growth rate of the import billin Australia starting in December 2010 and ending in May2017 The data are available online at the CEInet StatisticsDatabase site httpdbceicnpageDefaultaspx The meanand variance of the data are found to be 02069 and 153347The data are denoted as 1199101 1199102 11991078 Figure 1 is the samplepath plot for the real data 119910119905 119905 = 1 2 78
For such a process the mean function of 119910119905 is constantand we may assume that the process mean is subtracted outto produce a process 119883119905 = 119910119905 minus 119864(119910119905) with zero mean Theplot of the sample path autocorrelation function (ACF) andthe partial autocorrelation function (PACF) for series 119883119905 aregiven in Figures 1 and 2 respectively From Figure 1 we can
Mathematical Problems in Engineering 7
0 20 40 60 80
minus10
minus50
510
15
Time
Dat
a Val
ue
Time
Dat
a Val
ue
0 20 40 60 80
minus10
minus50
510
15
９Ｎ ８Ｎ
Figure 1 The left one is the sample path of original series 119910119905 and the right one is the sample plot of centering series119883119905 for the real data
0 5 10 15
minus05
00
05
10
Lag
ACF
5 10 15
minus04
minus02
00
02
Lag
PACF
８Ｎ８Ｎ
Figure 2 The sample autocorrelation function (ACF) and partial autocorrelation function (PACF) plots of centering series 119883119905 for the realdata
see119883119905may come from a stationary autoregressive time seriesprocess From Figure 2 we conclude that 119883119905 is from first-order autoregressive process Moreover we test the normalityof the residual data 119883119905 according to the Normal Q-Q plotintroduced by Henry CThode [21] From Figure 3 we noticethat the scatter points on theNormal Q-Q plot are close to thereference line so the residual data119883119905 can basically be seen asfollowing the normal distribution Hence we derive that themodel N1 would be more suitable to model these data
From the time series plot given in Figure 1 the constancyof the parameter may be suspected and therefore we areinterested in testing the hypothesis of constancy of thecoefficient parameter We carry out the test for 1198670 1205902120572 = 0against 1198671 1205902120572 gt 0 The value of the test statistic in (8)turned out to be 21151 with 119901 value 00344 which indicatesthe rejection of the null hypothesis at 5 level of significanceThus in this case the RCAR(1) model would be much moreappropriate as opposed to the AR(1) model
5 Conclusions
In this article we propose a locally most powerful-typetest for testing the constancy of the random coefficientparameter in autoregressive model and derive their limitingnull distributions under regularity conditions It is clear fromthe applications that the coefficient need not remain constantthroughout the time Therefore it is essential to have such a
Normal QminusQ Plot
minus10
minus50
510
15Sa
mpl
e Qua
ntile
s
minus1 0 1 2minus2Theoretical Quantiles
Figure 3 The normal Q-Q plot of centering series 119883119905 for the realdata
test conducted whenever the random variation is suspectedThrough our simulation study and a real-data analysis wedemonstrate that our test succeeds and it performs better thanthe competitor Finally we anticipate that our locally mostpowerful-type test can be extended to other types of timeseries model
8 Mathematical Problems in Engineering
Appendix
A Process of Establish the Test Statistic
In this Appendix A we present the detailed steps to obtainthe test statistics
Proof Therefore we can get the likelihood function
1198711198671 (1199091 119909119905) = 1198911198830 (1199090) 119899prod119905=1
119891119883119905|119883119905minus1 (119909119905 | 119909119905minus1)= 1198911198830 (1199090) 119899prod
119905=1
int119891120576 (119909119905 minus 120572119905119909119905minus1) 119889119865120572 (A1)
Now the Taylor series expansion of 1198711198671 around 120572 which isthe mean of 120572119905 gives that
1198711198671 (1199091 119909119905) = 1198911198830 (1199090) 119899prod119905=1
int119891120576 (119909119905minus 120572119905119909119905minus1) 119889119865120572
= 1198911198830 (1199090) 119899prod119905=1
int[119891120576 (119909119905 minus 120572119909119905minus1)
+ (120572119905 minus 120572) 1198911015840120576 (119909119905 minus 120572119909119905minus1)+ (120572119905 minus 120572)22 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)+ (120572119905 minus 120572)33 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) + sdot sdot sdot] 119889119865120572
= 1198911198830 (1199090) 119899prod119905=1
[119891120576 (119909119905 minus 120572119909119905minus1) + 12120590212057211989110158401015840120576 (119909119905minus 120572119909119905minus1) + 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) int (120572119905 minus 120572)33 119889119865120572+ sdot sdot sdot]
(A2)
Obviously we can conclude that 119875(120572119905 minus 120572 = 0) = 1 if 1205902120572 = 0That is to say the distribution function of random variable 120572119905is degenerated Furthermore we derive that
int (120572119905 minus 120572)119896119896 119889119865120572 = 0 119896 = 1 2 3 sdot sdot sdot (A3)
By taking the derivative of log 1198711198671 with respect to120573 at1205902120572 = 0we have the following equation
120597 log 11987111986711205971205902120572 1003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 120597sum119899119905=1 log [119891120576 (119909119905 minus 120572119909119905minus1) + (12) 120590212057211989110158401015840120576 (119909119905 minus 120572119909119905minus1) + 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) int ((120572119905 minus 120572)3 3) 119889119865120572 + sdot sdot sdot ]1205971205902120572 100381610038161003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 120597sum119899119905=1 log [119891120576 (119909119905 minus 120572119909119905minus1) + (12) 120590212057211989110158401015840120576 (119909119905 minus 120572119909119905minus1)]1205971205902120572 100381610038161003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 12 119899sum119905=111989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) (A4)
Thus the LMP test statistic has the form119876119899 (120573) = 119899sum119905=1
12 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) (A5)
This completes the proof
B Proof of Theorem 1
Proof of Theorem 1 In the following we can write down thelog-likelihood function under the null hypothesis
log 119871 = 119899sum119905=1
log119891120576 (119909119905 minus 120572119909119905minus1) (B1)
According to the maximum likelihood principle the maxi-mum likelihood estimate can be obtained by maximizing
the log-likelihood function log 119871 with respect to 120573 Applyingthe Taylor series expansion to 120597 log119891120576120597120573 gives
0 = 120597log119891120576120597120573 = 120597log119891120576120597120573 + ( minus 120573) 1205972log119891120576 (120573lowast)1205971205731205971205731015840 (B2)
where 120573lowast is on the line segment between 120573 and so 120573lowast119901997888rarr
120573(119899 997888rarr infin) Under the null hypothesis we can write
minus 120573 = minus 119899sum119905=1
1205972 log1198911205761205971205731205971205731015840 + 119900119901 (119899)minus1 119899sum119905=1120597 log119891120576120597120573 (B3)
Mathematical Problems in Engineering 9
Using the similar method as in the above expanding(1radic119899)sum119899119905=1 119877119905() around 120573 gives that1radic119899 119899sum119905=1119877119905 () = 1radic119899 119899sum119905=1119877119905 (120573)+ ( minus 120573)119879 1radic119899 119899sum119905=1120597119877119905 (120573)120597120573 + 119900119901 (1) (B4)
with 120597119877119905(120573)120597120573 = (120597119877119905120597120572 1205971198771199051205971205902)119879 The remainder termis 119900119901(1) since 119901997888rarr 120573Under (C3) and (C4) we conclude that1205972119877119905(120573)1205971205731205971205731015840 = 119874119901(119899) Therefore
( minus 120573)119879 ( 1radic119899) 119899sum119905=1120597119877119905 (120573)120597120573= minus1119899 119899sum119905=1[120597119877119905 (120573)120597120573 ]119879times 1119899 119899sum119905=11205972 log1198911205761205971205731205971205731015840 + 119900119901 (1)minus1times 1radic119899 119899sum119905=1120597 log119891120576120597120573
(B5)
Now the first term of (B5) converges to1119899 119899sum119905=1120597119877119905 (120573)120597120573 = 1119899 119899sum119905=1 120597120597120573 120597 log1198911205761205971205902120572 1205902120572=0
= (1119899 119899sum119905=11205972 log11989112057612059712057212059712059021205721119899 119899sum119905=11205972 log11989112057612059712059021205761205971205902120572 )1205902120572=0
997888rarr 119882as 119899 997888rarr infin
(B6)
with119882 = (1198821111988212)119879This is because
11988211 = 119864(120597 log1198911205761205971205721205971205902120572 ) = 119864(minus120597 log119891120576120597120572 120597 log1198911205761205971205902120572 )= minus12119864[120597 log119891120576120597120572 1205972 log1198911205761205971205722 + (120597 log119891120576120597120572 )2]
at 1205902120572 = 011988212 = 119864(120597 log1198911205761205971205902120576 1205971205902120572) = 119864(minus120597 log1198911205761205971205902120576 120597 log1198911205761205971205902120572 )= minus12119864[120597 log1198911205761205971205902120576 1205972 log1198911205761205971205722 + (120597 log119891120576120597120572 )2]
at 1205902120572 = 0
(B7)
Next we can obtain the elements of Fisher informationmatrix
Σ = 1119899 119899sum119905=1119864[120597 log119891120576120597120573 (120597 log119891120576120597120573 )119879 | F119910119905minus1]= [[[[[
minus119864[1205972 log1198911205761205971205722 ] minus119864[1205972 log1198911205761205971205721205971205902120576 ]minus119864[1205972 log1198911205761205971205721205971205902120576 ] minus119864[1205972 log11989112057612059712059021205761205971205902120576 ]]]]]]
(B8)
Thus (B5) is asymptotically equivalent to
minus119882Σminus1 1radic119899 119899sum119905=1120597 log119891120576120597120573 (B9)
We present the following lemma recommended by Billingsley[22] to establish the asymptotic normality of the test statistic
Lemma B1Under assumption of (A1) we have as 119899 997888rarr infin(i) lim119899997888rarrinfin(1119899)sum119899119905=1 119864(1198772119905 (120573) | F119877119905minus1) = 1205902 119886119904 (ii)(1radic119899)sum119899119905=1 119877119905(120573) 119889997888rarr 119873(0 1205902) where 1205902 = (14)1198641205972 log1198911205761205971205722 + (120597 log119891120576120597120572)2
Proof By the ergodic theorem the first conclusion holdsNow let F119905 = 1205901199090 1199091 119909119905 119905 ge 1 and F0 is a sigma
field Note that 119876119899(120573) = sum119899119905=1 119877119905(120573) It is easy to see that[119877119899(120573)|F119899minus1] = 0 Then we have
119864 [119876119899 (120573) | F119899minus1] = 119864 [119876119899minus1 (120573) + 119877119899 (120573) | F119899minus1]= 119876119899minus1 (120573) (B10)
Thus 119876119899F119899 119899 ge 0 is a martingale We have shownsup119905119864|119877119905(120573)|2+120578 lt infin which means 119864[1198772119905 (120573)] is uniformlyintegrable ByTheorem 11 of Billingsley [22] we have that as119899 997888rarr infin1119899 119899sum119905=11198772119905 (120573) 119886119904997888997888rarr 119864[1198772119899 (120573) | F119899minus1] = 1205902 (B11)
Hence using the following version of Martingale Cen-tral Limit Theorem from Hall and Heyde [23] we have(1radic119899)sum119899119905=1 119877119905(120573) 119889997888rarr 119873(0 1205902) The proof of Lemma B1 isthus completed ◻
Similarly we can verify that 119878119899 = sum119899119905=1(120597log119891120576120597120573) is amartingale By ergodic and stationary properties we obtainthat as 119899 997888rarr infin
1119899 119899sum119905=11205972 log119891120576120597120573120597120573119879 119886119904997888997888rarr 119864[120597 log119891120576120597120573 (120597 log119891120576120597120573 )119879] = Σ (B12)
Therefore (1radic119899)119878119899 119889997888rarr 119873(0 Σ)
10 Mathematical Problems in Engineering
In the same way for any vector 119888 = (1198881 1198882 1198883)119879 isin 1198773 (0 0 0) we have1radic119899119888119879(
119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) = 1radic119899sdot 119899sum119905=1
(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 ) 119889997888rarr 119873(0 119864(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 )2)
(B13)
By the Cramer-Wold device we obtain
1radic119899 (119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) 119889997888rarr 119873[(00) (1205902 119882119879119882 Σ )] (B14)
where119882 = (1198821111988212)119879 Then we can make the conclusionthat 1radic119899 119899sum119905=1119877119905 () 119889997888rarr 119873(0 1205962) (B15)
where 1205962 = 1205902 minus 119882119879Σminus1119882 The proof of the theorem iscompleted
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that they have no conflicts of interest
Acknowledgments
This work is supported by National Natural Science Foun-dation of China (Nos 11871028 11731015 11571051 and11501241) Natural Science Foundation of Jilin Province (Nos20180101216JC 20170101057JC and 20150520053JH) Pro-gram for Changbaishan Scholars of Jilin Province (2015010)and Science and Technology Program of Jilin EducationalDepartment during the ldquo13th Five-Yearrdquo Plan Period (No2016316)
References
[1] A Hoque ldquoFinite sample analysis of the first order autoregres-sive modelrdquoCalcutta Statistical Association Bulletin vol 34 no1-2 pp 51ndash63 1985
[2] T Ogawa H Sonoda S Ishiwa and Y Shigeta ldquoAn applicationof autoregressive model to pattern discrimination of brainelectrical activity mappingrdquo Brain Topography vol 6 no 1 pp3ndash11 1992
[3] A Subasi A Alkan E Koklukaya and M K Kiymik ldquoWaveletneural network classification of EEG signals by using ARmodelwith MLE preprocessingrdquo Journal of the International NeuralNetwork Society vol 18 pp 985ndash997 2005
[4] M Maleki and A R Nematollahi ldquoAutoregressive models withmixture of scale mixtures of gaussian innovationsrdquo IranianJournal of Science amp Technology Transactions A Science vol 41no 4 pp 1099ndash1107 2017
[5] J Conlisk ldquoStability in a random coefficient modelrdquo Interna-tional Economic Review vol 15 no 2 pp 529ndash533 1974
[6] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models Irdquo Journal of Time SeriesAnalysis vol 1 no 1 pp 37ndash46 1980
[7] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models IIrdquo Journal of Time SeriesAnalysis vol 2 no 3 pp 185ndash203 1981
[8] D F Nicholls and B G Quinn Random Coefficient Autore-gressive Models An Introduction Lecture Notes in StatisticsSpringer-Verlag New York NY USA 1982
[9] A Brandt ldquoThe stochastic equation Y119899+1 = A119899Y119899 + B119899 withstationary coefficientsrdquo Advances in Applied Probability vol 18no 1 pp 211ndash220 1986
[10] D Wang and S K Ghosh ldquoBayesian estimation and unitroot tests for random coefficient autoregressive modelsrdquoModelAssisted Statistics and Applications vol 3 no 4 pp 281ndash2952008
[11] DWang S K Ghosh and S G Pantula ldquoMaximum likelihoodestimation and unit root test for first order random coefficientautoregressive modelsrdquo Journal of Statistical Theory and Prac-tice vol 4 no 2 pp 261ndash278 2010
[12] T V Ramanathan and M B Rajarshi ldquoRank tests for testingthe randomness of autoregressive coefficientsrdquo Statistics ampProbability Letters vol 21 no 2 pp 115ndash120 1994
[13] S Lee J Ha O Na and S Na ldquoThe cusum test for parameterchange in time seriesmodelsrdquo Scandinavian Journal of Statisticsvol 30 no 4 pp 781ndash796 2003
[14] M Moreno and J Romo ldquoRobust unit root tests with autore-gressive errorsrdquo Communications in StatisticsmdashTheory andMethods vol 45 no 20 pp 5997ndash6021 2016
[15] V K Rohatgi A K Md Ehsanes Sleh R Ahluwalia and P JildquoNull distribution of locally most powerful tests for the twosample problemwhen the combined sample is type II censoredrdquoCommunications in Statistics - Theory and Methods vol 19 pp2337ndash2355 1992
[16] M S Chikkagoudar and B S Biradar ldquoLocally most powerfulrank tests for comparison of two failure rates based on multipletype-ii censored datardquoCommunications in Statistics Theory andMethods vol 41 no 23 pp 4315ndash4331 2012
[17] A Manik N Balakrishna and T V Ramanathan ldquoTesting theconstancy of the thinning parameter in a random coefficientinteger autoregressive modelrdquo Statistical Papers pp 1ndash25 2017
[18] P J Huber Robust Statistics JohnWiley amp Sons New York NYUSA 1981
[19] T P Hettmansperger Statistical Inference Based on Ranks JohnWiley amp Sons Inc New York NY USA 1984
[20] Z-W Zhao D-H Wang and C-X Peng ldquoCoefficient con-stancy test in generalized random coefficient autoregressivemodelrdquoApplied Mathematics and Computation vol 219 no 20pp 10283ndash10292 2013
[21] H C Thode Testing for Normality CRC Press New York NYUSA 2002
Mathematical Problems in Engineering 11
[22] P Billingsley ldquoThe lindeberg-Lvy theorem for martingalesrdquoProceedings of the American Mathematical Society vol 12 no1 pp 788ndash792 1961
[23] P Hall and C C Heyde Martingale Limit Theory and ItsApplication Academic Press New York NY USA 1980
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2 Mathematical Problems in Engineering
multiplier test is derived for the problem However thislikelihood ratio approach has several weaknesses as the truevalue of the parameter under the null hypothesis lies on theboundary of the parameter space the asymptotics will not beeasily established Ramanathan and Rajarshi [12] considereda non-Gaussian test method to handle the change problemfor parameters in RCAR(1) case Their signed rank testhowever require a symmetry assumption on the innovationdensity which is highly unnatural in this context Still forthe RCAR(1) model Lee [13] proposed the cumulative sumtest for parameter change in RCAR(1) model But it detectsa change only in a function of the parameter rather thanthe parameter themselves Recently Moreno and Romo [14]studied robust unit root test with autoregressive errors Theunit root test shows some size distortions and requires manyestimates in the construction of test
To deal with these drawbacks we construct a locallymost powerful-type (LMP) test for testing the constancy ofparameters in the RCAR(1) model The LMP test has beeninvestigated by several authors Rohatgi et al [15] Chikk-agoudar and Biradar [16] and Manik et al [17] consideredthe LMP test for parameter constancy The test has meritin that its calculation process is less cumbersome and itproduces stable sizes especially when parameter nearby thetrue value In contrast the traditional tests mentioned aboveshow some size distortions and require many estimates in theconstruction of tests see Ramanathan and Rajarshi [12] Ina general way our test can conventionally discard correlationeffects and enhance the performance of the test In this paperwe illustrate how our proposed method can be implementedfor finite samples under normal studentrsquos t and symmetriccontamination errors (see Huber [18] and Hettmansperger[19]) Through numerical simulation studies we can see thatour test has a stronger power in terms of maintaining theempirical level and power than the empirical likelihood (EL)test suggested by Zhao et al [20]
The outline of this paper is organized as follows InSection 2 we introduce our test statistic and derive its limitingdistribution under the null hypothesis Numerical simula-tions to evaluate the empirical size and power of our testtechnique are discussed in Section 3 Section 4 presents a reallife data example to illustrate the superior of ourmethodsWeprovide brief concluding remarks in Conclusions Appendixprovides the proofs of the main results
2 Methodology and Main Results
In this section we will construct a test statistic to test whether120572119905 is a constant To achieve this task we set up the null andalternative hypotheses 1198670 1205902120572 = 0
v119904 1198671 1205902120572 gt 0 (3)
In what follows we give our main results Suppose that thetime series data 1199091 1199092 119909119899 are generated from (2) Let120572119905 be an iid sequence of random variables with commonprobability distribution 119865120572 120576119905 is an iid sequence of randomvariableswith density function119891120576 In this article we regard1198830
as a givennumber alternatively if1198830 is a randomvariable weshall consider only inferences conditionally on 1198830 are fixedNote that 1198830 120572119905 and 120576119905 are independent
Next we assume the following conditions to establish theasymptotic properties of the test statistic
(C1) 1205722 + 1205902120572 lt 1(C2)The distribution 119865120572 of 120572119905 is such that 119864|120572119905|3 lt infin(C3) The third and mixed derivatives of log119891120576 with
respect to 120572 and 1205902120576 are uniformly bounded on (120572 1205902120576 )(C4) Differentiation thrice with respect to (120572 1205902120576 ) of 119891120576
under the integration is boundedIt is easy to derive that119883119905 is aMarkov chain on 0 1 2 sdot sdot sdot
with the following transition probabilities
119891119883119905|119883119905minus1 (119909119905 | 119909119905minus1) = int119891119883119905|119883119905minus1 (119909119905 | 119909119905minus1 120572119905) 119889119865120572= int119891120576 (119909119905 minus 120572119905119909119905minus1) 119889119865120572 (4)
The Markovity follows from (2) and the fact that 120572119905 is aniid sequence The conditional density 119891119883119905|119883119905minus1(119909119905 | 119909119905minus1)is obtained by noting that conditional on 120572119905 Conditionalon (119883119905minus1 120572119905) and 120576119905 is an iid sequence with the density119891120576 hence the probability density function of 119883119905 is given by119891120576(119909119905 minus 120572119905119909119905minus1) So we get the result in the above equation
Therefore we can write down the likelihood function 1198711198671for RCAR(1) model
1198711198671 (1199091 119909119905) = 1198911198830 (1199090) 119899prod119905=1
int119891120576 (119909119905 minus 120572119905119909119905minus1) 119889119865120572 (5)
Furthermore employing this spirit of LMP approach intro-duced by Manik et al [17] we can obtain our test statistic
119876119899 (120573) = 119899sum119905=1
12 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) = 119899sum119905=1119877119905 (120573) (6)
where
119877119905 (120573) = 12 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) (7)
And the detailed procedures can be found in Appendix ALet120573 = (120572 1205902120576 )119879 denote vector of all unknown parameters
of the RCAR(1) process Hereafter we use the notation = ( 1205902120576 ) to represent the maximum likelihood estimator(MLE) of 120573 In addition the actual test statistic can beobtained by replacing 120573 with their MLE It can be easilyderived that 119877119905(120573)F119877119905 is a zero mean martingale
Before we state our main results the following assump-tions will be made
(A1) sup119905119864|119877119905(120573)|2+120578 lt infin for some 120578 gt 0Then about the asymptotic distribution of119876119899() we have
the following theorem
Theorem 1 Let 119909119905 be a sequence of strictly stationaryergodic and F119905-measurable solutions to equation (2) under
Mathematical Problems in Engineering 3
Table 1 Empirical sizes of LMP and EL tests at nominal level 005 for model N1
119873 (120590119873 120572)(05 03) (05 05) (05 07) (05 09) (1 03) (1 05) (1 07) (1 09)100 MP 0095 0105 0109 0101 0116 0103 0089 0098
EL 0008 0004 0002 0003 0003 0002 0003 0002200 LMP 0089 0082 0085 0086 0085 0082 0083 0087
EL 0006 0007 0004 0005 0004 0004 0003 0003300 LMP 0074 0077 0082 0069 0076 0071 0076 0074
EL 0006 0009 0007 0006 0005 0007 0008 0009400 LMP 0063 0070 0075 0068 0067 0063 0070 0072
EL 0007 0009 0006 0004 0007 0008 0005 0004500 LMP 0066 0060 0077 0060 0065 0062 0060 0055
EL 0009 0005 0006 0005 0006 0009 0014 0008600 LMP 0062 0057 0067 0053 0065 0060 0059 0068
EL 0009 0005 0011 0007 0009 0007 0008 0007700 LMP 0064 0059 0068 0056 0057 0062 0058 0063
EL 0005 0010 0008 0007 0007 0008 0005 0003800 LMP 0060 0061 0060 0056 0058 0055 0055 0063
EL 0009 0007 0007 0010 0011 0012 0008 0009900 LMP 0055 0062 0054 0048 0053 0053 0062 0062
EL 0015 0015 0008 0010 0011 0007 0010 00091000 LMP 0063 0051 0057 0052 0058 0062 0057 0054
EL 0019 0009 0012 0009 0011 0012 0007 00095000 LMP 0051 0051 0051 0053 0050 0048 0049 0053
EL 0020 0022 0018 0020 0016 0021 0024 0022
conditions of (C1) minus (C4) and above (A1) Then under 1198670we have asymptotic normality
(radic119899)minus1119876119899 () 119889997888rarr 119873(0 2) (8)
here 2 = 1205902 minus119882119879Σminus1119882(The expressions for 1205902 and Σminus1 are derived in
Appendix B)
3 Simulation
In this section we carry out some simulation studies tocompare performances of the locally most powerful-typetest and the empirical likelihood test in terms of empiricalsize and power The empirical size and power for the twotests in Tables 1ndash6 are based on 1000 repetitions with thehelp of R software Within each study we set the initialvalues 1198830 equiv 1 and employ the significance level at 005Throughout this simulation we used notation LMP for locallymost powerful-type test by our algorithm EL for empiricallikelihood method
31 Empirical Size To calculate empirical sizes we pay ourattention to three kinds of model (1) defined as follows
(N1) 120572119905 = 120572 120576119905 sim 119873(0 1205902119873)(N2) 120572119905 = 120572 120576119905 sim 119905(119898)
(N3) 120572119905 = 120572 120576119905 sim F120576(119909)The distribution function of 120585-contamination is
F120576 (119909) = 120585Φ( 1199091205901) + (1 minus 120585)Φ( 1199091205902) (9)
where 120585 is a fixed constant satisfying 0 lt 120585 lt 1 andΦ(119909) is thedistribution function of standard normal random variable120590119894 gt 0 119894 = 1 2
Let us now describe how the simulated results areobtained First of all we use the model N1 to generate datawhen the 120590119873 = 05 1 and 120572 = 03 05 07 09 Secondly wegenerate data from model N2 with the degrees of freedom119898 = 8 10 and 120572 = 03 05 07 09 For model N1 and N2we take the sample size 119873 = 100 200 300 400 500 600700 800 900 1000 and 5000 Thirdly we simulate samplesfrommodel N3 with the (120585 1205901 1205902 ) = (08 1 3) (120585 1205901 1205902 ) =(09 1 3) and 120572 = 03 05 07 09 For model N3 we set thesample size119873 = 100 300 500 1000 and 5000 The empiricallevels of threemodels are presented inTables 1ndash3 respectivelyAs seen from Tables 1ndash3 the LMP test give similar resultsunder normal studentrsquos t and symmetric contaminationerrors In addition the LMP test produces sizes closer to thenominal significance level 005 quite satisfactorily especiallyfor larger sample sizes However the EL method has lowerlevels for each sample size so one may make wrong decisionsin tests based on them Looking at three models results wecan conclude that our method has a greater effect than the ELtest in terms of the empirical level
4 Mathematical Problems in Engineering
Table 2 Empirical sizes of LMP and EL tests at nominal level 005 for model N2
119873 (119898 120572)(8 03) (8 05) (8 07) (8 09) (10 03) (10 05) (10 07) (10 09)100 MP 0060 0065 0066 0061 0065 0068 0075 0067
EL 0003 0003 0002 0003 0004 0003 0004 0001200 LMP 0059 0061 0065 0059 0066 0063 0068 0061
EL 0002 0004 0002 0002 0004 0004 0002 0004300 LMP 0058 0057 0059 0055 0063 0058 0063 0055
EL 0005 0004 0003 0009 0006 0006 0004 0002400 LMP 0058 0055 0058 0055 0062 0054 0061 0058
EL 0004 0003 0005 0003 0004 0003 0005 0006500 LMP 0056 0055 0056 0052 0057 0059 0055 0054
EL 0006 0004 0005 0005 0007 0005 0006 0007600 LMP 0060 0057 0058 0060 0062 0059 0056 0053
EL 0003 0005 0005 0005 0008 0007 0009 0006700 LMP 0055 0056 0058 0057 0060 0060 0056 0058
EL 0006 0005 0005 0008 0006 0006 0004 0008800 LMP 0050 0054 0054 0052 0056 0054 0054 0054
EL 0005 0007 0008 0006 0006 0008 0008 0008900 LMP 0051 0052 0058 0056 0056 0060 0049 0051
EL 0005 0006 0006 0009 0006 0008 0010 00051000 LMP 0052 0056 0057 0055 0050 0053 0050 0054
EL 0007 0008 0005 0006 0012 0009 0009 00075000 LMP 0052 0050 0050 0050 0051 0049 0054 0052
EL 0015 0014 0012 0016 0017 0014 0012 0013
Table 3 Empirical sizes of LMP and EL tests at nominal level 005 for model N3
119873 (1205901 1205902 120572)(1 3 01) (1 3 02) (1 3 03) (1 3 04) (1 3 05) (1 3 06) (1 3 07) (1 3 08) (1 3 09)120585 = 08100 LMP 0064 0068 0065 0072 0053 0078 0071 0053 0047
EL 0000 0000 0001 0000 0000 0001 0000 0000 0001300 LMP 0045 0066 0053 0038 0054 0081 0051 0043 0040
EL 0000 0001 0000 0001 0001 0002 0000 0001 0000500 LMP 0058 0057 0057 0049 0050 0060 0043 0049 0044
EL 0001 0000 0001 0000 0000 0001 0000 0000 00011000 LMP 0050 0059 0052 0051 0045 0054 0036 0043 0036
EL 0002 0001 0001 0001 0001 0001 0003 0001 00015000 LMP 0053 0050 0054 0050 0048 0050 0047 0050 0046
EL 0003 0004 0003 0007 0007 0006 0008 0006 0011120585 = 09100 LMP 0078 0081 0089 0069 0089 0095 0083 0091 0086
EL 0000 0002 0001 0000 0000 0000 0000 0000 0001300 LMP 0080 0070 0074 0070 0070 0056 0063 0057 0055
EL 0000 0000 0001 0000 0000 0001 0002 0000 0001500 LMP 0072 0081 0068 0067 0058 0059 0060 0046 0053
EL 0001 0000 0001 0000 0000 0001 0001 0000 00041000 LMP 0064 0061 0067 0067 0055 0055 0052 0051 0045
EL 0000 0000 0000 0001 0002 0003 0003 0000 00015000 LMP 0055 0053 0050 0050 0055 0046 0050 0047 0050
EL 0004 0002 0003 0002 0002 0006 0003 0003 0009
Mathematical Problems in Engineering 5
Table 4 Empirical powers of LMP and EL tests at nominal level 005 for model A1
119873 (120590119873 119886 119887)(05 05 05) (05 05 1) (05 1 05) (05 1 1) (1 05 05) (1 05 1) (1 1 05) (1 1 1)100 LMP 0165 0102 0171 0110 0171 0107 0174 0090
EL 0055 0024 0040 0023 0050 0022 0041 0025200 LMP 0318 0143 0294 0147 0320 0140 0311 0160
EL 0191 0070 0156 0073 0202 0049 0155 0067300 LMP 0513 0209 0494 0253 0511 0175 0449 0241
EL 0405 0144 0332 0173 0397 0124 0294 0168400 LMP 0674 0232 0561 0330 0663 0264 0590 0338
EL 0578 0176 0456 0272 0559 0189 0450 0260500 LMP 0791 0377 0720 0427 0770 0374 0725 0438
EL 0722 0298 0629 0366 0705 0303 0625 0374600 LMP 0842 0415 0792 0521 0832 0424 0790 0508
EL 0798 0352 0717 0460 0793 0365 0712 0458700 LMP 0912 0483 0862 0598 0919 0478 0855 0598
EL 0882 0429 0809 0541 0885 0423 0794 0536800 LMP 0964 0560 0915 0645 0947 0568 0918 0643
EL 0938 0512 0877 0604 0928 0516 0872 0599900 LMP 0967 0633 0936 0727 0971 0623 0955 0703
EL 0952 0586 0910 0696 0959 0586 0933 06871000 LMP 0988 0712 0963 0799 0978 0702 0962 0776
EL 0981 0678 0942 0774 0969 0676 0938 07495000 LMP 1000 1000 1000 1000 1000 1000 1000 1000
EL 1000 1000 1000 1000 1000 1000 1000 1000
Table 5 Empirical powers of LMP and EL tests at nominal level 005 for model A2119873 (119898 119886 119887)(6 05 05) (6 05 1) (6 1 05) (6 1 1) (10 05 05) (10 05 1) (10 1 05) (10 1 1)100 LMP 0289 0110 0220 0149 0241 0096 0183 0131
EL 0045 0016 0027 0019 0047 0018 0027 0029200 LMP 0596 0252 0519 0310 0525 0199 0452 0272
EL 0161 0052 0118 0055 0186 0057 0146 0063300 LMP 0790 0400 0722 0475 0762 0351 0686 0429
EL 0336 0111 0276 0124 0373 0124 0293 0152400 LMP 0899 0538 0861 0644 0862 0462 0815 0564
EL 0486 0147 0399 0196 0520 0173 0440 0223500 LMP 0964 0624 0933 0757 0957 0596 0906 0699
EL 0622 0218 0529 0285 0695 0262 0590 0345600 LMP 0979 0701 0951 0819 0965 0657 0947 0754
EL 0708 0237 0606 0351 0767 0301 0650 0395700 LMP 0995 0791 0974 0862 0983 0727 0978 0824
EL 0814 0351 0669 0404 0839 0389 0773 0483800 LMP 0996 0837 0990 0927 0994 0809 0984 0882
EL 0845 0404 0767 0512 0895 0488 0839 0585900 LMP 0999 0874 0995 0934 0996 0831 0993 0924
EL 0879 0430 0801 0548 0929 0514 0877 06221000 LMP 1000 0936 1000 0971 1000 0889 0996 0951
EL 0913 0515 0887 0626 0946 0601 0930 07065000 LMP 1000 1000 1000 1000 1000 1000 1000 1000
EL 0992 0967 0994 0985 1000 1000 1000 0999
6 Mathematical Problems in Engineering
Table 6 Empirical powers of LMP and EL tests at nominal level 005 for model A3 with (1205901 1205902) = (1 3)119873 (119886 119887)(05 05) (05 1) (05 15) (1 05) (1 1) (1 15) (15 05) (15 1) (15 15)120585 = 08100 LMP 0141 0054 0034 0120 0073 0038 0084 0082 0046
EL 0039 0009 0006 0026 0012 0008 0018 0012 0009300 LMP 0775 0388 0183 0628 0493 0299 0475 0448 0331
EL 0325 0102 0030 0231 0097 0052 0143 0114 0061500 LMP 0959 0675 0381 0880 0791 0594 0754 0729 0583
EL 0656 0236 0070 0500 0268 0143 0347 0222 01311000 LMP 1000 0967 0753 0996 0992 0934 0982 0982 0939
EL 0945 0599 0255 0891 0673 0399 0720 0609 04025000 LMP 1000 1000 1000 1000 1000 1000 1000 1000 1000
EL 1000 1000 0981 1000 1000 0997 0999 1000 0997120585 = 09100 LMP 0118 0043 0035 0071 0041 0025 0049 0052 0037
EL 0042 0013 0015 0032 0021 0014 0020 0013 0011300 LMP 0673 0278 0101 0545 0372 0190 0388 0313 0220
EL 0301 0077 0021 0199 0101 0043 0128 0093 0055500 LMP 0914 0558 0277 0868 0681 0423 0700 0622 0467
EL 0571 0183 0054 0486 0239 0128 0353 0212 01131000 LMP 1000 0929 0627 0995 0965 0849 0970 0954 0865
EL 0927 0503 0204 0852 0637 0375 0702 0548 03685000 LMP 1000 1000 1000 1000 1000 1000 1000 1000 1000
EL 1000 1000 0963 1000 1000 0996 1000 1000 0994
32 Empirical Power In order to investigate the empiricalpowers we consider the alternatives under three versions ofmodel (2)
(A1) 120572119905 sim 119861119890119905119886(119886 119887) 120576119905 sim 119873(0 1205902119873)(A2) 120572119905 sim 119861119890119905119886(119886 119887) 120576119905 sim 119905(119898)(A3)120572119905 sim 119861119890119905119886(119886 119887) 120576119905 sim F120576(119909)Thedistribution function
of 120585-contamination is
F120576 (119909) = 120585Φ( 1199091205901) + (1 minus 120585)Φ( 1199091205902) (10)
where 120585 is a fixed constant satisfying 0 lt 120585 lt 1 andΦ(119909) is thedistribution function of standard normal random variable120590119894 gt 0 119894 = 1 2
To calculate the empirical powers of the two tests inthe first we generate samples from the model A1 in whichthe parameters 119886 = 05 1 119887 = 05 1 and 120590119873 = 05 1In the second setting we simulate samples from model A2with the degrees of freedom 119898 = 6 10 the parameters119886 = 05 1 and 119887 = 05 1 For model A1 and A2 we takethe sample size 119873 = 100 200 300 400 500 600 700800 900 1000 and 5000 In the last case we generate datafrom the model A3 in which the (120585 1205901 1205902 ) = (08 1 3)(120585 1205901 1205902 ) = (09 1 3) and the parameters 119886 = 05 1 15119887 = 05 1 15 For model A3 we employ the sample size119873 = 100 300 500 1000 and 5000 The empirical power ofthe above three models are summed up in Tables 4ndash6 Inthree cases the power of two test statistics increase with thesample size while our test produces relatively better powers
than the EL test Furthermore both tests powers are closeto 1 at the sample size 119873 = 5000 Overall from thesesimulations we conclude that the accuracy of using LMP is67 288 261 higher than that of EL test under normalstudentrsquos 119905 and symmetric contamination errors respectivelyAs anticipated our finding shows that the LMP test is afunctional tool to detect a parameter change for RCAR(1)model Therefore we recommend the LMP for practicaluse because its computation is not difficult and the overallperformance is better than that EL under normal studentrsquos tand symmetric contamination errors
4 A Real Life Data Analysis
In this section we illustrate how the LMP method can beapplied to a practical application This data consist of 78monthly number of annulus growth rate of the import billin Australia starting in December 2010 and ending in May2017 The data are available online at the CEInet StatisticsDatabase site httpdbceicnpageDefaultaspx The meanand variance of the data are found to be 02069 and 153347The data are denoted as 1199101 1199102 11991078 Figure 1 is the samplepath plot for the real data 119910119905 119905 = 1 2 78
For such a process the mean function of 119910119905 is constantand we may assume that the process mean is subtracted outto produce a process 119883119905 = 119910119905 minus 119864(119910119905) with zero mean Theplot of the sample path autocorrelation function (ACF) andthe partial autocorrelation function (PACF) for series 119883119905 aregiven in Figures 1 and 2 respectively From Figure 1 we can
Mathematical Problems in Engineering 7
0 20 40 60 80
minus10
minus50
510
15
Time
Dat
a Val
ue
Time
Dat
a Val
ue
0 20 40 60 80
minus10
minus50
510
15
９Ｎ ８Ｎ
Figure 1 The left one is the sample path of original series 119910119905 and the right one is the sample plot of centering series119883119905 for the real data
0 5 10 15
minus05
00
05
10
Lag
ACF
5 10 15
minus04
minus02
00
02
Lag
PACF
８Ｎ８Ｎ
Figure 2 The sample autocorrelation function (ACF) and partial autocorrelation function (PACF) plots of centering series 119883119905 for the realdata
see119883119905may come from a stationary autoregressive time seriesprocess From Figure 2 we conclude that 119883119905 is from first-order autoregressive process Moreover we test the normalityof the residual data 119883119905 according to the Normal Q-Q plotintroduced by Henry CThode [21] From Figure 3 we noticethat the scatter points on theNormal Q-Q plot are close to thereference line so the residual data119883119905 can basically be seen asfollowing the normal distribution Hence we derive that themodel N1 would be more suitable to model these data
From the time series plot given in Figure 1 the constancyof the parameter may be suspected and therefore we areinterested in testing the hypothesis of constancy of thecoefficient parameter We carry out the test for 1198670 1205902120572 = 0against 1198671 1205902120572 gt 0 The value of the test statistic in (8)turned out to be 21151 with 119901 value 00344 which indicatesthe rejection of the null hypothesis at 5 level of significanceThus in this case the RCAR(1) model would be much moreappropriate as opposed to the AR(1) model
5 Conclusions
In this article we propose a locally most powerful-typetest for testing the constancy of the random coefficientparameter in autoregressive model and derive their limitingnull distributions under regularity conditions It is clear fromthe applications that the coefficient need not remain constantthroughout the time Therefore it is essential to have such a
Normal QminusQ Plot
minus10
minus50
510
15Sa
mpl
e Qua
ntile
s
minus1 0 1 2minus2Theoretical Quantiles
Figure 3 The normal Q-Q plot of centering series 119883119905 for the realdata
test conducted whenever the random variation is suspectedThrough our simulation study and a real-data analysis wedemonstrate that our test succeeds and it performs better thanthe competitor Finally we anticipate that our locally mostpowerful-type test can be extended to other types of timeseries model
8 Mathematical Problems in Engineering
Appendix
A Process of Establish the Test Statistic
In this Appendix A we present the detailed steps to obtainthe test statistics
Proof Therefore we can get the likelihood function
1198711198671 (1199091 119909119905) = 1198911198830 (1199090) 119899prod119905=1
119891119883119905|119883119905minus1 (119909119905 | 119909119905minus1)= 1198911198830 (1199090) 119899prod
119905=1
int119891120576 (119909119905 minus 120572119905119909119905minus1) 119889119865120572 (A1)
Now the Taylor series expansion of 1198711198671 around 120572 which isthe mean of 120572119905 gives that
1198711198671 (1199091 119909119905) = 1198911198830 (1199090) 119899prod119905=1
int119891120576 (119909119905minus 120572119905119909119905minus1) 119889119865120572
= 1198911198830 (1199090) 119899prod119905=1
int[119891120576 (119909119905 minus 120572119909119905minus1)
+ (120572119905 minus 120572) 1198911015840120576 (119909119905 minus 120572119909119905minus1)+ (120572119905 minus 120572)22 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)+ (120572119905 minus 120572)33 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) + sdot sdot sdot] 119889119865120572
= 1198911198830 (1199090) 119899prod119905=1
[119891120576 (119909119905 minus 120572119909119905minus1) + 12120590212057211989110158401015840120576 (119909119905minus 120572119909119905minus1) + 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) int (120572119905 minus 120572)33 119889119865120572+ sdot sdot sdot]
(A2)
Obviously we can conclude that 119875(120572119905 minus 120572 = 0) = 1 if 1205902120572 = 0That is to say the distribution function of random variable 120572119905is degenerated Furthermore we derive that
int (120572119905 minus 120572)119896119896 119889119865120572 = 0 119896 = 1 2 3 sdot sdot sdot (A3)
By taking the derivative of log 1198711198671 with respect to120573 at1205902120572 = 0we have the following equation
120597 log 11987111986711205971205902120572 1003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 120597sum119899119905=1 log [119891120576 (119909119905 minus 120572119909119905minus1) + (12) 120590212057211989110158401015840120576 (119909119905 minus 120572119909119905minus1) + 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) int ((120572119905 minus 120572)3 3) 119889119865120572 + sdot sdot sdot ]1205971205902120572 100381610038161003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 120597sum119899119905=1 log [119891120576 (119909119905 minus 120572119909119905minus1) + (12) 120590212057211989110158401015840120576 (119909119905 minus 120572119909119905minus1)]1205971205902120572 100381610038161003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 12 119899sum119905=111989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) (A4)
Thus the LMP test statistic has the form119876119899 (120573) = 119899sum119905=1
12 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) (A5)
This completes the proof
B Proof of Theorem 1
Proof of Theorem 1 In the following we can write down thelog-likelihood function under the null hypothesis
log 119871 = 119899sum119905=1
log119891120576 (119909119905 minus 120572119909119905minus1) (B1)
According to the maximum likelihood principle the maxi-mum likelihood estimate can be obtained by maximizing
the log-likelihood function log 119871 with respect to 120573 Applyingthe Taylor series expansion to 120597 log119891120576120597120573 gives
0 = 120597log119891120576120597120573 = 120597log119891120576120597120573 + ( minus 120573) 1205972log119891120576 (120573lowast)1205971205731205971205731015840 (B2)
where 120573lowast is on the line segment between 120573 and so 120573lowast119901997888rarr
120573(119899 997888rarr infin) Under the null hypothesis we can write
minus 120573 = minus 119899sum119905=1
1205972 log1198911205761205971205731205971205731015840 + 119900119901 (119899)minus1 119899sum119905=1120597 log119891120576120597120573 (B3)
Mathematical Problems in Engineering 9
Using the similar method as in the above expanding(1radic119899)sum119899119905=1 119877119905() around 120573 gives that1radic119899 119899sum119905=1119877119905 () = 1radic119899 119899sum119905=1119877119905 (120573)+ ( minus 120573)119879 1radic119899 119899sum119905=1120597119877119905 (120573)120597120573 + 119900119901 (1) (B4)
with 120597119877119905(120573)120597120573 = (120597119877119905120597120572 1205971198771199051205971205902)119879 The remainder termis 119900119901(1) since 119901997888rarr 120573Under (C3) and (C4) we conclude that1205972119877119905(120573)1205971205731205971205731015840 = 119874119901(119899) Therefore
( minus 120573)119879 ( 1radic119899) 119899sum119905=1120597119877119905 (120573)120597120573= minus1119899 119899sum119905=1[120597119877119905 (120573)120597120573 ]119879times 1119899 119899sum119905=11205972 log1198911205761205971205731205971205731015840 + 119900119901 (1)minus1times 1radic119899 119899sum119905=1120597 log119891120576120597120573
(B5)
Now the first term of (B5) converges to1119899 119899sum119905=1120597119877119905 (120573)120597120573 = 1119899 119899sum119905=1 120597120597120573 120597 log1198911205761205971205902120572 1205902120572=0
= (1119899 119899sum119905=11205972 log11989112057612059712057212059712059021205721119899 119899sum119905=11205972 log11989112057612059712059021205761205971205902120572 )1205902120572=0
997888rarr 119882as 119899 997888rarr infin
(B6)
with119882 = (1198821111988212)119879This is because
11988211 = 119864(120597 log1198911205761205971205721205971205902120572 ) = 119864(minus120597 log119891120576120597120572 120597 log1198911205761205971205902120572 )= minus12119864[120597 log119891120576120597120572 1205972 log1198911205761205971205722 + (120597 log119891120576120597120572 )2]
at 1205902120572 = 011988212 = 119864(120597 log1198911205761205971205902120576 1205971205902120572) = 119864(minus120597 log1198911205761205971205902120576 120597 log1198911205761205971205902120572 )= minus12119864[120597 log1198911205761205971205902120576 1205972 log1198911205761205971205722 + (120597 log119891120576120597120572 )2]
at 1205902120572 = 0
(B7)
Next we can obtain the elements of Fisher informationmatrix
Σ = 1119899 119899sum119905=1119864[120597 log119891120576120597120573 (120597 log119891120576120597120573 )119879 | F119910119905minus1]= [[[[[
minus119864[1205972 log1198911205761205971205722 ] minus119864[1205972 log1198911205761205971205721205971205902120576 ]minus119864[1205972 log1198911205761205971205721205971205902120576 ] minus119864[1205972 log11989112057612059712059021205761205971205902120576 ]]]]]]
(B8)
Thus (B5) is asymptotically equivalent to
minus119882Σminus1 1radic119899 119899sum119905=1120597 log119891120576120597120573 (B9)
We present the following lemma recommended by Billingsley[22] to establish the asymptotic normality of the test statistic
Lemma B1Under assumption of (A1) we have as 119899 997888rarr infin(i) lim119899997888rarrinfin(1119899)sum119899119905=1 119864(1198772119905 (120573) | F119877119905minus1) = 1205902 119886119904 (ii)(1radic119899)sum119899119905=1 119877119905(120573) 119889997888rarr 119873(0 1205902) where 1205902 = (14)1198641205972 log1198911205761205971205722 + (120597 log119891120576120597120572)2
Proof By the ergodic theorem the first conclusion holdsNow let F119905 = 1205901199090 1199091 119909119905 119905 ge 1 and F0 is a sigma
field Note that 119876119899(120573) = sum119899119905=1 119877119905(120573) It is easy to see that[119877119899(120573)|F119899minus1] = 0 Then we have
119864 [119876119899 (120573) | F119899minus1] = 119864 [119876119899minus1 (120573) + 119877119899 (120573) | F119899minus1]= 119876119899minus1 (120573) (B10)
Thus 119876119899F119899 119899 ge 0 is a martingale We have shownsup119905119864|119877119905(120573)|2+120578 lt infin which means 119864[1198772119905 (120573)] is uniformlyintegrable ByTheorem 11 of Billingsley [22] we have that as119899 997888rarr infin1119899 119899sum119905=11198772119905 (120573) 119886119904997888997888rarr 119864[1198772119899 (120573) | F119899minus1] = 1205902 (B11)
Hence using the following version of Martingale Cen-tral Limit Theorem from Hall and Heyde [23] we have(1radic119899)sum119899119905=1 119877119905(120573) 119889997888rarr 119873(0 1205902) The proof of Lemma B1 isthus completed ◻
Similarly we can verify that 119878119899 = sum119899119905=1(120597log119891120576120597120573) is amartingale By ergodic and stationary properties we obtainthat as 119899 997888rarr infin
1119899 119899sum119905=11205972 log119891120576120597120573120597120573119879 119886119904997888997888rarr 119864[120597 log119891120576120597120573 (120597 log119891120576120597120573 )119879] = Σ (B12)
Therefore (1radic119899)119878119899 119889997888rarr 119873(0 Σ)
10 Mathematical Problems in Engineering
In the same way for any vector 119888 = (1198881 1198882 1198883)119879 isin 1198773 (0 0 0) we have1radic119899119888119879(
119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) = 1radic119899sdot 119899sum119905=1
(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 ) 119889997888rarr 119873(0 119864(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 )2)
(B13)
By the Cramer-Wold device we obtain
1radic119899 (119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) 119889997888rarr 119873[(00) (1205902 119882119879119882 Σ )] (B14)
where119882 = (1198821111988212)119879 Then we can make the conclusionthat 1radic119899 119899sum119905=1119877119905 () 119889997888rarr 119873(0 1205962) (B15)
where 1205962 = 1205902 minus 119882119879Σminus1119882 The proof of the theorem iscompleted
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that they have no conflicts of interest
Acknowledgments
This work is supported by National Natural Science Foun-dation of China (Nos 11871028 11731015 11571051 and11501241) Natural Science Foundation of Jilin Province (Nos20180101216JC 20170101057JC and 20150520053JH) Pro-gram for Changbaishan Scholars of Jilin Province (2015010)and Science and Technology Program of Jilin EducationalDepartment during the ldquo13th Five-Yearrdquo Plan Period (No2016316)
References
[1] A Hoque ldquoFinite sample analysis of the first order autoregres-sive modelrdquoCalcutta Statistical Association Bulletin vol 34 no1-2 pp 51ndash63 1985
[2] T Ogawa H Sonoda S Ishiwa and Y Shigeta ldquoAn applicationof autoregressive model to pattern discrimination of brainelectrical activity mappingrdquo Brain Topography vol 6 no 1 pp3ndash11 1992
[3] A Subasi A Alkan E Koklukaya and M K Kiymik ldquoWaveletneural network classification of EEG signals by using ARmodelwith MLE preprocessingrdquo Journal of the International NeuralNetwork Society vol 18 pp 985ndash997 2005
[4] M Maleki and A R Nematollahi ldquoAutoregressive models withmixture of scale mixtures of gaussian innovationsrdquo IranianJournal of Science amp Technology Transactions A Science vol 41no 4 pp 1099ndash1107 2017
[5] J Conlisk ldquoStability in a random coefficient modelrdquo Interna-tional Economic Review vol 15 no 2 pp 529ndash533 1974
[6] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models Irdquo Journal of Time SeriesAnalysis vol 1 no 1 pp 37ndash46 1980
[7] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models IIrdquo Journal of Time SeriesAnalysis vol 2 no 3 pp 185ndash203 1981
[8] D F Nicholls and B G Quinn Random Coefficient Autore-gressive Models An Introduction Lecture Notes in StatisticsSpringer-Verlag New York NY USA 1982
[9] A Brandt ldquoThe stochastic equation Y119899+1 = A119899Y119899 + B119899 withstationary coefficientsrdquo Advances in Applied Probability vol 18no 1 pp 211ndash220 1986
[10] D Wang and S K Ghosh ldquoBayesian estimation and unitroot tests for random coefficient autoregressive modelsrdquoModelAssisted Statistics and Applications vol 3 no 4 pp 281ndash2952008
[11] DWang S K Ghosh and S G Pantula ldquoMaximum likelihoodestimation and unit root test for first order random coefficientautoregressive modelsrdquo Journal of Statistical Theory and Prac-tice vol 4 no 2 pp 261ndash278 2010
[12] T V Ramanathan and M B Rajarshi ldquoRank tests for testingthe randomness of autoregressive coefficientsrdquo Statistics ampProbability Letters vol 21 no 2 pp 115ndash120 1994
[13] S Lee J Ha O Na and S Na ldquoThe cusum test for parameterchange in time seriesmodelsrdquo Scandinavian Journal of Statisticsvol 30 no 4 pp 781ndash796 2003
[14] M Moreno and J Romo ldquoRobust unit root tests with autore-gressive errorsrdquo Communications in StatisticsmdashTheory andMethods vol 45 no 20 pp 5997ndash6021 2016
[15] V K Rohatgi A K Md Ehsanes Sleh R Ahluwalia and P JildquoNull distribution of locally most powerful tests for the twosample problemwhen the combined sample is type II censoredrdquoCommunications in Statistics - Theory and Methods vol 19 pp2337ndash2355 1992
[16] M S Chikkagoudar and B S Biradar ldquoLocally most powerfulrank tests for comparison of two failure rates based on multipletype-ii censored datardquoCommunications in Statistics Theory andMethods vol 41 no 23 pp 4315ndash4331 2012
[17] A Manik N Balakrishna and T V Ramanathan ldquoTesting theconstancy of the thinning parameter in a random coefficientinteger autoregressive modelrdquo Statistical Papers pp 1ndash25 2017
[18] P J Huber Robust Statistics JohnWiley amp Sons New York NYUSA 1981
[19] T P Hettmansperger Statistical Inference Based on Ranks JohnWiley amp Sons Inc New York NY USA 1984
[20] Z-W Zhao D-H Wang and C-X Peng ldquoCoefficient con-stancy test in generalized random coefficient autoregressivemodelrdquoApplied Mathematics and Computation vol 219 no 20pp 10283ndash10292 2013
[21] H C Thode Testing for Normality CRC Press New York NYUSA 2002
Mathematical Problems in Engineering 11
[22] P Billingsley ldquoThe lindeberg-Lvy theorem for martingalesrdquoProceedings of the American Mathematical Society vol 12 no1 pp 788ndash792 1961
[23] P Hall and C C Heyde Martingale Limit Theory and ItsApplication Academic Press New York NY USA 1980
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Mathematical Problems in Engineering 3
Table 1 Empirical sizes of LMP and EL tests at nominal level 005 for model N1
119873 (120590119873 120572)(05 03) (05 05) (05 07) (05 09) (1 03) (1 05) (1 07) (1 09)100 MP 0095 0105 0109 0101 0116 0103 0089 0098
EL 0008 0004 0002 0003 0003 0002 0003 0002200 LMP 0089 0082 0085 0086 0085 0082 0083 0087
EL 0006 0007 0004 0005 0004 0004 0003 0003300 LMP 0074 0077 0082 0069 0076 0071 0076 0074
EL 0006 0009 0007 0006 0005 0007 0008 0009400 LMP 0063 0070 0075 0068 0067 0063 0070 0072
EL 0007 0009 0006 0004 0007 0008 0005 0004500 LMP 0066 0060 0077 0060 0065 0062 0060 0055
EL 0009 0005 0006 0005 0006 0009 0014 0008600 LMP 0062 0057 0067 0053 0065 0060 0059 0068
EL 0009 0005 0011 0007 0009 0007 0008 0007700 LMP 0064 0059 0068 0056 0057 0062 0058 0063
EL 0005 0010 0008 0007 0007 0008 0005 0003800 LMP 0060 0061 0060 0056 0058 0055 0055 0063
EL 0009 0007 0007 0010 0011 0012 0008 0009900 LMP 0055 0062 0054 0048 0053 0053 0062 0062
EL 0015 0015 0008 0010 0011 0007 0010 00091000 LMP 0063 0051 0057 0052 0058 0062 0057 0054
EL 0019 0009 0012 0009 0011 0012 0007 00095000 LMP 0051 0051 0051 0053 0050 0048 0049 0053
EL 0020 0022 0018 0020 0016 0021 0024 0022
conditions of (C1) minus (C4) and above (A1) Then under 1198670we have asymptotic normality
(radic119899)minus1119876119899 () 119889997888rarr 119873(0 2) (8)
here 2 = 1205902 minus119882119879Σminus1119882(The expressions for 1205902 and Σminus1 are derived in
Appendix B)
3 Simulation
In this section we carry out some simulation studies tocompare performances of the locally most powerful-typetest and the empirical likelihood test in terms of empiricalsize and power The empirical size and power for the twotests in Tables 1ndash6 are based on 1000 repetitions with thehelp of R software Within each study we set the initialvalues 1198830 equiv 1 and employ the significance level at 005Throughout this simulation we used notation LMP for locallymost powerful-type test by our algorithm EL for empiricallikelihood method
31 Empirical Size To calculate empirical sizes we pay ourattention to three kinds of model (1) defined as follows
(N1) 120572119905 = 120572 120576119905 sim 119873(0 1205902119873)(N2) 120572119905 = 120572 120576119905 sim 119905(119898)
(N3) 120572119905 = 120572 120576119905 sim F120576(119909)The distribution function of 120585-contamination is
F120576 (119909) = 120585Φ( 1199091205901) + (1 minus 120585)Φ( 1199091205902) (9)
where 120585 is a fixed constant satisfying 0 lt 120585 lt 1 andΦ(119909) is thedistribution function of standard normal random variable120590119894 gt 0 119894 = 1 2
Let us now describe how the simulated results areobtained First of all we use the model N1 to generate datawhen the 120590119873 = 05 1 and 120572 = 03 05 07 09 Secondly wegenerate data from model N2 with the degrees of freedom119898 = 8 10 and 120572 = 03 05 07 09 For model N1 and N2we take the sample size 119873 = 100 200 300 400 500 600700 800 900 1000 and 5000 Thirdly we simulate samplesfrommodel N3 with the (120585 1205901 1205902 ) = (08 1 3) (120585 1205901 1205902 ) =(09 1 3) and 120572 = 03 05 07 09 For model N3 we set thesample size119873 = 100 300 500 1000 and 5000 The empiricallevels of threemodels are presented inTables 1ndash3 respectivelyAs seen from Tables 1ndash3 the LMP test give similar resultsunder normal studentrsquos t and symmetric contaminationerrors In addition the LMP test produces sizes closer to thenominal significance level 005 quite satisfactorily especiallyfor larger sample sizes However the EL method has lowerlevels for each sample size so one may make wrong decisionsin tests based on them Looking at three models results wecan conclude that our method has a greater effect than the ELtest in terms of the empirical level
4 Mathematical Problems in Engineering
Table 2 Empirical sizes of LMP and EL tests at nominal level 005 for model N2
119873 (119898 120572)(8 03) (8 05) (8 07) (8 09) (10 03) (10 05) (10 07) (10 09)100 MP 0060 0065 0066 0061 0065 0068 0075 0067
EL 0003 0003 0002 0003 0004 0003 0004 0001200 LMP 0059 0061 0065 0059 0066 0063 0068 0061
EL 0002 0004 0002 0002 0004 0004 0002 0004300 LMP 0058 0057 0059 0055 0063 0058 0063 0055
EL 0005 0004 0003 0009 0006 0006 0004 0002400 LMP 0058 0055 0058 0055 0062 0054 0061 0058
EL 0004 0003 0005 0003 0004 0003 0005 0006500 LMP 0056 0055 0056 0052 0057 0059 0055 0054
EL 0006 0004 0005 0005 0007 0005 0006 0007600 LMP 0060 0057 0058 0060 0062 0059 0056 0053
EL 0003 0005 0005 0005 0008 0007 0009 0006700 LMP 0055 0056 0058 0057 0060 0060 0056 0058
EL 0006 0005 0005 0008 0006 0006 0004 0008800 LMP 0050 0054 0054 0052 0056 0054 0054 0054
EL 0005 0007 0008 0006 0006 0008 0008 0008900 LMP 0051 0052 0058 0056 0056 0060 0049 0051
EL 0005 0006 0006 0009 0006 0008 0010 00051000 LMP 0052 0056 0057 0055 0050 0053 0050 0054
EL 0007 0008 0005 0006 0012 0009 0009 00075000 LMP 0052 0050 0050 0050 0051 0049 0054 0052
EL 0015 0014 0012 0016 0017 0014 0012 0013
Table 3 Empirical sizes of LMP and EL tests at nominal level 005 for model N3
119873 (1205901 1205902 120572)(1 3 01) (1 3 02) (1 3 03) (1 3 04) (1 3 05) (1 3 06) (1 3 07) (1 3 08) (1 3 09)120585 = 08100 LMP 0064 0068 0065 0072 0053 0078 0071 0053 0047
EL 0000 0000 0001 0000 0000 0001 0000 0000 0001300 LMP 0045 0066 0053 0038 0054 0081 0051 0043 0040
EL 0000 0001 0000 0001 0001 0002 0000 0001 0000500 LMP 0058 0057 0057 0049 0050 0060 0043 0049 0044
EL 0001 0000 0001 0000 0000 0001 0000 0000 00011000 LMP 0050 0059 0052 0051 0045 0054 0036 0043 0036
EL 0002 0001 0001 0001 0001 0001 0003 0001 00015000 LMP 0053 0050 0054 0050 0048 0050 0047 0050 0046
EL 0003 0004 0003 0007 0007 0006 0008 0006 0011120585 = 09100 LMP 0078 0081 0089 0069 0089 0095 0083 0091 0086
EL 0000 0002 0001 0000 0000 0000 0000 0000 0001300 LMP 0080 0070 0074 0070 0070 0056 0063 0057 0055
EL 0000 0000 0001 0000 0000 0001 0002 0000 0001500 LMP 0072 0081 0068 0067 0058 0059 0060 0046 0053
EL 0001 0000 0001 0000 0000 0001 0001 0000 00041000 LMP 0064 0061 0067 0067 0055 0055 0052 0051 0045
EL 0000 0000 0000 0001 0002 0003 0003 0000 00015000 LMP 0055 0053 0050 0050 0055 0046 0050 0047 0050
EL 0004 0002 0003 0002 0002 0006 0003 0003 0009
Mathematical Problems in Engineering 5
Table 4 Empirical powers of LMP and EL tests at nominal level 005 for model A1
119873 (120590119873 119886 119887)(05 05 05) (05 05 1) (05 1 05) (05 1 1) (1 05 05) (1 05 1) (1 1 05) (1 1 1)100 LMP 0165 0102 0171 0110 0171 0107 0174 0090
EL 0055 0024 0040 0023 0050 0022 0041 0025200 LMP 0318 0143 0294 0147 0320 0140 0311 0160
EL 0191 0070 0156 0073 0202 0049 0155 0067300 LMP 0513 0209 0494 0253 0511 0175 0449 0241
EL 0405 0144 0332 0173 0397 0124 0294 0168400 LMP 0674 0232 0561 0330 0663 0264 0590 0338
EL 0578 0176 0456 0272 0559 0189 0450 0260500 LMP 0791 0377 0720 0427 0770 0374 0725 0438
EL 0722 0298 0629 0366 0705 0303 0625 0374600 LMP 0842 0415 0792 0521 0832 0424 0790 0508
EL 0798 0352 0717 0460 0793 0365 0712 0458700 LMP 0912 0483 0862 0598 0919 0478 0855 0598
EL 0882 0429 0809 0541 0885 0423 0794 0536800 LMP 0964 0560 0915 0645 0947 0568 0918 0643
EL 0938 0512 0877 0604 0928 0516 0872 0599900 LMP 0967 0633 0936 0727 0971 0623 0955 0703
EL 0952 0586 0910 0696 0959 0586 0933 06871000 LMP 0988 0712 0963 0799 0978 0702 0962 0776
EL 0981 0678 0942 0774 0969 0676 0938 07495000 LMP 1000 1000 1000 1000 1000 1000 1000 1000
EL 1000 1000 1000 1000 1000 1000 1000 1000
Table 5 Empirical powers of LMP and EL tests at nominal level 005 for model A2119873 (119898 119886 119887)(6 05 05) (6 05 1) (6 1 05) (6 1 1) (10 05 05) (10 05 1) (10 1 05) (10 1 1)100 LMP 0289 0110 0220 0149 0241 0096 0183 0131
EL 0045 0016 0027 0019 0047 0018 0027 0029200 LMP 0596 0252 0519 0310 0525 0199 0452 0272
EL 0161 0052 0118 0055 0186 0057 0146 0063300 LMP 0790 0400 0722 0475 0762 0351 0686 0429
EL 0336 0111 0276 0124 0373 0124 0293 0152400 LMP 0899 0538 0861 0644 0862 0462 0815 0564
EL 0486 0147 0399 0196 0520 0173 0440 0223500 LMP 0964 0624 0933 0757 0957 0596 0906 0699
EL 0622 0218 0529 0285 0695 0262 0590 0345600 LMP 0979 0701 0951 0819 0965 0657 0947 0754
EL 0708 0237 0606 0351 0767 0301 0650 0395700 LMP 0995 0791 0974 0862 0983 0727 0978 0824
EL 0814 0351 0669 0404 0839 0389 0773 0483800 LMP 0996 0837 0990 0927 0994 0809 0984 0882
EL 0845 0404 0767 0512 0895 0488 0839 0585900 LMP 0999 0874 0995 0934 0996 0831 0993 0924
EL 0879 0430 0801 0548 0929 0514 0877 06221000 LMP 1000 0936 1000 0971 1000 0889 0996 0951
EL 0913 0515 0887 0626 0946 0601 0930 07065000 LMP 1000 1000 1000 1000 1000 1000 1000 1000
EL 0992 0967 0994 0985 1000 1000 1000 0999
6 Mathematical Problems in Engineering
Table 6 Empirical powers of LMP and EL tests at nominal level 005 for model A3 with (1205901 1205902) = (1 3)119873 (119886 119887)(05 05) (05 1) (05 15) (1 05) (1 1) (1 15) (15 05) (15 1) (15 15)120585 = 08100 LMP 0141 0054 0034 0120 0073 0038 0084 0082 0046
EL 0039 0009 0006 0026 0012 0008 0018 0012 0009300 LMP 0775 0388 0183 0628 0493 0299 0475 0448 0331
EL 0325 0102 0030 0231 0097 0052 0143 0114 0061500 LMP 0959 0675 0381 0880 0791 0594 0754 0729 0583
EL 0656 0236 0070 0500 0268 0143 0347 0222 01311000 LMP 1000 0967 0753 0996 0992 0934 0982 0982 0939
EL 0945 0599 0255 0891 0673 0399 0720 0609 04025000 LMP 1000 1000 1000 1000 1000 1000 1000 1000 1000
EL 1000 1000 0981 1000 1000 0997 0999 1000 0997120585 = 09100 LMP 0118 0043 0035 0071 0041 0025 0049 0052 0037
EL 0042 0013 0015 0032 0021 0014 0020 0013 0011300 LMP 0673 0278 0101 0545 0372 0190 0388 0313 0220
EL 0301 0077 0021 0199 0101 0043 0128 0093 0055500 LMP 0914 0558 0277 0868 0681 0423 0700 0622 0467
EL 0571 0183 0054 0486 0239 0128 0353 0212 01131000 LMP 1000 0929 0627 0995 0965 0849 0970 0954 0865
EL 0927 0503 0204 0852 0637 0375 0702 0548 03685000 LMP 1000 1000 1000 1000 1000 1000 1000 1000 1000
EL 1000 1000 0963 1000 1000 0996 1000 1000 0994
32 Empirical Power In order to investigate the empiricalpowers we consider the alternatives under three versions ofmodel (2)
(A1) 120572119905 sim 119861119890119905119886(119886 119887) 120576119905 sim 119873(0 1205902119873)(A2) 120572119905 sim 119861119890119905119886(119886 119887) 120576119905 sim 119905(119898)(A3)120572119905 sim 119861119890119905119886(119886 119887) 120576119905 sim F120576(119909)Thedistribution function
of 120585-contamination is
F120576 (119909) = 120585Φ( 1199091205901) + (1 minus 120585)Φ( 1199091205902) (10)
where 120585 is a fixed constant satisfying 0 lt 120585 lt 1 andΦ(119909) is thedistribution function of standard normal random variable120590119894 gt 0 119894 = 1 2
To calculate the empirical powers of the two tests inthe first we generate samples from the model A1 in whichthe parameters 119886 = 05 1 119887 = 05 1 and 120590119873 = 05 1In the second setting we simulate samples from model A2with the degrees of freedom 119898 = 6 10 the parameters119886 = 05 1 and 119887 = 05 1 For model A1 and A2 we takethe sample size 119873 = 100 200 300 400 500 600 700800 900 1000 and 5000 In the last case we generate datafrom the model A3 in which the (120585 1205901 1205902 ) = (08 1 3)(120585 1205901 1205902 ) = (09 1 3) and the parameters 119886 = 05 1 15119887 = 05 1 15 For model A3 we employ the sample size119873 = 100 300 500 1000 and 5000 The empirical power ofthe above three models are summed up in Tables 4ndash6 Inthree cases the power of two test statistics increase with thesample size while our test produces relatively better powers
than the EL test Furthermore both tests powers are closeto 1 at the sample size 119873 = 5000 Overall from thesesimulations we conclude that the accuracy of using LMP is67 288 261 higher than that of EL test under normalstudentrsquos 119905 and symmetric contamination errors respectivelyAs anticipated our finding shows that the LMP test is afunctional tool to detect a parameter change for RCAR(1)model Therefore we recommend the LMP for practicaluse because its computation is not difficult and the overallperformance is better than that EL under normal studentrsquos tand symmetric contamination errors
4 A Real Life Data Analysis
In this section we illustrate how the LMP method can beapplied to a practical application This data consist of 78monthly number of annulus growth rate of the import billin Australia starting in December 2010 and ending in May2017 The data are available online at the CEInet StatisticsDatabase site httpdbceicnpageDefaultaspx The meanand variance of the data are found to be 02069 and 153347The data are denoted as 1199101 1199102 11991078 Figure 1 is the samplepath plot for the real data 119910119905 119905 = 1 2 78
For such a process the mean function of 119910119905 is constantand we may assume that the process mean is subtracted outto produce a process 119883119905 = 119910119905 minus 119864(119910119905) with zero mean Theplot of the sample path autocorrelation function (ACF) andthe partial autocorrelation function (PACF) for series 119883119905 aregiven in Figures 1 and 2 respectively From Figure 1 we can
Mathematical Problems in Engineering 7
0 20 40 60 80
minus10
minus50
510
15
Time
Dat
a Val
ue
Time
Dat
a Val
ue
0 20 40 60 80
minus10
minus50
510
15
９Ｎ ８Ｎ
Figure 1 The left one is the sample path of original series 119910119905 and the right one is the sample plot of centering series119883119905 for the real data
0 5 10 15
minus05
00
05
10
Lag
ACF
5 10 15
minus04
minus02
00
02
Lag
PACF
８Ｎ８Ｎ
Figure 2 The sample autocorrelation function (ACF) and partial autocorrelation function (PACF) plots of centering series 119883119905 for the realdata
see119883119905may come from a stationary autoregressive time seriesprocess From Figure 2 we conclude that 119883119905 is from first-order autoregressive process Moreover we test the normalityof the residual data 119883119905 according to the Normal Q-Q plotintroduced by Henry CThode [21] From Figure 3 we noticethat the scatter points on theNormal Q-Q plot are close to thereference line so the residual data119883119905 can basically be seen asfollowing the normal distribution Hence we derive that themodel N1 would be more suitable to model these data
From the time series plot given in Figure 1 the constancyof the parameter may be suspected and therefore we areinterested in testing the hypothesis of constancy of thecoefficient parameter We carry out the test for 1198670 1205902120572 = 0against 1198671 1205902120572 gt 0 The value of the test statistic in (8)turned out to be 21151 with 119901 value 00344 which indicatesthe rejection of the null hypothesis at 5 level of significanceThus in this case the RCAR(1) model would be much moreappropriate as opposed to the AR(1) model
5 Conclusions
In this article we propose a locally most powerful-typetest for testing the constancy of the random coefficientparameter in autoregressive model and derive their limitingnull distributions under regularity conditions It is clear fromthe applications that the coefficient need not remain constantthroughout the time Therefore it is essential to have such a
Normal QminusQ Plot
minus10
minus50
510
15Sa
mpl
e Qua
ntile
s
minus1 0 1 2minus2Theoretical Quantiles
Figure 3 The normal Q-Q plot of centering series 119883119905 for the realdata
test conducted whenever the random variation is suspectedThrough our simulation study and a real-data analysis wedemonstrate that our test succeeds and it performs better thanthe competitor Finally we anticipate that our locally mostpowerful-type test can be extended to other types of timeseries model
8 Mathematical Problems in Engineering
Appendix
A Process of Establish the Test Statistic
In this Appendix A we present the detailed steps to obtainthe test statistics
Proof Therefore we can get the likelihood function
1198711198671 (1199091 119909119905) = 1198911198830 (1199090) 119899prod119905=1
119891119883119905|119883119905minus1 (119909119905 | 119909119905minus1)= 1198911198830 (1199090) 119899prod
119905=1
int119891120576 (119909119905 minus 120572119905119909119905minus1) 119889119865120572 (A1)
Now the Taylor series expansion of 1198711198671 around 120572 which isthe mean of 120572119905 gives that
1198711198671 (1199091 119909119905) = 1198911198830 (1199090) 119899prod119905=1
int119891120576 (119909119905minus 120572119905119909119905minus1) 119889119865120572
= 1198911198830 (1199090) 119899prod119905=1
int[119891120576 (119909119905 minus 120572119909119905minus1)
+ (120572119905 minus 120572) 1198911015840120576 (119909119905 minus 120572119909119905minus1)+ (120572119905 minus 120572)22 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)+ (120572119905 minus 120572)33 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) + sdot sdot sdot] 119889119865120572
= 1198911198830 (1199090) 119899prod119905=1
[119891120576 (119909119905 minus 120572119909119905minus1) + 12120590212057211989110158401015840120576 (119909119905minus 120572119909119905minus1) + 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) int (120572119905 minus 120572)33 119889119865120572+ sdot sdot sdot]
(A2)
Obviously we can conclude that 119875(120572119905 minus 120572 = 0) = 1 if 1205902120572 = 0That is to say the distribution function of random variable 120572119905is degenerated Furthermore we derive that
int (120572119905 minus 120572)119896119896 119889119865120572 = 0 119896 = 1 2 3 sdot sdot sdot (A3)
By taking the derivative of log 1198711198671 with respect to120573 at1205902120572 = 0we have the following equation
120597 log 11987111986711205971205902120572 1003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 120597sum119899119905=1 log [119891120576 (119909119905 minus 120572119909119905minus1) + (12) 120590212057211989110158401015840120576 (119909119905 minus 120572119909119905minus1) + 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) int ((120572119905 minus 120572)3 3) 119889119865120572 + sdot sdot sdot ]1205971205902120572 100381610038161003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 120597sum119899119905=1 log [119891120576 (119909119905 minus 120572119909119905minus1) + (12) 120590212057211989110158401015840120576 (119909119905 minus 120572119909119905minus1)]1205971205902120572 100381610038161003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 12 119899sum119905=111989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) (A4)
Thus the LMP test statistic has the form119876119899 (120573) = 119899sum119905=1
12 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) (A5)
This completes the proof
B Proof of Theorem 1
Proof of Theorem 1 In the following we can write down thelog-likelihood function under the null hypothesis
log 119871 = 119899sum119905=1
log119891120576 (119909119905 minus 120572119909119905minus1) (B1)
According to the maximum likelihood principle the maxi-mum likelihood estimate can be obtained by maximizing
the log-likelihood function log 119871 with respect to 120573 Applyingthe Taylor series expansion to 120597 log119891120576120597120573 gives
0 = 120597log119891120576120597120573 = 120597log119891120576120597120573 + ( minus 120573) 1205972log119891120576 (120573lowast)1205971205731205971205731015840 (B2)
where 120573lowast is on the line segment between 120573 and so 120573lowast119901997888rarr
120573(119899 997888rarr infin) Under the null hypothesis we can write
minus 120573 = minus 119899sum119905=1
1205972 log1198911205761205971205731205971205731015840 + 119900119901 (119899)minus1 119899sum119905=1120597 log119891120576120597120573 (B3)
Mathematical Problems in Engineering 9
Using the similar method as in the above expanding(1radic119899)sum119899119905=1 119877119905() around 120573 gives that1radic119899 119899sum119905=1119877119905 () = 1radic119899 119899sum119905=1119877119905 (120573)+ ( minus 120573)119879 1radic119899 119899sum119905=1120597119877119905 (120573)120597120573 + 119900119901 (1) (B4)
with 120597119877119905(120573)120597120573 = (120597119877119905120597120572 1205971198771199051205971205902)119879 The remainder termis 119900119901(1) since 119901997888rarr 120573Under (C3) and (C4) we conclude that1205972119877119905(120573)1205971205731205971205731015840 = 119874119901(119899) Therefore
( minus 120573)119879 ( 1radic119899) 119899sum119905=1120597119877119905 (120573)120597120573= minus1119899 119899sum119905=1[120597119877119905 (120573)120597120573 ]119879times 1119899 119899sum119905=11205972 log1198911205761205971205731205971205731015840 + 119900119901 (1)minus1times 1radic119899 119899sum119905=1120597 log119891120576120597120573
(B5)
Now the first term of (B5) converges to1119899 119899sum119905=1120597119877119905 (120573)120597120573 = 1119899 119899sum119905=1 120597120597120573 120597 log1198911205761205971205902120572 1205902120572=0
= (1119899 119899sum119905=11205972 log11989112057612059712057212059712059021205721119899 119899sum119905=11205972 log11989112057612059712059021205761205971205902120572 )1205902120572=0
997888rarr 119882as 119899 997888rarr infin
(B6)
with119882 = (1198821111988212)119879This is because
11988211 = 119864(120597 log1198911205761205971205721205971205902120572 ) = 119864(minus120597 log119891120576120597120572 120597 log1198911205761205971205902120572 )= minus12119864[120597 log119891120576120597120572 1205972 log1198911205761205971205722 + (120597 log119891120576120597120572 )2]
at 1205902120572 = 011988212 = 119864(120597 log1198911205761205971205902120576 1205971205902120572) = 119864(minus120597 log1198911205761205971205902120576 120597 log1198911205761205971205902120572 )= minus12119864[120597 log1198911205761205971205902120576 1205972 log1198911205761205971205722 + (120597 log119891120576120597120572 )2]
at 1205902120572 = 0
(B7)
Next we can obtain the elements of Fisher informationmatrix
Σ = 1119899 119899sum119905=1119864[120597 log119891120576120597120573 (120597 log119891120576120597120573 )119879 | F119910119905minus1]= [[[[[
minus119864[1205972 log1198911205761205971205722 ] minus119864[1205972 log1198911205761205971205721205971205902120576 ]minus119864[1205972 log1198911205761205971205721205971205902120576 ] minus119864[1205972 log11989112057612059712059021205761205971205902120576 ]]]]]]
(B8)
Thus (B5) is asymptotically equivalent to
minus119882Σminus1 1radic119899 119899sum119905=1120597 log119891120576120597120573 (B9)
We present the following lemma recommended by Billingsley[22] to establish the asymptotic normality of the test statistic
Lemma B1Under assumption of (A1) we have as 119899 997888rarr infin(i) lim119899997888rarrinfin(1119899)sum119899119905=1 119864(1198772119905 (120573) | F119877119905minus1) = 1205902 119886119904 (ii)(1radic119899)sum119899119905=1 119877119905(120573) 119889997888rarr 119873(0 1205902) where 1205902 = (14)1198641205972 log1198911205761205971205722 + (120597 log119891120576120597120572)2
Proof By the ergodic theorem the first conclusion holdsNow let F119905 = 1205901199090 1199091 119909119905 119905 ge 1 and F0 is a sigma
field Note that 119876119899(120573) = sum119899119905=1 119877119905(120573) It is easy to see that[119877119899(120573)|F119899minus1] = 0 Then we have
119864 [119876119899 (120573) | F119899minus1] = 119864 [119876119899minus1 (120573) + 119877119899 (120573) | F119899minus1]= 119876119899minus1 (120573) (B10)
Thus 119876119899F119899 119899 ge 0 is a martingale We have shownsup119905119864|119877119905(120573)|2+120578 lt infin which means 119864[1198772119905 (120573)] is uniformlyintegrable ByTheorem 11 of Billingsley [22] we have that as119899 997888rarr infin1119899 119899sum119905=11198772119905 (120573) 119886119904997888997888rarr 119864[1198772119899 (120573) | F119899minus1] = 1205902 (B11)
Hence using the following version of Martingale Cen-tral Limit Theorem from Hall and Heyde [23] we have(1radic119899)sum119899119905=1 119877119905(120573) 119889997888rarr 119873(0 1205902) The proof of Lemma B1 isthus completed ◻
Similarly we can verify that 119878119899 = sum119899119905=1(120597log119891120576120597120573) is amartingale By ergodic and stationary properties we obtainthat as 119899 997888rarr infin
1119899 119899sum119905=11205972 log119891120576120597120573120597120573119879 119886119904997888997888rarr 119864[120597 log119891120576120597120573 (120597 log119891120576120597120573 )119879] = Σ (B12)
Therefore (1radic119899)119878119899 119889997888rarr 119873(0 Σ)
10 Mathematical Problems in Engineering
In the same way for any vector 119888 = (1198881 1198882 1198883)119879 isin 1198773 (0 0 0) we have1radic119899119888119879(
119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) = 1radic119899sdot 119899sum119905=1
(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 ) 119889997888rarr 119873(0 119864(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 )2)
(B13)
By the Cramer-Wold device we obtain
1radic119899 (119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) 119889997888rarr 119873[(00) (1205902 119882119879119882 Σ )] (B14)
where119882 = (1198821111988212)119879 Then we can make the conclusionthat 1radic119899 119899sum119905=1119877119905 () 119889997888rarr 119873(0 1205962) (B15)
where 1205962 = 1205902 minus 119882119879Σminus1119882 The proof of the theorem iscompleted
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that they have no conflicts of interest
Acknowledgments
This work is supported by National Natural Science Foun-dation of China (Nos 11871028 11731015 11571051 and11501241) Natural Science Foundation of Jilin Province (Nos20180101216JC 20170101057JC and 20150520053JH) Pro-gram for Changbaishan Scholars of Jilin Province (2015010)and Science and Technology Program of Jilin EducationalDepartment during the ldquo13th Five-Yearrdquo Plan Period (No2016316)
References
[1] A Hoque ldquoFinite sample analysis of the first order autoregres-sive modelrdquoCalcutta Statistical Association Bulletin vol 34 no1-2 pp 51ndash63 1985
[2] T Ogawa H Sonoda S Ishiwa and Y Shigeta ldquoAn applicationof autoregressive model to pattern discrimination of brainelectrical activity mappingrdquo Brain Topography vol 6 no 1 pp3ndash11 1992
[3] A Subasi A Alkan E Koklukaya and M K Kiymik ldquoWaveletneural network classification of EEG signals by using ARmodelwith MLE preprocessingrdquo Journal of the International NeuralNetwork Society vol 18 pp 985ndash997 2005
[4] M Maleki and A R Nematollahi ldquoAutoregressive models withmixture of scale mixtures of gaussian innovationsrdquo IranianJournal of Science amp Technology Transactions A Science vol 41no 4 pp 1099ndash1107 2017
[5] J Conlisk ldquoStability in a random coefficient modelrdquo Interna-tional Economic Review vol 15 no 2 pp 529ndash533 1974
[6] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models Irdquo Journal of Time SeriesAnalysis vol 1 no 1 pp 37ndash46 1980
[7] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models IIrdquo Journal of Time SeriesAnalysis vol 2 no 3 pp 185ndash203 1981
[8] D F Nicholls and B G Quinn Random Coefficient Autore-gressive Models An Introduction Lecture Notes in StatisticsSpringer-Verlag New York NY USA 1982
[9] A Brandt ldquoThe stochastic equation Y119899+1 = A119899Y119899 + B119899 withstationary coefficientsrdquo Advances in Applied Probability vol 18no 1 pp 211ndash220 1986
[10] D Wang and S K Ghosh ldquoBayesian estimation and unitroot tests for random coefficient autoregressive modelsrdquoModelAssisted Statistics and Applications vol 3 no 4 pp 281ndash2952008
[11] DWang S K Ghosh and S G Pantula ldquoMaximum likelihoodestimation and unit root test for first order random coefficientautoregressive modelsrdquo Journal of Statistical Theory and Prac-tice vol 4 no 2 pp 261ndash278 2010
[12] T V Ramanathan and M B Rajarshi ldquoRank tests for testingthe randomness of autoregressive coefficientsrdquo Statistics ampProbability Letters vol 21 no 2 pp 115ndash120 1994
[13] S Lee J Ha O Na and S Na ldquoThe cusum test for parameterchange in time seriesmodelsrdquo Scandinavian Journal of Statisticsvol 30 no 4 pp 781ndash796 2003
[14] M Moreno and J Romo ldquoRobust unit root tests with autore-gressive errorsrdquo Communications in StatisticsmdashTheory andMethods vol 45 no 20 pp 5997ndash6021 2016
[15] V K Rohatgi A K Md Ehsanes Sleh R Ahluwalia and P JildquoNull distribution of locally most powerful tests for the twosample problemwhen the combined sample is type II censoredrdquoCommunications in Statistics - Theory and Methods vol 19 pp2337ndash2355 1992
[16] M S Chikkagoudar and B S Biradar ldquoLocally most powerfulrank tests for comparison of two failure rates based on multipletype-ii censored datardquoCommunications in Statistics Theory andMethods vol 41 no 23 pp 4315ndash4331 2012
[17] A Manik N Balakrishna and T V Ramanathan ldquoTesting theconstancy of the thinning parameter in a random coefficientinteger autoregressive modelrdquo Statistical Papers pp 1ndash25 2017
[18] P J Huber Robust Statistics JohnWiley amp Sons New York NYUSA 1981
[19] T P Hettmansperger Statistical Inference Based on Ranks JohnWiley amp Sons Inc New York NY USA 1984
[20] Z-W Zhao D-H Wang and C-X Peng ldquoCoefficient con-stancy test in generalized random coefficient autoregressivemodelrdquoApplied Mathematics and Computation vol 219 no 20pp 10283ndash10292 2013
[21] H C Thode Testing for Normality CRC Press New York NYUSA 2002
Mathematical Problems in Engineering 11
[22] P Billingsley ldquoThe lindeberg-Lvy theorem for martingalesrdquoProceedings of the American Mathematical Society vol 12 no1 pp 788ndash792 1961
[23] P Hall and C C Heyde Martingale Limit Theory and ItsApplication Academic Press New York NY USA 1980
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4 Mathematical Problems in Engineering
Table 2 Empirical sizes of LMP and EL tests at nominal level 005 for model N2
119873 (119898 120572)(8 03) (8 05) (8 07) (8 09) (10 03) (10 05) (10 07) (10 09)100 MP 0060 0065 0066 0061 0065 0068 0075 0067
EL 0003 0003 0002 0003 0004 0003 0004 0001200 LMP 0059 0061 0065 0059 0066 0063 0068 0061
EL 0002 0004 0002 0002 0004 0004 0002 0004300 LMP 0058 0057 0059 0055 0063 0058 0063 0055
EL 0005 0004 0003 0009 0006 0006 0004 0002400 LMP 0058 0055 0058 0055 0062 0054 0061 0058
EL 0004 0003 0005 0003 0004 0003 0005 0006500 LMP 0056 0055 0056 0052 0057 0059 0055 0054
EL 0006 0004 0005 0005 0007 0005 0006 0007600 LMP 0060 0057 0058 0060 0062 0059 0056 0053
EL 0003 0005 0005 0005 0008 0007 0009 0006700 LMP 0055 0056 0058 0057 0060 0060 0056 0058
EL 0006 0005 0005 0008 0006 0006 0004 0008800 LMP 0050 0054 0054 0052 0056 0054 0054 0054
EL 0005 0007 0008 0006 0006 0008 0008 0008900 LMP 0051 0052 0058 0056 0056 0060 0049 0051
EL 0005 0006 0006 0009 0006 0008 0010 00051000 LMP 0052 0056 0057 0055 0050 0053 0050 0054
EL 0007 0008 0005 0006 0012 0009 0009 00075000 LMP 0052 0050 0050 0050 0051 0049 0054 0052
EL 0015 0014 0012 0016 0017 0014 0012 0013
Table 3 Empirical sizes of LMP and EL tests at nominal level 005 for model N3
119873 (1205901 1205902 120572)(1 3 01) (1 3 02) (1 3 03) (1 3 04) (1 3 05) (1 3 06) (1 3 07) (1 3 08) (1 3 09)120585 = 08100 LMP 0064 0068 0065 0072 0053 0078 0071 0053 0047
EL 0000 0000 0001 0000 0000 0001 0000 0000 0001300 LMP 0045 0066 0053 0038 0054 0081 0051 0043 0040
EL 0000 0001 0000 0001 0001 0002 0000 0001 0000500 LMP 0058 0057 0057 0049 0050 0060 0043 0049 0044
EL 0001 0000 0001 0000 0000 0001 0000 0000 00011000 LMP 0050 0059 0052 0051 0045 0054 0036 0043 0036
EL 0002 0001 0001 0001 0001 0001 0003 0001 00015000 LMP 0053 0050 0054 0050 0048 0050 0047 0050 0046
EL 0003 0004 0003 0007 0007 0006 0008 0006 0011120585 = 09100 LMP 0078 0081 0089 0069 0089 0095 0083 0091 0086
EL 0000 0002 0001 0000 0000 0000 0000 0000 0001300 LMP 0080 0070 0074 0070 0070 0056 0063 0057 0055
EL 0000 0000 0001 0000 0000 0001 0002 0000 0001500 LMP 0072 0081 0068 0067 0058 0059 0060 0046 0053
EL 0001 0000 0001 0000 0000 0001 0001 0000 00041000 LMP 0064 0061 0067 0067 0055 0055 0052 0051 0045
EL 0000 0000 0000 0001 0002 0003 0003 0000 00015000 LMP 0055 0053 0050 0050 0055 0046 0050 0047 0050
EL 0004 0002 0003 0002 0002 0006 0003 0003 0009
Mathematical Problems in Engineering 5
Table 4 Empirical powers of LMP and EL tests at nominal level 005 for model A1
119873 (120590119873 119886 119887)(05 05 05) (05 05 1) (05 1 05) (05 1 1) (1 05 05) (1 05 1) (1 1 05) (1 1 1)100 LMP 0165 0102 0171 0110 0171 0107 0174 0090
EL 0055 0024 0040 0023 0050 0022 0041 0025200 LMP 0318 0143 0294 0147 0320 0140 0311 0160
EL 0191 0070 0156 0073 0202 0049 0155 0067300 LMP 0513 0209 0494 0253 0511 0175 0449 0241
EL 0405 0144 0332 0173 0397 0124 0294 0168400 LMP 0674 0232 0561 0330 0663 0264 0590 0338
EL 0578 0176 0456 0272 0559 0189 0450 0260500 LMP 0791 0377 0720 0427 0770 0374 0725 0438
EL 0722 0298 0629 0366 0705 0303 0625 0374600 LMP 0842 0415 0792 0521 0832 0424 0790 0508
EL 0798 0352 0717 0460 0793 0365 0712 0458700 LMP 0912 0483 0862 0598 0919 0478 0855 0598
EL 0882 0429 0809 0541 0885 0423 0794 0536800 LMP 0964 0560 0915 0645 0947 0568 0918 0643
EL 0938 0512 0877 0604 0928 0516 0872 0599900 LMP 0967 0633 0936 0727 0971 0623 0955 0703
EL 0952 0586 0910 0696 0959 0586 0933 06871000 LMP 0988 0712 0963 0799 0978 0702 0962 0776
EL 0981 0678 0942 0774 0969 0676 0938 07495000 LMP 1000 1000 1000 1000 1000 1000 1000 1000
EL 1000 1000 1000 1000 1000 1000 1000 1000
Table 5 Empirical powers of LMP and EL tests at nominal level 005 for model A2119873 (119898 119886 119887)(6 05 05) (6 05 1) (6 1 05) (6 1 1) (10 05 05) (10 05 1) (10 1 05) (10 1 1)100 LMP 0289 0110 0220 0149 0241 0096 0183 0131
EL 0045 0016 0027 0019 0047 0018 0027 0029200 LMP 0596 0252 0519 0310 0525 0199 0452 0272
EL 0161 0052 0118 0055 0186 0057 0146 0063300 LMP 0790 0400 0722 0475 0762 0351 0686 0429
EL 0336 0111 0276 0124 0373 0124 0293 0152400 LMP 0899 0538 0861 0644 0862 0462 0815 0564
EL 0486 0147 0399 0196 0520 0173 0440 0223500 LMP 0964 0624 0933 0757 0957 0596 0906 0699
EL 0622 0218 0529 0285 0695 0262 0590 0345600 LMP 0979 0701 0951 0819 0965 0657 0947 0754
EL 0708 0237 0606 0351 0767 0301 0650 0395700 LMP 0995 0791 0974 0862 0983 0727 0978 0824
EL 0814 0351 0669 0404 0839 0389 0773 0483800 LMP 0996 0837 0990 0927 0994 0809 0984 0882
EL 0845 0404 0767 0512 0895 0488 0839 0585900 LMP 0999 0874 0995 0934 0996 0831 0993 0924
EL 0879 0430 0801 0548 0929 0514 0877 06221000 LMP 1000 0936 1000 0971 1000 0889 0996 0951
EL 0913 0515 0887 0626 0946 0601 0930 07065000 LMP 1000 1000 1000 1000 1000 1000 1000 1000
EL 0992 0967 0994 0985 1000 1000 1000 0999
6 Mathematical Problems in Engineering
Table 6 Empirical powers of LMP and EL tests at nominal level 005 for model A3 with (1205901 1205902) = (1 3)119873 (119886 119887)(05 05) (05 1) (05 15) (1 05) (1 1) (1 15) (15 05) (15 1) (15 15)120585 = 08100 LMP 0141 0054 0034 0120 0073 0038 0084 0082 0046
EL 0039 0009 0006 0026 0012 0008 0018 0012 0009300 LMP 0775 0388 0183 0628 0493 0299 0475 0448 0331
EL 0325 0102 0030 0231 0097 0052 0143 0114 0061500 LMP 0959 0675 0381 0880 0791 0594 0754 0729 0583
EL 0656 0236 0070 0500 0268 0143 0347 0222 01311000 LMP 1000 0967 0753 0996 0992 0934 0982 0982 0939
EL 0945 0599 0255 0891 0673 0399 0720 0609 04025000 LMP 1000 1000 1000 1000 1000 1000 1000 1000 1000
EL 1000 1000 0981 1000 1000 0997 0999 1000 0997120585 = 09100 LMP 0118 0043 0035 0071 0041 0025 0049 0052 0037
EL 0042 0013 0015 0032 0021 0014 0020 0013 0011300 LMP 0673 0278 0101 0545 0372 0190 0388 0313 0220
EL 0301 0077 0021 0199 0101 0043 0128 0093 0055500 LMP 0914 0558 0277 0868 0681 0423 0700 0622 0467
EL 0571 0183 0054 0486 0239 0128 0353 0212 01131000 LMP 1000 0929 0627 0995 0965 0849 0970 0954 0865
EL 0927 0503 0204 0852 0637 0375 0702 0548 03685000 LMP 1000 1000 1000 1000 1000 1000 1000 1000 1000
EL 1000 1000 0963 1000 1000 0996 1000 1000 0994
32 Empirical Power In order to investigate the empiricalpowers we consider the alternatives under three versions ofmodel (2)
(A1) 120572119905 sim 119861119890119905119886(119886 119887) 120576119905 sim 119873(0 1205902119873)(A2) 120572119905 sim 119861119890119905119886(119886 119887) 120576119905 sim 119905(119898)(A3)120572119905 sim 119861119890119905119886(119886 119887) 120576119905 sim F120576(119909)Thedistribution function
of 120585-contamination is
F120576 (119909) = 120585Φ( 1199091205901) + (1 minus 120585)Φ( 1199091205902) (10)
where 120585 is a fixed constant satisfying 0 lt 120585 lt 1 andΦ(119909) is thedistribution function of standard normal random variable120590119894 gt 0 119894 = 1 2
To calculate the empirical powers of the two tests inthe first we generate samples from the model A1 in whichthe parameters 119886 = 05 1 119887 = 05 1 and 120590119873 = 05 1In the second setting we simulate samples from model A2with the degrees of freedom 119898 = 6 10 the parameters119886 = 05 1 and 119887 = 05 1 For model A1 and A2 we takethe sample size 119873 = 100 200 300 400 500 600 700800 900 1000 and 5000 In the last case we generate datafrom the model A3 in which the (120585 1205901 1205902 ) = (08 1 3)(120585 1205901 1205902 ) = (09 1 3) and the parameters 119886 = 05 1 15119887 = 05 1 15 For model A3 we employ the sample size119873 = 100 300 500 1000 and 5000 The empirical power ofthe above three models are summed up in Tables 4ndash6 Inthree cases the power of two test statistics increase with thesample size while our test produces relatively better powers
than the EL test Furthermore both tests powers are closeto 1 at the sample size 119873 = 5000 Overall from thesesimulations we conclude that the accuracy of using LMP is67 288 261 higher than that of EL test under normalstudentrsquos 119905 and symmetric contamination errors respectivelyAs anticipated our finding shows that the LMP test is afunctional tool to detect a parameter change for RCAR(1)model Therefore we recommend the LMP for practicaluse because its computation is not difficult and the overallperformance is better than that EL under normal studentrsquos tand symmetric contamination errors
4 A Real Life Data Analysis
In this section we illustrate how the LMP method can beapplied to a practical application This data consist of 78monthly number of annulus growth rate of the import billin Australia starting in December 2010 and ending in May2017 The data are available online at the CEInet StatisticsDatabase site httpdbceicnpageDefaultaspx The meanand variance of the data are found to be 02069 and 153347The data are denoted as 1199101 1199102 11991078 Figure 1 is the samplepath plot for the real data 119910119905 119905 = 1 2 78
For such a process the mean function of 119910119905 is constantand we may assume that the process mean is subtracted outto produce a process 119883119905 = 119910119905 minus 119864(119910119905) with zero mean Theplot of the sample path autocorrelation function (ACF) andthe partial autocorrelation function (PACF) for series 119883119905 aregiven in Figures 1 and 2 respectively From Figure 1 we can
Mathematical Problems in Engineering 7
0 20 40 60 80
minus10
minus50
510
15
Time
Dat
a Val
ue
Time
Dat
a Val
ue
0 20 40 60 80
minus10
minus50
510
15
９Ｎ ８Ｎ
Figure 1 The left one is the sample path of original series 119910119905 and the right one is the sample plot of centering series119883119905 for the real data
0 5 10 15
minus05
00
05
10
Lag
ACF
5 10 15
minus04
minus02
00
02
Lag
PACF
８Ｎ８Ｎ
Figure 2 The sample autocorrelation function (ACF) and partial autocorrelation function (PACF) plots of centering series 119883119905 for the realdata
see119883119905may come from a stationary autoregressive time seriesprocess From Figure 2 we conclude that 119883119905 is from first-order autoregressive process Moreover we test the normalityof the residual data 119883119905 according to the Normal Q-Q plotintroduced by Henry CThode [21] From Figure 3 we noticethat the scatter points on theNormal Q-Q plot are close to thereference line so the residual data119883119905 can basically be seen asfollowing the normal distribution Hence we derive that themodel N1 would be more suitable to model these data
From the time series plot given in Figure 1 the constancyof the parameter may be suspected and therefore we areinterested in testing the hypothesis of constancy of thecoefficient parameter We carry out the test for 1198670 1205902120572 = 0against 1198671 1205902120572 gt 0 The value of the test statistic in (8)turned out to be 21151 with 119901 value 00344 which indicatesthe rejection of the null hypothesis at 5 level of significanceThus in this case the RCAR(1) model would be much moreappropriate as opposed to the AR(1) model
5 Conclusions
In this article we propose a locally most powerful-typetest for testing the constancy of the random coefficientparameter in autoregressive model and derive their limitingnull distributions under regularity conditions It is clear fromthe applications that the coefficient need not remain constantthroughout the time Therefore it is essential to have such a
Normal QminusQ Plot
minus10
minus50
510
15Sa
mpl
e Qua
ntile
s
minus1 0 1 2minus2Theoretical Quantiles
Figure 3 The normal Q-Q plot of centering series 119883119905 for the realdata
test conducted whenever the random variation is suspectedThrough our simulation study and a real-data analysis wedemonstrate that our test succeeds and it performs better thanthe competitor Finally we anticipate that our locally mostpowerful-type test can be extended to other types of timeseries model
8 Mathematical Problems in Engineering
Appendix
A Process of Establish the Test Statistic
In this Appendix A we present the detailed steps to obtainthe test statistics
Proof Therefore we can get the likelihood function
1198711198671 (1199091 119909119905) = 1198911198830 (1199090) 119899prod119905=1
119891119883119905|119883119905minus1 (119909119905 | 119909119905minus1)= 1198911198830 (1199090) 119899prod
119905=1
int119891120576 (119909119905 minus 120572119905119909119905minus1) 119889119865120572 (A1)
Now the Taylor series expansion of 1198711198671 around 120572 which isthe mean of 120572119905 gives that
1198711198671 (1199091 119909119905) = 1198911198830 (1199090) 119899prod119905=1
int119891120576 (119909119905minus 120572119905119909119905minus1) 119889119865120572
= 1198911198830 (1199090) 119899prod119905=1
int[119891120576 (119909119905 minus 120572119909119905minus1)
+ (120572119905 minus 120572) 1198911015840120576 (119909119905 minus 120572119909119905minus1)+ (120572119905 minus 120572)22 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)+ (120572119905 minus 120572)33 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) + sdot sdot sdot] 119889119865120572
= 1198911198830 (1199090) 119899prod119905=1
[119891120576 (119909119905 minus 120572119909119905minus1) + 12120590212057211989110158401015840120576 (119909119905minus 120572119909119905minus1) + 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) int (120572119905 minus 120572)33 119889119865120572+ sdot sdot sdot]
(A2)
Obviously we can conclude that 119875(120572119905 minus 120572 = 0) = 1 if 1205902120572 = 0That is to say the distribution function of random variable 120572119905is degenerated Furthermore we derive that
int (120572119905 minus 120572)119896119896 119889119865120572 = 0 119896 = 1 2 3 sdot sdot sdot (A3)
By taking the derivative of log 1198711198671 with respect to120573 at1205902120572 = 0we have the following equation
120597 log 11987111986711205971205902120572 1003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 120597sum119899119905=1 log [119891120576 (119909119905 minus 120572119909119905minus1) + (12) 120590212057211989110158401015840120576 (119909119905 minus 120572119909119905minus1) + 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) int ((120572119905 minus 120572)3 3) 119889119865120572 + sdot sdot sdot ]1205971205902120572 100381610038161003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 120597sum119899119905=1 log [119891120576 (119909119905 minus 120572119909119905minus1) + (12) 120590212057211989110158401015840120576 (119909119905 minus 120572119909119905minus1)]1205971205902120572 100381610038161003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 12 119899sum119905=111989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) (A4)
Thus the LMP test statistic has the form119876119899 (120573) = 119899sum119905=1
12 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) (A5)
This completes the proof
B Proof of Theorem 1
Proof of Theorem 1 In the following we can write down thelog-likelihood function under the null hypothesis
log 119871 = 119899sum119905=1
log119891120576 (119909119905 minus 120572119909119905minus1) (B1)
According to the maximum likelihood principle the maxi-mum likelihood estimate can be obtained by maximizing
the log-likelihood function log 119871 with respect to 120573 Applyingthe Taylor series expansion to 120597 log119891120576120597120573 gives
0 = 120597log119891120576120597120573 = 120597log119891120576120597120573 + ( minus 120573) 1205972log119891120576 (120573lowast)1205971205731205971205731015840 (B2)
where 120573lowast is on the line segment between 120573 and so 120573lowast119901997888rarr
120573(119899 997888rarr infin) Under the null hypothesis we can write
minus 120573 = minus 119899sum119905=1
1205972 log1198911205761205971205731205971205731015840 + 119900119901 (119899)minus1 119899sum119905=1120597 log119891120576120597120573 (B3)
Mathematical Problems in Engineering 9
Using the similar method as in the above expanding(1radic119899)sum119899119905=1 119877119905() around 120573 gives that1radic119899 119899sum119905=1119877119905 () = 1radic119899 119899sum119905=1119877119905 (120573)+ ( minus 120573)119879 1radic119899 119899sum119905=1120597119877119905 (120573)120597120573 + 119900119901 (1) (B4)
with 120597119877119905(120573)120597120573 = (120597119877119905120597120572 1205971198771199051205971205902)119879 The remainder termis 119900119901(1) since 119901997888rarr 120573Under (C3) and (C4) we conclude that1205972119877119905(120573)1205971205731205971205731015840 = 119874119901(119899) Therefore
( minus 120573)119879 ( 1radic119899) 119899sum119905=1120597119877119905 (120573)120597120573= minus1119899 119899sum119905=1[120597119877119905 (120573)120597120573 ]119879times 1119899 119899sum119905=11205972 log1198911205761205971205731205971205731015840 + 119900119901 (1)minus1times 1radic119899 119899sum119905=1120597 log119891120576120597120573
(B5)
Now the first term of (B5) converges to1119899 119899sum119905=1120597119877119905 (120573)120597120573 = 1119899 119899sum119905=1 120597120597120573 120597 log1198911205761205971205902120572 1205902120572=0
= (1119899 119899sum119905=11205972 log11989112057612059712057212059712059021205721119899 119899sum119905=11205972 log11989112057612059712059021205761205971205902120572 )1205902120572=0
997888rarr 119882as 119899 997888rarr infin
(B6)
with119882 = (1198821111988212)119879This is because
11988211 = 119864(120597 log1198911205761205971205721205971205902120572 ) = 119864(minus120597 log119891120576120597120572 120597 log1198911205761205971205902120572 )= minus12119864[120597 log119891120576120597120572 1205972 log1198911205761205971205722 + (120597 log119891120576120597120572 )2]
at 1205902120572 = 011988212 = 119864(120597 log1198911205761205971205902120576 1205971205902120572) = 119864(minus120597 log1198911205761205971205902120576 120597 log1198911205761205971205902120572 )= minus12119864[120597 log1198911205761205971205902120576 1205972 log1198911205761205971205722 + (120597 log119891120576120597120572 )2]
at 1205902120572 = 0
(B7)
Next we can obtain the elements of Fisher informationmatrix
Σ = 1119899 119899sum119905=1119864[120597 log119891120576120597120573 (120597 log119891120576120597120573 )119879 | F119910119905minus1]= [[[[[
minus119864[1205972 log1198911205761205971205722 ] minus119864[1205972 log1198911205761205971205721205971205902120576 ]minus119864[1205972 log1198911205761205971205721205971205902120576 ] minus119864[1205972 log11989112057612059712059021205761205971205902120576 ]]]]]]
(B8)
Thus (B5) is asymptotically equivalent to
minus119882Σminus1 1radic119899 119899sum119905=1120597 log119891120576120597120573 (B9)
We present the following lemma recommended by Billingsley[22] to establish the asymptotic normality of the test statistic
Lemma B1Under assumption of (A1) we have as 119899 997888rarr infin(i) lim119899997888rarrinfin(1119899)sum119899119905=1 119864(1198772119905 (120573) | F119877119905minus1) = 1205902 119886119904 (ii)(1radic119899)sum119899119905=1 119877119905(120573) 119889997888rarr 119873(0 1205902) where 1205902 = (14)1198641205972 log1198911205761205971205722 + (120597 log119891120576120597120572)2
Proof By the ergodic theorem the first conclusion holdsNow let F119905 = 1205901199090 1199091 119909119905 119905 ge 1 and F0 is a sigma
field Note that 119876119899(120573) = sum119899119905=1 119877119905(120573) It is easy to see that[119877119899(120573)|F119899minus1] = 0 Then we have
119864 [119876119899 (120573) | F119899minus1] = 119864 [119876119899minus1 (120573) + 119877119899 (120573) | F119899minus1]= 119876119899minus1 (120573) (B10)
Thus 119876119899F119899 119899 ge 0 is a martingale We have shownsup119905119864|119877119905(120573)|2+120578 lt infin which means 119864[1198772119905 (120573)] is uniformlyintegrable ByTheorem 11 of Billingsley [22] we have that as119899 997888rarr infin1119899 119899sum119905=11198772119905 (120573) 119886119904997888997888rarr 119864[1198772119899 (120573) | F119899minus1] = 1205902 (B11)
Hence using the following version of Martingale Cen-tral Limit Theorem from Hall and Heyde [23] we have(1radic119899)sum119899119905=1 119877119905(120573) 119889997888rarr 119873(0 1205902) The proof of Lemma B1 isthus completed ◻
Similarly we can verify that 119878119899 = sum119899119905=1(120597log119891120576120597120573) is amartingale By ergodic and stationary properties we obtainthat as 119899 997888rarr infin
1119899 119899sum119905=11205972 log119891120576120597120573120597120573119879 119886119904997888997888rarr 119864[120597 log119891120576120597120573 (120597 log119891120576120597120573 )119879] = Σ (B12)
Therefore (1radic119899)119878119899 119889997888rarr 119873(0 Σ)
10 Mathematical Problems in Engineering
In the same way for any vector 119888 = (1198881 1198882 1198883)119879 isin 1198773 (0 0 0) we have1radic119899119888119879(
119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) = 1radic119899sdot 119899sum119905=1
(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 ) 119889997888rarr 119873(0 119864(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 )2)
(B13)
By the Cramer-Wold device we obtain
1radic119899 (119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) 119889997888rarr 119873[(00) (1205902 119882119879119882 Σ )] (B14)
where119882 = (1198821111988212)119879 Then we can make the conclusionthat 1radic119899 119899sum119905=1119877119905 () 119889997888rarr 119873(0 1205962) (B15)
where 1205962 = 1205902 minus 119882119879Σminus1119882 The proof of the theorem iscompleted
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that they have no conflicts of interest
Acknowledgments
This work is supported by National Natural Science Foun-dation of China (Nos 11871028 11731015 11571051 and11501241) Natural Science Foundation of Jilin Province (Nos20180101216JC 20170101057JC and 20150520053JH) Pro-gram for Changbaishan Scholars of Jilin Province (2015010)and Science and Technology Program of Jilin EducationalDepartment during the ldquo13th Five-Yearrdquo Plan Period (No2016316)
References
[1] A Hoque ldquoFinite sample analysis of the first order autoregres-sive modelrdquoCalcutta Statistical Association Bulletin vol 34 no1-2 pp 51ndash63 1985
[2] T Ogawa H Sonoda S Ishiwa and Y Shigeta ldquoAn applicationof autoregressive model to pattern discrimination of brainelectrical activity mappingrdquo Brain Topography vol 6 no 1 pp3ndash11 1992
[3] A Subasi A Alkan E Koklukaya and M K Kiymik ldquoWaveletneural network classification of EEG signals by using ARmodelwith MLE preprocessingrdquo Journal of the International NeuralNetwork Society vol 18 pp 985ndash997 2005
[4] M Maleki and A R Nematollahi ldquoAutoregressive models withmixture of scale mixtures of gaussian innovationsrdquo IranianJournal of Science amp Technology Transactions A Science vol 41no 4 pp 1099ndash1107 2017
[5] J Conlisk ldquoStability in a random coefficient modelrdquo Interna-tional Economic Review vol 15 no 2 pp 529ndash533 1974
[6] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models Irdquo Journal of Time SeriesAnalysis vol 1 no 1 pp 37ndash46 1980
[7] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models IIrdquo Journal of Time SeriesAnalysis vol 2 no 3 pp 185ndash203 1981
[8] D F Nicholls and B G Quinn Random Coefficient Autore-gressive Models An Introduction Lecture Notes in StatisticsSpringer-Verlag New York NY USA 1982
[9] A Brandt ldquoThe stochastic equation Y119899+1 = A119899Y119899 + B119899 withstationary coefficientsrdquo Advances in Applied Probability vol 18no 1 pp 211ndash220 1986
[10] D Wang and S K Ghosh ldquoBayesian estimation and unitroot tests for random coefficient autoregressive modelsrdquoModelAssisted Statistics and Applications vol 3 no 4 pp 281ndash2952008
[11] DWang S K Ghosh and S G Pantula ldquoMaximum likelihoodestimation and unit root test for first order random coefficientautoregressive modelsrdquo Journal of Statistical Theory and Prac-tice vol 4 no 2 pp 261ndash278 2010
[12] T V Ramanathan and M B Rajarshi ldquoRank tests for testingthe randomness of autoregressive coefficientsrdquo Statistics ampProbability Letters vol 21 no 2 pp 115ndash120 1994
[13] S Lee J Ha O Na and S Na ldquoThe cusum test for parameterchange in time seriesmodelsrdquo Scandinavian Journal of Statisticsvol 30 no 4 pp 781ndash796 2003
[14] M Moreno and J Romo ldquoRobust unit root tests with autore-gressive errorsrdquo Communications in StatisticsmdashTheory andMethods vol 45 no 20 pp 5997ndash6021 2016
[15] V K Rohatgi A K Md Ehsanes Sleh R Ahluwalia and P JildquoNull distribution of locally most powerful tests for the twosample problemwhen the combined sample is type II censoredrdquoCommunications in Statistics - Theory and Methods vol 19 pp2337ndash2355 1992
[16] M S Chikkagoudar and B S Biradar ldquoLocally most powerfulrank tests for comparison of two failure rates based on multipletype-ii censored datardquoCommunications in Statistics Theory andMethods vol 41 no 23 pp 4315ndash4331 2012
[17] A Manik N Balakrishna and T V Ramanathan ldquoTesting theconstancy of the thinning parameter in a random coefficientinteger autoregressive modelrdquo Statistical Papers pp 1ndash25 2017
[18] P J Huber Robust Statistics JohnWiley amp Sons New York NYUSA 1981
[19] T P Hettmansperger Statistical Inference Based on Ranks JohnWiley amp Sons Inc New York NY USA 1984
[20] Z-W Zhao D-H Wang and C-X Peng ldquoCoefficient con-stancy test in generalized random coefficient autoregressivemodelrdquoApplied Mathematics and Computation vol 219 no 20pp 10283ndash10292 2013
[21] H C Thode Testing for Normality CRC Press New York NYUSA 2002
Mathematical Problems in Engineering 11
[22] P Billingsley ldquoThe lindeberg-Lvy theorem for martingalesrdquoProceedings of the American Mathematical Society vol 12 no1 pp 788ndash792 1961
[23] P Hall and C C Heyde Martingale Limit Theory and ItsApplication Academic Press New York NY USA 1980
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Mathematical Problems in Engineering 5
Table 4 Empirical powers of LMP and EL tests at nominal level 005 for model A1
119873 (120590119873 119886 119887)(05 05 05) (05 05 1) (05 1 05) (05 1 1) (1 05 05) (1 05 1) (1 1 05) (1 1 1)100 LMP 0165 0102 0171 0110 0171 0107 0174 0090
EL 0055 0024 0040 0023 0050 0022 0041 0025200 LMP 0318 0143 0294 0147 0320 0140 0311 0160
EL 0191 0070 0156 0073 0202 0049 0155 0067300 LMP 0513 0209 0494 0253 0511 0175 0449 0241
EL 0405 0144 0332 0173 0397 0124 0294 0168400 LMP 0674 0232 0561 0330 0663 0264 0590 0338
EL 0578 0176 0456 0272 0559 0189 0450 0260500 LMP 0791 0377 0720 0427 0770 0374 0725 0438
EL 0722 0298 0629 0366 0705 0303 0625 0374600 LMP 0842 0415 0792 0521 0832 0424 0790 0508
EL 0798 0352 0717 0460 0793 0365 0712 0458700 LMP 0912 0483 0862 0598 0919 0478 0855 0598
EL 0882 0429 0809 0541 0885 0423 0794 0536800 LMP 0964 0560 0915 0645 0947 0568 0918 0643
EL 0938 0512 0877 0604 0928 0516 0872 0599900 LMP 0967 0633 0936 0727 0971 0623 0955 0703
EL 0952 0586 0910 0696 0959 0586 0933 06871000 LMP 0988 0712 0963 0799 0978 0702 0962 0776
EL 0981 0678 0942 0774 0969 0676 0938 07495000 LMP 1000 1000 1000 1000 1000 1000 1000 1000
EL 1000 1000 1000 1000 1000 1000 1000 1000
Table 5 Empirical powers of LMP and EL tests at nominal level 005 for model A2119873 (119898 119886 119887)(6 05 05) (6 05 1) (6 1 05) (6 1 1) (10 05 05) (10 05 1) (10 1 05) (10 1 1)100 LMP 0289 0110 0220 0149 0241 0096 0183 0131
EL 0045 0016 0027 0019 0047 0018 0027 0029200 LMP 0596 0252 0519 0310 0525 0199 0452 0272
EL 0161 0052 0118 0055 0186 0057 0146 0063300 LMP 0790 0400 0722 0475 0762 0351 0686 0429
EL 0336 0111 0276 0124 0373 0124 0293 0152400 LMP 0899 0538 0861 0644 0862 0462 0815 0564
EL 0486 0147 0399 0196 0520 0173 0440 0223500 LMP 0964 0624 0933 0757 0957 0596 0906 0699
EL 0622 0218 0529 0285 0695 0262 0590 0345600 LMP 0979 0701 0951 0819 0965 0657 0947 0754
EL 0708 0237 0606 0351 0767 0301 0650 0395700 LMP 0995 0791 0974 0862 0983 0727 0978 0824
EL 0814 0351 0669 0404 0839 0389 0773 0483800 LMP 0996 0837 0990 0927 0994 0809 0984 0882
EL 0845 0404 0767 0512 0895 0488 0839 0585900 LMP 0999 0874 0995 0934 0996 0831 0993 0924
EL 0879 0430 0801 0548 0929 0514 0877 06221000 LMP 1000 0936 1000 0971 1000 0889 0996 0951
EL 0913 0515 0887 0626 0946 0601 0930 07065000 LMP 1000 1000 1000 1000 1000 1000 1000 1000
EL 0992 0967 0994 0985 1000 1000 1000 0999
6 Mathematical Problems in Engineering
Table 6 Empirical powers of LMP and EL tests at nominal level 005 for model A3 with (1205901 1205902) = (1 3)119873 (119886 119887)(05 05) (05 1) (05 15) (1 05) (1 1) (1 15) (15 05) (15 1) (15 15)120585 = 08100 LMP 0141 0054 0034 0120 0073 0038 0084 0082 0046
EL 0039 0009 0006 0026 0012 0008 0018 0012 0009300 LMP 0775 0388 0183 0628 0493 0299 0475 0448 0331
EL 0325 0102 0030 0231 0097 0052 0143 0114 0061500 LMP 0959 0675 0381 0880 0791 0594 0754 0729 0583
EL 0656 0236 0070 0500 0268 0143 0347 0222 01311000 LMP 1000 0967 0753 0996 0992 0934 0982 0982 0939
EL 0945 0599 0255 0891 0673 0399 0720 0609 04025000 LMP 1000 1000 1000 1000 1000 1000 1000 1000 1000
EL 1000 1000 0981 1000 1000 0997 0999 1000 0997120585 = 09100 LMP 0118 0043 0035 0071 0041 0025 0049 0052 0037
EL 0042 0013 0015 0032 0021 0014 0020 0013 0011300 LMP 0673 0278 0101 0545 0372 0190 0388 0313 0220
EL 0301 0077 0021 0199 0101 0043 0128 0093 0055500 LMP 0914 0558 0277 0868 0681 0423 0700 0622 0467
EL 0571 0183 0054 0486 0239 0128 0353 0212 01131000 LMP 1000 0929 0627 0995 0965 0849 0970 0954 0865
EL 0927 0503 0204 0852 0637 0375 0702 0548 03685000 LMP 1000 1000 1000 1000 1000 1000 1000 1000 1000
EL 1000 1000 0963 1000 1000 0996 1000 1000 0994
32 Empirical Power In order to investigate the empiricalpowers we consider the alternatives under three versions ofmodel (2)
(A1) 120572119905 sim 119861119890119905119886(119886 119887) 120576119905 sim 119873(0 1205902119873)(A2) 120572119905 sim 119861119890119905119886(119886 119887) 120576119905 sim 119905(119898)(A3)120572119905 sim 119861119890119905119886(119886 119887) 120576119905 sim F120576(119909)Thedistribution function
of 120585-contamination is
F120576 (119909) = 120585Φ( 1199091205901) + (1 minus 120585)Φ( 1199091205902) (10)
where 120585 is a fixed constant satisfying 0 lt 120585 lt 1 andΦ(119909) is thedistribution function of standard normal random variable120590119894 gt 0 119894 = 1 2
To calculate the empirical powers of the two tests inthe first we generate samples from the model A1 in whichthe parameters 119886 = 05 1 119887 = 05 1 and 120590119873 = 05 1In the second setting we simulate samples from model A2with the degrees of freedom 119898 = 6 10 the parameters119886 = 05 1 and 119887 = 05 1 For model A1 and A2 we takethe sample size 119873 = 100 200 300 400 500 600 700800 900 1000 and 5000 In the last case we generate datafrom the model A3 in which the (120585 1205901 1205902 ) = (08 1 3)(120585 1205901 1205902 ) = (09 1 3) and the parameters 119886 = 05 1 15119887 = 05 1 15 For model A3 we employ the sample size119873 = 100 300 500 1000 and 5000 The empirical power ofthe above three models are summed up in Tables 4ndash6 Inthree cases the power of two test statistics increase with thesample size while our test produces relatively better powers
than the EL test Furthermore both tests powers are closeto 1 at the sample size 119873 = 5000 Overall from thesesimulations we conclude that the accuracy of using LMP is67 288 261 higher than that of EL test under normalstudentrsquos 119905 and symmetric contamination errors respectivelyAs anticipated our finding shows that the LMP test is afunctional tool to detect a parameter change for RCAR(1)model Therefore we recommend the LMP for practicaluse because its computation is not difficult and the overallperformance is better than that EL under normal studentrsquos tand symmetric contamination errors
4 A Real Life Data Analysis
In this section we illustrate how the LMP method can beapplied to a practical application This data consist of 78monthly number of annulus growth rate of the import billin Australia starting in December 2010 and ending in May2017 The data are available online at the CEInet StatisticsDatabase site httpdbceicnpageDefaultaspx The meanand variance of the data are found to be 02069 and 153347The data are denoted as 1199101 1199102 11991078 Figure 1 is the samplepath plot for the real data 119910119905 119905 = 1 2 78
For such a process the mean function of 119910119905 is constantand we may assume that the process mean is subtracted outto produce a process 119883119905 = 119910119905 minus 119864(119910119905) with zero mean Theplot of the sample path autocorrelation function (ACF) andthe partial autocorrelation function (PACF) for series 119883119905 aregiven in Figures 1 and 2 respectively From Figure 1 we can
Mathematical Problems in Engineering 7
0 20 40 60 80
minus10
minus50
510
15
Time
Dat
a Val
ue
Time
Dat
a Val
ue
0 20 40 60 80
minus10
minus50
510
15
９Ｎ ８Ｎ
Figure 1 The left one is the sample path of original series 119910119905 and the right one is the sample plot of centering series119883119905 for the real data
0 5 10 15
minus05
00
05
10
Lag
ACF
5 10 15
minus04
minus02
00
02
Lag
PACF
８Ｎ８Ｎ
Figure 2 The sample autocorrelation function (ACF) and partial autocorrelation function (PACF) plots of centering series 119883119905 for the realdata
see119883119905may come from a stationary autoregressive time seriesprocess From Figure 2 we conclude that 119883119905 is from first-order autoregressive process Moreover we test the normalityof the residual data 119883119905 according to the Normal Q-Q plotintroduced by Henry CThode [21] From Figure 3 we noticethat the scatter points on theNormal Q-Q plot are close to thereference line so the residual data119883119905 can basically be seen asfollowing the normal distribution Hence we derive that themodel N1 would be more suitable to model these data
From the time series plot given in Figure 1 the constancyof the parameter may be suspected and therefore we areinterested in testing the hypothesis of constancy of thecoefficient parameter We carry out the test for 1198670 1205902120572 = 0against 1198671 1205902120572 gt 0 The value of the test statistic in (8)turned out to be 21151 with 119901 value 00344 which indicatesthe rejection of the null hypothesis at 5 level of significanceThus in this case the RCAR(1) model would be much moreappropriate as opposed to the AR(1) model
5 Conclusions
In this article we propose a locally most powerful-typetest for testing the constancy of the random coefficientparameter in autoregressive model and derive their limitingnull distributions under regularity conditions It is clear fromthe applications that the coefficient need not remain constantthroughout the time Therefore it is essential to have such a
Normal QminusQ Plot
minus10
minus50
510
15Sa
mpl
e Qua
ntile
s
minus1 0 1 2minus2Theoretical Quantiles
Figure 3 The normal Q-Q plot of centering series 119883119905 for the realdata
test conducted whenever the random variation is suspectedThrough our simulation study and a real-data analysis wedemonstrate that our test succeeds and it performs better thanthe competitor Finally we anticipate that our locally mostpowerful-type test can be extended to other types of timeseries model
8 Mathematical Problems in Engineering
Appendix
A Process of Establish the Test Statistic
In this Appendix A we present the detailed steps to obtainthe test statistics
Proof Therefore we can get the likelihood function
1198711198671 (1199091 119909119905) = 1198911198830 (1199090) 119899prod119905=1
119891119883119905|119883119905minus1 (119909119905 | 119909119905minus1)= 1198911198830 (1199090) 119899prod
119905=1
int119891120576 (119909119905 minus 120572119905119909119905minus1) 119889119865120572 (A1)
Now the Taylor series expansion of 1198711198671 around 120572 which isthe mean of 120572119905 gives that
1198711198671 (1199091 119909119905) = 1198911198830 (1199090) 119899prod119905=1
int119891120576 (119909119905minus 120572119905119909119905minus1) 119889119865120572
= 1198911198830 (1199090) 119899prod119905=1
int[119891120576 (119909119905 minus 120572119909119905minus1)
+ (120572119905 minus 120572) 1198911015840120576 (119909119905 minus 120572119909119905minus1)+ (120572119905 minus 120572)22 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)+ (120572119905 minus 120572)33 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) + sdot sdot sdot] 119889119865120572
= 1198911198830 (1199090) 119899prod119905=1
[119891120576 (119909119905 minus 120572119909119905minus1) + 12120590212057211989110158401015840120576 (119909119905minus 120572119909119905minus1) + 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) int (120572119905 minus 120572)33 119889119865120572+ sdot sdot sdot]
(A2)
Obviously we can conclude that 119875(120572119905 minus 120572 = 0) = 1 if 1205902120572 = 0That is to say the distribution function of random variable 120572119905is degenerated Furthermore we derive that
int (120572119905 minus 120572)119896119896 119889119865120572 = 0 119896 = 1 2 3 sdot sdot sdot (A3)
By taking the derivative of log 1198711198671 with respect to120573 at1205902120572 = 0we have the following equation
120597 log 11987111986711205971205902120572 1003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 120597sum119899119905=1 log [119891120576 (119909119905 minus 120572119909119905minus1) + (12) 120590212057211989110158401015840120576 (119909119905 minus 120572119909119905minus1) + 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) int ((120572119905 minus 120572)3 3) 119889119865120572 + sdot sdot sdot ]1205971205902120572 100381610038161003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 120597sum119899119905=1 log [119891120576 (119909119905 minus 120572119909119905minus1) + (12) 120590212057211989110158401015840120576 (119909119905 minus 120572119909119905minus1)]1205971205902120572 100381610038161003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 12 119899sum119905=111989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) (A4)
Thus the LMP test statistic has the form119876119899 (120573) = 119899sum119905=1
12 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) (A5)
This completes the proof
B Proof of Theorem 1
Proof of Theorem 1 In the following we can write down thelog-likelihood function under the null hypothesis
log 119871 = 119899sum119905=1
log119891120576 (119909119905 minus 120572119909119905minus1) (B1)
According to the maximum likelihood principle the maxi-mum likelihood estimate can be obtained by maximizing
the log-likelihood function log 119871 with respect to 120573 Applyingthe Taylor series expansion to 120597 log119891120576120597120573 gives
0 = 120597log119891120576120597120573 = 120597log119891120576120597120573 + ( minus 120573) 1205972log119891120576 (120573lowast)1205971205731205971205731015840 (B2)
where 120573lowast is on the line segment between 120573 and so 120573lowast119901997888rarr
120573(119899 997888rarr infin) Under the null hypothesis we can write
minus 120573 = minus 119899sum119905=1
1205972 log1198911205761205971205731205971205731015840 + 119900119901 (119899)minus1 119899sum119905=1120597 log119891120576120597120573 (B3)
Mathematical Problems in Engineering 9
Using the similar method as in the above expanding(1radic119899)sum119899119905=1 119877119905() around 120573 gives that1radic119899 119899sum119905=1119877119905 () = 1radic119899 119899sum119905=1119877119905 (120573)+ ( minus 120573)119879 1radic119899 119899sum119905=1120597119877119905 (120573)120597120573 + 119900119901 (1) (B4)
with 120597119877119905(120573)120597120573 = (120597119877119905120597120572 1205971198771199051205971205902)119879 The remainder termis 119900119901(1) since 119901997888rarr 120573Under (C3) and (C4) we conclude that1205972119877119905(120573)1205971205731205971205731015840 = 119874119901(119899) Therefore
( minus 120573)119879 ( 1radic119899) 119899sum119905=1120597119877119905 (120573)120597120573= minus1119899 119899sum119905=1[120597119877119905 (120573)120597120573 ]119879times 1119899 119899sum119905=11205972 log1198911205761205971205731205971205731015840 + 119900119901 (1)minus1times 1radic119899 119899sum119905=1120597 log119891120576120597120573
(B5)
Now the first term of (B5) converges to1119899 119899sum119905=1120597119877119905 (120573)120597120573 = 1119899 119899sum119905=1 120597120597120573 120597 log1198911205761205971205902120572 1205902120572=0
= (1119899 119899sum119905=11205972 log11989112057612059712057212059712059021205721119899 119899sum119905=11205972 log11989112057612059712059021205761205971205902120572 )1205902120572=0
997888rarr 119882as 119899 997888rarr infin
(B6)
with119882 = (1198821111988212)119879This is because
11988211 = 119864(120597 log1198911205761205971205721205971205902120572 ) = 119864(minus120597 log119891120576120597120572 120597 log1198911205761205971205902120572 )= minus12119864[120597 log119891120576120597120572 1205972 log1198911205761205971205722 + (120597 log119891120576120597120572 )2]
at 1205902120572 = 011988212 = 119864(120597 log1198911205761205971205902120576 1205971205902120572) = 119864(minus120597 log1198911205761205971205902120576 120597 log1198911205761205971205902120572 )= minus12119864[120597 log1198911205761205971205902120576 1205972 log1198911205761205971205722 + (120597 log119891120576120597120572 )2]
at 1205902120572 = 0
(B7)
Next we can obtain the elements of Fisher informationmatrix
Σ = 1119899 119899sum119905=1119864[120597 log119891120576120597120573 (120597 log119891120576120597120573 )119879 | F119910119905minus1]= [[[[[
minus119864[1205972 log1198911205761205971205722 ] minus119864[1205972 log1198911205761205971205721205971205902120576 ]minus119864[1205972 log1198911205761205971205721205971205902120576 ] minus119864[1205972 log11989112057612059712059021205761205971205902120576 ]]]]]]
(B8)
Thus (B5) is asymptotically equivalent to
minus119882Σminus1 1radic119899 119899sum119905=1120597 log119891120576120597120573 (B9)
We present the following lemma recommended by Billingsley[22] to establish the asymptotic normality of the test statistic
Lemma B1Under assumption of (A1) we have as 119899 997888rarr infin(i) lim119899997888rarrinfin(1119899)sum119899119905=1 119864(1198772119905 (120573) | F119877119905minus1) = 1205902 119886119904 (ii)(1radic119899)sum119899119905=1 119877119905(120573) 119889997888rarr 119873(0 1205902) where 1205902 = (14)1198641205972 log1198911205761205971205722 + (120597 log119891120576120597120572)2
Proof By the ergodic theorem the first conclusion holdsNow let F119905 = 1205901199090 1199091 119909119905 119905 ge 1 and F0 is a sigma
field Note that 119876119899(120573) = sum119899119905=1 119877119905(120573) It is easy to see that[119877119899(120573)|F119899minus1] = 0 Then we have
119864 [119876119899 (120573) | F119899minus1] = 119864 [119876119899minus1 (120573) + 119877119899 (120573) | F119899minus1]= 119876119899minus1 (120573) (B10)
Thus 119876119899F119899 119899 ge 0 is a martingale We have shownsup119905119864|119877119905(120573)|2+120578 lt infin which means 119864[1198772119905 (120573)] is uniformlyintegrable ByTheorem 11 of Billingsley [22] we have that as119899 997888rarr infin1119899 119899sum119905=11198772119905 (120573) 119886119904997888997888rarr 119864[1198772119899 (120573) | F119899minus1] = 1205902 (B11)
Hence using the following version of Martingale Cen-tral Limit Theorem from Hall and Heyde [23] we have(1radic119899)sum119899119905=1 119877119905(120573) 119889997888rarr 119873(0 1205902) The proof of Lemma B1 isthus completed ◻
Similarly we can verify that 119878119899 = sum119899119905=1(120597log119891120576120597120573) is amartingale By ergodic and stationary properties we obtainthat as 119899 997888rarr infin
1119899 119899sum119905=11205972 log119891120576120597120573120597120573119879 119886119904997888997888rarr 119864[120597 log119891120576120597120573 (120597 log119891120576120597120573 )119879] = Σ (B12)
Therefore (1radic119899)119878119899 119889997888rarr 119873(0 Σ)
10 Mathematical Problems in Engineering
In the same way for any vector 119888 = (1198881 1198882 1198883)119879 isin 1198773 (0 0 0) we have1radic119899119888119879(
119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) = 1radic119899sdot 119899sum119905=1
(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 ) 119889997888rarr 119873(0 119864(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 )2)
(B13)
By the Cramer-Wold device we obtain
1radic119899 (119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) 119889997888rarr 119873[(00) (1205902 119882119879119882 Σ )] (B14)
where119882 = (1198821111988212)119879 Then we can make the conclusionthat 1radic119899 119899sum119905=1119877119905 () 119889997888rarr 119873(0 1205962) (B15)
where 1205962 = 1205902 minus 119882119879Σminus1119882 The proof of the theorem iscompleted
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that they have no conflicts of interest
Acknowledgments
This work is supported by National Natural Science Foun-dation of China (Nos 11871028 11731015 11571051 and11501241) Natural Science Foundation of Jilin Province (Nos20180101216JC 20170101057JC and 20150520053JH) Pro-gram for Changbaishan Scholars of Jilin Province (2015010)and Science and Technology Program of Jilin EducationalDepartment during the ldquo13th Five-Yearrdquo Plan Period (No2016316)
References
[1] A Hoque ldquoFinite sample analysis of the first order autoregres-sive modelrdquoCalcutta Statistical Association Bulletin vol 34 no1-2 pp 51ndash63 1985
[2] T Ogawa H Sonoda S Ishiwa and Y Shigeta ldquoAn applicationof autoregressive model to pattern discrimination of brainelectrical activity mappingrdquo Brain Topography vol 6 no 1 pp3ndash11 1992
[3] A Subasi A Alkan E Koklukaya and M K Kiymik ldquoWaveletneural network classification of EEG signals by using ARmodelwith MLE preprocessingrdquo Journal of the International NeuralNetwork Society vol 18 pp 985ndash997 2005
[4] M Maleki and A R Nematollahi ldquoAutoregressive models withmixture of scale mixtures of gaussian innovationsrdquo IranianJournal of Science amp Technology Transactions A Science vol 41no 4 pp 1099ndash1107 2017
[5] J Conlisk ldquoStability in a random coefficient modelrdquo Interna-tional Economic Review vol 15 no 2 pp 529ndash533 1974
[6] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models Irdquo Journal of Time SeriesAnalysis vol 1 no 1 pp 37ndash46 1980
[7] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models IIrdquo Journal of Time SeriesAnalysis vol 2 no 3 pp 185ndash203 1981
[8] D F Nicholls and B G Quinn Random Coefficient Autore-gressive Models An Introduction Lecture Notes in StatisticsSpringer-Verlag New York NY USA 1982
[9] A Brandt ldquoThe stochastic equation Y119899+1 = A119899Y119899 + B119899 withstationary coefficientsrdquo Advances in Applied Probability vol 18no 1 pp 211ndash220 1986
[10] D Wang and S K Ghosh ldquoBayesian estimation and unitroot tests for random coefficient autoregressive modelsrdquoModelAssisted Statistics and Applications vol 3 no 4 pp 281ndash2952008
[11] DWang S K Ghosh and S G Pantula ldquoMaximum likelihoodestimation and unit root test for first order random coefficientautoregressive modelsrdquo Journal of Statistical Theory and Prac-tice vol 4 no 2 pp 261ndash278 2010
[12] T V Ramanathan and M B Rajarshi ldquoRank tests for testingthe randomness of autoregressive coefficientsrdquo Statistics ampProbability Letters vol 21 no 2 pp 115ndash120 1994
[13] S Lee J Ha O Na and S Na ldquoThe cusum test for parameterchange in time seriesmodelsrdquo Scandinavian Journal of Statisticsvol 30 no 4 pp 781ndash796 2003
[14] M Moreno and J Romo ldquoRobust unit root tests with autore-gressive errorsrdquo Communications in StatisticsmdashTheory andMethods vol 45 no 20 pp 5997ndash6021 2016
[15] V K Rohatgi A K Md Ehsanes Sleh R Ahluwalia and P JildquoNull distribution of locally most powerful tests for the twosample problemwhen the combined sample is type II censoredrdquoCommunications in Statistics - Theory and Methods vol 19 pp2337ndash2355 1992
[16] M S Chikkagoudar and B S Biradar ldquoLocally most powerfulrank tests for comparison of two failure rates based on multipletype-ii censored datardquoCommunications in Statistics Theory andMethods vol 41 no 23 pp 4315ndash4331 2012
[17] A Manik N Balakrishna and T V Ramanathan ldquoTesting theconstancy of the thinning parameter in a random coefficientinteger autoregressive modelrdquo Statistical Papers pp 1ndash25 2017
[18] P J Huber Robust Statistics JohnWiley amp Sons New York NYUSA 1981
[19] T P Hettmansperger Statistical Inference Based on Ranks JohnWiley amp Sons Inc New York NY USA 1984
[20] Z-W Zhao D-H Wang and C-X Peng ldquoCoefficient con-stancy test in generalized random coefficient autoregressivemodelrdquoApplied Mathematics and Computation vol 219 no 20pp 10283ndash10292 2013
[21] H C Thode Testing for Normality CRC Press New York NYUSA 2002
Mathematical Problems in Engineering 11
[22] P Billingsley ldquoThe lindeberg-Lvy theorem for martingalesrdquoProceedings of the American Mathematical Society vol 12 no1 pp 788ndash792 1961
[23] P Hall and C C Heyde Martingale Limit Theory and ItsApplication Academic Press New York NY USA 1980
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6 Mathematical Problems in Engineering
Table 6 Empirical powers of LMP and EL tests at nominal level 005 for model A3 with (1205901 1205902) = (1 3)119873 (119886 119887)(05 05) (05 1) (05 15) (1 05) (1 1) (1 15) (15 05) (15 1) (15 15)120585 = 08100 LMP 0141 0054 0034 0120 0073 0038 0084 0082 0046
EL 0039 0009 0006 0026 0012 0008 0018 0012 0009300 LMP 0775 0388 0183 0628 0493 0299 0475 0448 0331
EL 0325 0102 0030 0231 0097 0052 0143 0114 0061500 LMP 0959 0675 0381 0880 0791 0594 0754 0729 0583
EL 0656 0236 0070 0500 0268 0143 0347 0222 01311000 LMP 1000 0967 0753 0996 0992 0934 0982 0982 0939
EL 0945 0599 0255 0891 0673 0399 0720 0609 04025000 LMP 1000 1000 1000 1000 1000 1000 1000 1000 1000
EL 1000 1000 0981 1000 1000 0997 0999 1000 0997120585 = 09100 LMP 0118 0043 0035 0071 0041 0025 0049 0052 0037
EL 0042 0013 0015 0032 0021 0014 0020 0013 0011300 LMP 0673 0278 0101 0545 0372 0190 0388 0313 0220
EL 0301 0077 0021 0199 0101 0043 0128 0093 0055500 LMP 0914 0558 0277 0868 0681 0423 0700 0622 0467
EL 0571 0183 0054 0486 0239 0128 0353 0212 01131000 LMP 1000 0929 0627 0995 0965 0849 0970 0954 0865
EL 0927 0503 0204 0852 0637 0375 0702 0548 03685000 LMP 1000 1000 1000 1000 1000 1000 1000 1000 1000
EL 1000 1000 0963 1000 1000 0996 1000 1000 0994
32 Empirical Power In order to investigate the empiricalpowers we consider the alternatives under three versions ofmodel (2)
(A1) 120572119905 sim 119861119890119905119886(119886 119887) 120576119905 sim 119873(0 1205902119873)(A2) 120572119905 sim 119861119890119905119886(119886 119887) 120576119905 sim 119905(119898)(A3)120572119905 sim 119861119890119905119886(119886 119887) 120576119905 sim F120576(119909)Thedistribution function
of 120585-contamination is
F120576 (119909) = 120585Φ( 1199091205901) + (1 minus 120585)Φ( 1199091205902) (10)
where 120585 is a fixed constant satisfying 0 lt 120585 lt 1 andΦ(119909) is thedistribution function of standard normal random variable120590119894 gt 0 119894 = 1 2
To calculate the empirical powers of the two tests inthe first we generate samples from the model A1 in whichthe parameters 119886 = 05 1 119887 = 05 1 and 120590119873 = 05 1In the second setting we simulate samples from model A2with the degrees of freedom 119898 = 6 10 the parameters119886 = 05 1 and 119887 = 05 1 For model A1 and A2 we takethe sample size 119873 = 100 200 300 400 500 600 700800 900 1000 and 5000 In the last case we generate datafrom the model A3 in which the (120585 1205901 1205902 ) = (08 1 3)(120585 1205901 1205902 ) = (09 1 3) and the parameters 119886 = 05 1 15119887 = 05 1 15 For model A3 we employ the sample size119873 = 100 300 500 1000 and 5000 The empirical power ofthe above three models are summed up in Tables 4ndash6 Inthree cases the power of two test statistics increase with thesample size while our test produces relatively better powers
than the EL test Furthermore both tests powers are closeto 1 at the sample size 119873 = 5000 Overall from thesesimulations we conclude that the accuracy of using LMP is67 288 261 higher than that of EL test under normalstudentrsquos 119905 and symmetric contamination errors respectivelyAs anticipated our finding shows that the LMP test is afunctional tool to detect a parameter change for RCAR(1)model Therefore we recommend the LMP for practicaluse because its computation is not difficult and the overallperformance is better than that EL under normal studentrsquos tand symmetric contamination errors
4 A Real Life Data Analysis
In this section we illustrate how the LMP method can beapplied to a practical application This data consist of 78monthly number of annulus growth rate of the import billin Australia starting in December 2010 and ending in May2017 The data are available online at the CEInet StatisticsDatabase site httpdbceicnpageDefaultaspx The meanand variance of the data are found to be 02069 and 153347The data are denoted as 1199101 1199102 11991078 Figure 1 is the samplepath plot for the real data 119910119905 119905 = 1 2 78
For such a process the mean function of 119910119905 is constantand we may assume that the process mean is subtracted outto produce a process 119883119905 = 119910119905 minus 119864(119910119905) with zero mean Theplot of the sample path autocorrelation function (ACF) andthe partial autocorrelation function (PACF) for series 119883119905 aregiven in Figures 1 and 2 respectively From Figure 1 we can
Mathematical Problems in Engineering 7
0 20 40 60 80
minus10
minus50
510
15
Time
Dat
a Val
ue
Time
Dat
a Val
ue
0 20 40 60 80
minus10
minus50
510
15
９Ｎ ８Ｎ
Figure 1 The left one is the sample path of original series 119910119905 and the right one is the sample plot of centering series119883119905 for the real data
0 5 10 15
minus05
00
05
10
Lag
ACF
5 10 15
minus04
minus02
00
02
Lag
PACF
８Ｎ８Ｎ
Figure 2 The sample autocorrelation function (ACF) and partial autocorrelation function (PACF) plots of centering series 119883119905 for the realdata
see119883119905may come from a stationary autoregressive time seriesprocess From Figure 2 we conclude that 119883119905 is from first-order autoregressive process Moreover we test the normalityof the residual data 119883119905 according to the Normal Q-Q plotintroduced by Henry CThode [21] From Figure 3 we noticethat the scatter points on theNormal Q-Q plot are close to thereference line so the residual data119883119905 can basically be seen asfollowing the normal distribution Hence we derive that themodel N1 would be more suitable to model these data
From the time series plot given in Figure 1 the constancyof the parameter may be suspected and therefore we areinterested in testing the hypothesis of constancy of thecoefficient parameter We carry out the test for 1198670 1205902120572 = 0against 1198671 1205902120572 gt 0 The value of the test statistic in (8)turned out to be 21151 with 119901 value 00344 which indicatesthe rejection of the null hypothesis at 5 level of significanceThus in this case the RCAR(1) model would be much moreappropriate as opposed to the AR(1) model
5 Conclusions
In this article we propose a locally most powerful-typetest for testing the constancy of the random coefficientparameter in autoregressive model and derive their limitingnull distributions under regularity conditions It is clear fromthe applications that the coefficient need not remain constantthroughout the time Therefore it is essential to have such a
Normal QminusQ Plot
minus10
minus50
510
15Sa
mpl
e Qua
ntile
s
minus1 0 1 2minus2Theoretical Quantiles
Figure 3 The normal Q-Q plot of centering series 119883119905 for the realdata
test conducted whenever the random variation is suspectedThrough our simulation study and a real-data analysis wedemonstrate that our test succeeds and it performs better thanthe competitor Finally we anticipate that our locally mostpowerful-type test can be extended to other types of timeseries model
8 Mathematical Problems in Engineering
Appendix
A Process of Establish the Test Statistic
In this Appendix A we present the detailed steps to obtainthe test statistics
Proof Therefore we can get the likelihood function
1198711198671 (1199091 119909119905) = 1198911198830 (1199090) 119899prod119905=1
119891119883119905|119883119905minus1 (119909119905 | 119909119905minus1)= 1198911198830 (1199090) 119899prod
119905=1
int119891120576 (119909119905 minus 120572119905119909119905minus1) 119889119865120572 (A1)
Now the Taylor series expansion of 1198711198671 around 120572 which isthe mean of 120572119905 gives that
1198711198671 (1199091 119909119905) = 1198911198830 (1199090) 119899prod119905=1
int119891120576 (119909119905minus 120572119905119909119905minus1) 119889119865120572
= 1198911198830 (1199090) 119899prod119905=1
int[119891120576 (119909119905 minus 120572119909119905minus1)
+ (120572119905 minus 120572) 1198911015840120576 (119909119905 minus 120572119909119905minus1)+ (120572119905 minus 120572)22 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)+ (120572119905 minus 120572)33 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) + sdot sdot sdot] 119889119865120572
= 1198911198830 (1199090) 119899prod119905=1
[119891120576 (119909119905 minus 120572119909119905minus1) + 12120590212057211989110158401015840120576 (119909119905minus 120572119909119905minus1) + 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) int (120572119905 minus 120572)33 119889119865120572+ sdot sdot sdot]
(A2)
Obviously we can conclude that 119875(120572119905 minus 120572 = 0) = 1 if 1205902120572 = 0That is to say the distribution function of random variable 120572119905is degenerated Furthermore we derive that
int (120572119905 minus 120572)119896119896 119889119865120572 = 0 119896 = 1 2 3 sdot sdot sdot (A3)
By taking the derivative of log 1198711198671 with respect to120573 at1205902120572 = 0we have the following equation
120597 log 11987111986711205971205902120572 1003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 120597sum119899119905=1 log [119891120576 (119909119905 minus 120572119909119905minus1) + (12) 120590212057211989110158401015840120576 (119909119905 minus 120572119909119905minus1) + 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) int ((120572119905 minus 120572)3 3) 119889119865120572 + sdot sdot sdot ]1205971205902120572 100381610038161003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 120597sum119899119905=1 log [119891120576 (119909119905 minus 120572119909119905minus1) + (12) 120590212057211989110158401015840120576 (119909119905 minus 120572119909119905minus1)]1205971205902120572 100381610038161003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 12 119899sum119905=111989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) (A4)
Thus the LMP test statistic has the form119876119899 (120573) = 119899sum119905=1
12 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) (A5)
This completes the proof
B Proof of Theorem 1
Proof of Theorem 1 In the following we can write down thelog-likelihood function under the null hypothesis
log 119871 = 119899sum119905=1
log119891120576 (119909119905 minus 120572119909119905minus1) (B1)
According to the maximum likelihood principle the maxi-mum likelihood estimate can be obtained by maximizing
the log-likelihood function log 119871 with respect to 120573 Applyingthe Taylor series expansion to 120597 log119891120576120597120573 gives
0 = 120597log119891120576120597120573 = 120597log119891120576120597120573 + ( minus 120573) 1205972log119891120576 (120573lowast)1205971205731205971205731015840 (B2)
where 120573lowast is on the line segment between 120573 and so 120573lowast119901997888rarr
120573(119899 997888rarr infin) Under the null hypothesis we can write
minus 120573 = minus 119899sum119905=1
1205972 log1198911205761205971205731205971205731015840 + 119900119901 (119899)minus1 119899sum119905=1120597 log119891120576120597120573 (B3)
Mathematical Problems in Engineering 9
Using the similar method as in the above expanding(1radic119899)sum119899119905=1 119877119905() around 120573 gives that1radic119899 119899sum119905=1119877119905 () = 1radic119899 119899sum119905=1119877119905 (120573)+ ( minus 120573)119879 1radic119899 119899sum119905=1120597119877119905 (120573)120597120573 + 119900119901 (1) (B4)
with 120597119877119905(120573)120597120573 = (120597119877119905120597120572 1205971198771199051205971205902)119879 The remainder termis 119900119901(1) since 119901997888rarr 120573Under (C3) and (C4) we conclude that1205972119877119905(120573)1205971205731205971205731015840 = 119874119901(119899) Therefore
( minus 120573)119879 ( 1radic119899) 119899sum119905=1120597119877119905 (120573)120597120573= minus1119899 119899sum119905=1[120597119877119905 (120573)120597120573 ]119879times 1119899 119899sum119905=11205972 log1198911205761205971205731205971205731015840 + 119900119901 (1)minus1times 1radic119899 119899sum119905=1120597 log119891120576120597120573
(B5)
Now the first term of (B5) converges to1119899 119899sum119905=1120597119877119905 (120573)120597120573 = 1119899 119899sum119905=1 120597120597120573 120597 log1198911205761205971205902120572 1205902120572=0
= (1119899 119899sum119905=11205972 log11989112057612059712057212059712059021205721119899 119899sum119905=11205972 log11989112057612059712059021205761205971205902120572 )1205902120572=0
997888rarr 119882as 119899 997888rarr infin
(B6)
with119882 = (1198821111988212)119879This is because
11988211 = 119864(120597 log1198911205761205971205721205971205902120572 ) = 119864(minus120597 log119891120576120597120572 120597 log1198911205761205971205902120572 )= minus12119864[120597 log119891120576120597120572 1205972 log1198911205761205971205722 + (120597 log119891120576120597120572 )2]
at 1205902120572 = 011988212 = 119864(120597 log1198911205761205971205902120576 1205971205902120572) = 119864(minus120597 log1198911205761205971205902120576 120597 log1198911205761205971205902120572 )= minus12119864[120597 log1198911205761205971205902120576 1205972 log1198911205761205971205722 + (120597 log119891120576120597120572 )2]
at 1205902120572 = 0
(B7)
Next we can obtain the elements of Fisher informationmatrix
Σ = 1119899 119899sum119905=1119864[120597 log119891120576120597120573 (120597 log119891120576120597120573 )119879 | F119910119905minus1]= [[[[[
minus119864[1205972 log1198911205761205971205722 ] minus119864[1205972 log1198911205761205971205721205971205902120576 ]minus119864[1205972 log1198911205761205971205721205971205902120576 ] minus119864[1205972 log11989112057612059712059021205761205971205902120576 ]]]]]]
(B8)
Thus (B5) is asymptotically equivalent to
minus119882Σminus1 1radic119899 119899sum119905=1120597 log119891120576120597120573 (B9)
We present the following lemma recommended by Billingsley[22] to establish the asymptotic normality of the test statistic
Lemma B1Under assumption of (A1) we have as 119899 997888rarr infin(i) lim119899997888rarrinfin(1119899)sum119899119905=1 119864(1198772119905 (120573) | F119877119905minus1) = 1205902 119886119904 (ii)(1radic119899)sum119899119905=1 119877119905(120573) 119889997888rarr 119873(0 1205902) where 1205902 = (14)1198641205972 log1198911205761205971205722 + (120597 log119891120576120597120572)2
Proof By the ergodic theorem the first conclusion holdsNow let F119905 = 1205901199090 1199091 119909119905 119905 ge 1 and F0 is a sigma
field Note that 119876119899(120573) = sum119899119905=1 119877119905(120573) It is easy to see that[119877119899(120573)|F119899minus1] = 0 Then we have
119864 [119876119899 (120573) | F119899minus1] = 119864 [119876119899minus1 (120573) + 119877119899 (120573) | F119899minus1]= 119876119899minus1 (120573) (B10)
Thus 119876119899F119899 119899 ge 0 is a martingale We have shownsup119905119864|119877119905(120573)|2+120578 lt infin which means 119864[1198772119905 (120573)] is uniformlyintegrable ByTheorem 11 of Billingsley [22] we have that as119899 997888rarr infin1119899 119899sum119905=11198772119905 (120573) 119886119904997888997888rarr 119864[1198772119899 (120573) | F119899minus1] = 1205902 (B11)
Hence using the following version of Martingale Cen-tral Limit Theorem from Hall and Heyde [23] we have(1radic119899)sum119899119905=1 119877119905(120573) 119889997888rarr 119873(0 1205902) The proof of Lemma B1 isthus completed ◻
Similarly we can verify that 119878119899 = sum119899119905=1(120597log119891120576120597120573) is amartingale By ergodic and stationary properties we obtainthat as 119899 997888rarr infin
1119899 119899sum119905=11205972 log119891120576120597120573120597120573119879 119886119904997888997888rarr 119864[120597 log119891120576120597120573 (120597 log119891120576120597120573 )119879] = Σ (B12)
Therefore (1radic119899)119878119899 119889997888rarr 119873(0 Σ)
10 Mathematical Problems in Engineering
In the same way for any vector 119888 = (1198881 1198882 1198883)119879 isin 1198773 (0 0 0) we have1radic119899119888119879(
119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) = 1radic119899sdot 119899sum119905=1
(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 ) 119889997888rarr 119873(0 119864(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 )2)
(B13)
By the Cramer-Wold device we obtain
1radic119899 (119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) 119889997888rarr 119873[(00) (1205902 119882119879119882 Σ )] (B14)
where119882 = (1198821111988212)119879 Then we can make the conclusionthat 1radic119899 119899sum119905=1119877119905 () 119889997888rarr 119873(0 1205962) (B15)
where 1205962 = 1205902 minus 119882119879Σminus1119882 The proof of the theorem iscompleted
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that they have no conflicts of interest
Acknowledgments
This work is supported by National Natural Science Foun-dation of China (Nos 11871028 11731015 11571051 and11501241) Natural Science Foundation of Jilin Province (Nos20180101216JC 20170101057JC and 20150520053JH) Pro-gram for Changbaishan Scholars of Jilin Province (2015010)and Science and Technology Program of Jilin EducationalDepartment during the ldquo13th Five-Yearrdquo Plan Period (No2016316)
References
[1] A Hoque ldquoFinite sample analysis of the first order autoregres-sive modelrdquoCalcutta Statistical Association Bulletin vol 34 no1-2 pp 51ndash63 1985
[2] T Ogawa H Sonoda S Ishiwa and Y Shigeta ldquoAn applicationof autoregressive model to pattern discrimination of brainelectrical activity mappingrdquo Brain Topography vol 6 no 1 pp3ndash11 1992
[3] A Subasi A Alkan E Koklukaya and M K Kiymik ldquoWaveletneural network classification of EEG signals by using ARmodelwith MLE preprocessingrdquo Journal of the International NeuralNetwork Society vol 18 pp 985ndash997 2005
[4] M Maleki and A R Nematollahi ldquoAutoregressive models withmixture of scale mixtures of gaussian innovationsrdquo IranianJournal of Science amp Technology Transactions A Science vol 41no 4 pp 1099ndash1107 2017
[5] J Conlisk ldquoStability in a random coefficient modelrdquo Interna-tional Economic Review vol 15 no 2 pp 529ndash533 1974
[6] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models Irdquo Journal of Time SeriesAnalysis vol 1 no 1 pp 37ndash46 1980
[7] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models IIrdquo Journal of Time SeriesAnalysis vol 2 no 3 pp 185ndash203 1981
[8] D F Nicholls and B G Quinn Random Coefficient Autore-gressive Models An Introduction Lecture Notes in StatisticsSpringer-Verlag New York NY USA 1982
[9] A Brandt ldquoThe stochastic equation Y119899+1 = A119899Y119899 + B119899 withstationary coefficientsrdquo Advances in Applied Probability vol 18no 1 pp 211ndash220 1986
[10] D Wang and S K Ghosh ldquoBayesian estimation and unitroot tests for random coefficient autoregressive modelsrdquoModelAssisted Statistics and Applications vol 3 no 4 pp 281ndash2952008
[11] DWang S K Ghosh and S G Pantula ldquoMaximum likelihoodestimation and unit root test for first order random coefficientautoregressive modelsrdquo Journal of Statistical Theory and Prac-tice vol 4 no 2 pp 261ndash278 2010
[12] T V Ramanathan and M B Rajarshi ldquoRank tests for testingthe randomness of autoregressive coefficientsrdquo Statistics ampProbability Letters vol 21 no 2 pp 115ndash120 1994
[13] S Lee J Ha O Na and S Na ldquoThe cusum test for parameterchange in time seriesmodelsrdquo Scandinavian Journal of Statisticsvol 30 no 4 pp 781ndash796 2003
[14] M Moreno and J Romo ldquoRobust unit root tests with autore-gressive errorsrdquo Communications in StatisticsmdashTheory andMethods vol 45 no 20 pp 5997ndash6021 2016
[15] V K Rohatgi A K Md Ehsanes Sleh R Ahluwalia and P JildquoNull distribution of locally most powerful tests for the twosample problemwhen the combined sample is type II censoredrdquoCommunications in Statistics - Theory and Methods vol 19 pp2337ndash2355 1992
[16] M S Chikkagoudar and B S Biradar ldquoLocally most powerfulrank tests for comparison of two failure rates based on multipletype-ii censored datardquoCommunications in Statistics Theory andMethods vol 41 no 23 pp 4315ndash4331 2012
[17] A Manik N Balakrishna and T V Ramanathan ldquoTesting theconstancy of the thinning parameter in a random coefficientinteger autoregressive modelrdquo Statistical Papers pp 1ndash25 2017
[18] P J Huber Robust Statistics JohnWiley amp Sons New York NYUSA 1981
[19] T P Hettmansperger Statistical Inference Based on Ranks JohnWiley amp Sons Inc New York NY USA 1984
[20] Z-W Zhao D-H Wang and C-X Peng ldquoCoefficient con-stancy test in generalized random coefficient autoregressivemodelrdquoApplied Mathematics and Computation vol 219 no 20pp 10283ndash10292 2013
[21] H C Thode Testing for Normality CRC Press New York NYUSA 2002
Mathematical Problems in Engineering 11
[22] P Billingsley ldquoThe lindeberg-Lvy theorem for martingalesrdquoProceedings of the American Mathematical Society vol 12 no1 pp 788ndash792 1961
[23] P Hall and C C Heyde Martingale Limit Theory and ItsApplication Academic Press New York NY USA 1980
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Submit your manuscripts atwwwhindawicom
Mathematical Problems in Engineering 7
0 20 40 60 80
minus10
minus50
510
15
Time
Dat
a Val
ue
Time
Dat
a Val
ue
0 20 40 60 80
minus10
minus50
510
15
９Ｎ ８Ｎ
Figure 1 The left one is the sample path of original series 119910119905 and the right one is the sample plot of centering series119883119905 for the real data
0 5 10 15
minus05
00
05
10
Lag
ACF
5 10 15
minus04
minus02
00
02
Lag
PACF
８Ｎ８Ｎ
Figure 2 The sample autocorrelation function (ACF) and partial autocorrelation function (PACF) plots of centering series 119883119905 for the realdata
see119883119905may come from a stationary autoregressive time seriesprocess From Figure 2 we conclude that 119883119905 is from first-order autoregressive process Moreover we test the normalityof the residual data 119883119905 according to the Normal Q-Q plotintroduced by Henry CThode [21] From Figure 3 we noticethat the scatter points on theNormal Q-Q plot are close to thereference line so the residual data119883119905 can basically be seen asfollowing the normal distribution Hence we derive that themodel N1 would be more suitable to model these data
From the time series plot given in Figure 1 the constancyof the parameter may be suspected and therefore we areinterested in testing the hypothesis of constancy of thecoefficient parameter We carry out the test for 1198670 1205902120572 = 0against 1198671 1205902120572 gt 0 The value of the test statistic in (8)turned out to be 21151 with 119901 value 00344 which indicatesthe rejection of the null hypothesis at 5 level of significanceThus in this case the RCAR(1) model would be much moreappropriate as opposed to the AR(1) model
5 Conclusions
In this article we propose a locally most powerful-typetest for testing the constancy of the random coefficientparameter in autoregressive model and derive their limitingnull distributions under regularity conditions It is clear fromthe applications that the coefficient need not remain constantthroughout the time Therefore it is essential to have such a
Normal QminusQ Plot
minus10
minus50
510
15Sa
mpl
e Qua
ntile
s
minus1 0 1 2minus2Theoretical Quantiles
Figure 3 The normal Q-Q plot of centering series 119883119905 for the realdata
test conducted whenever the random variation is suspectedThrough our simulation study and a real-data analysis wedemonstrate that our test succeeds and it performs better thanthe competitor Finally we anticipate that our locally mostpowerful-type test can be extended to other types of timeseries model
8 Mathematical Problems in Engineering
Appendix
A Process of Establish the Test Statistic
In this Appendix A we present the detailed steps to obtainthe test statistics
Proof Therefore we can get the likelihood function
1198711198671 (1199091 119909119905) = 1198911198830 (1199090) 119899prod119905=1
119891119883119905|119883119905minus1 (119909119905 | 119909119905minus1)= 1198911198830 (1199090) 119899prod
119905=1
int119891120576 (119909119905 minus 120572119905119909119905minus1) 119889119865120572 (A1)
Now the Taylor series expansion of 1198711198671 around 120572 which isthe mean of 120572119905 gives that
1198711198671 (1199091 119909119905) = 1198911198830 (1199090) 119899prod119905=1
int119891120576 (119909119905minus 120572119905119909119905minus1) 119889119865120572
= 1198911198830 (1199090) 119899prod119905=1
int[119891120576 (119909119905 minus 120572119909119905minus1)
+ (120572119905 minus 120572) 1198911015840120576 (119909119905 minus 120572119909119905minus1)+ (120572119905 minus 120572)22 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)+ (120572119905 minus 120572)33 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) + sdot sdot sdot] 119889119865120572
= 1198911198830 (1199090) 119899prod119905=1
[119891120576 (119909119905 minus 120572119909119905minus1) + 12120590212057211989110158401015840120576 (119909119905minus 120572119909119905minus1) + 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) int (120572119905 minus 120572)33 119889119865120572+ sdot sdot sdot]
(A2)
Obviously we can conclude that 119875(120572119905 minus 120572 = 0) = 1 if 1205902120572 = 0That is to say the distribution function of random variable 120572119905is degenerated Furthermore we derive that
int (120572119905 minus 120572)119896119896 119889119865120572 = 0 119896 = 1 2 3 sdot sdot sdot (A3)
By taking the derivative of log 1198711198671 with respect to120573 at1205902120572 = 0we have the following equation
120597 log 11987111986711205971205902120572 1003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 120597sum119899119905=1 log [119891120576 (119909119905 minus 120572119909119905minus1) + (12) 120590212057211989110158401015840120576 (119909119905 minus 120572119909119905minus1) + 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) int ((120572119905 minus 120572)3 3) 119889119865120572 + sdot sdot sdot ]1205971205902120572 100381610038161003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 120597sum119899119905=1 log [119891120576 (119909119905 minus 120572119909119905minus1) + (12) 120590212057211989110158401015840120576 (119909119905 minus 120572119909119905minus1)]1205971205902120572 100381610038161003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 12 119899sum119905=111989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) (A4)
Thus the LMP test statistic has the form119876119899 (120573) = 119899sum119905=1
12 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) (A5)
This completes the proof
B Proof of Theorem 1
Proof of Theorem 1 In the following we can write down thelog-likelihood function under the null hypothesis
log 119871 = 119899sum119905=1
log119891120576 (119909119905 minus 120572119909119905minus1) (B1)
According to the maximum likelihood principle the maxi-mum likelihood estimate can be obtained by maximizing
the log-likelihood function log 119871 with respect to 120573 Applyingthe Taylor series expansion to 120597 log119891120576120597120573 gives
0 = 120597log119891120576120597120573 = 120597log119891120576120597120573 + ( minus 120573) 1205972log119891120576 (120573lowast)1205971205731205971205731015840 (B2)
where 120573lowast is on the line segment between 120573 and so 120573lowast119901997888rarr
120573(119899 997888rarr infin) Under the null hypothesis we can write
minus 120573 = minus 119899sum119905=1
1205972 log1198911205761205971205731205971205731015840 + 119900119901 (119899)minus1 119899sum119905=1120597 log119891120576120597120573 (B3)
Mathematical Problems in Engineering 9
Using the similar method as in the above expanding(1radic119899)sum119899119905=1 119877119905() around 120573 gives that1radic119899 119899sum119905=1119877119905 () = 1radic119899 119899sum119905=1119877119905 (120573)+ ( minus 120573)119879 1radic119899 119899sum119905=1120597119877119905 (120573)120597120573 + 119900119901 (1) (B4)
with 120597119877119905(120573)120597120573 = (120597119877119905120597120572 1205971198771199051205971205902)119879 The remainder termis 119900119901(1) since 119901997888rarr 120573Under (C3) and (C4) we conclude that1205972119877119905(120573)1205971205731205971205731015840 = 119874119901(119899) Therefore
( minus 120573)119879 ( 1radic119899) 119899sum119905=1120597119877119905 (120573)120597120573= minus1119899 119899sum119905=1[120597119877119905 (120573)120597120573 ]119879times 1119899 119899sum119905=11205972 log1198911205761205971205731205971205731015840 + 119900119901 (1)minus1times 1radic119899 119899sum119905=1120597 log119891120576120597120573
(B5)
Now the first term of (B5) converges to1119899 119899sum119905=1120597119877119905 (120573)120597120573 = 1119899 119899sum119905=1 120597120597120573 120597 log1198911205761205971205902120572 1205902120572=0
= (1119899 119899sum119905=11205972 log11989112057612059712057212059712059021205721119899 119899sum119905=11205972 log11989112057612059712059021205761205971205902120572 )1205902120572=0
997888rarr 119882as 119899 997888rarr infin
(B6)
with119882 = (1198821111988212)119879This is because
11988211 = 119864(120597 log1198911205761205971205721205971205902120572 ) = 119864(minus120597 log119891120576120597120572 120597 log1198911205761205971205902120572 )= minus12119864[120597 log119891120576120597120572 1205972 log1198911205761205971205722 + (120597 log119891120576120597120572 )2]
at 1205902120572 = 011988212 = 119864(120597 log1198911205761205971205902120576 1205971205902120572) = 119864(minus120597 log1198911205761205971205902120576 120597 log1198911205761205971205902120572 )= minus12119864[120597 log1198911205761205971205902120576 1205972 log1198911205761205971205722 + (120597 log119891120576120597120572 )2]
at 1205902120572 = 0
(B7)
Next we can obtain the elements of Fisher informationmatrix
Σ = 1119899 119899sum119905=1119864[120597 log119891120576120597120573 (120597 log119891120576120597120573 )119879 | F119910119905minus1]= [[[[[
minus119864[1205972 log1198911205761205971205722 ] minus119864[1205972 log1198911205761205971205721205971205902120576 ]minus119864[1205972 log1198911205761205971205721205971205902120576 ] minus119864[1205972 log11989112057612059712059021205761205971205902120576 ]]]]]]
(B8)
Thus (B5) is asymptotically equivalent to
minus119882Σminus1 1radic119899 119899sum119905=1120597 log119891120576120597120573 (B9)
We present the following lemma recommended by Billingsley[22] to establish the asymptotic normality of the test statistic
Lemma B1Under assumption of (A1) we have as 119899 997888rarr infin(i) lim119899997888rarrinfin(1119899)sum119899119905=1 119864(1198772119905 (120573) | F119877119905minus1) = 1205902 119886119904 (ii)(1radic119899)sum119899119905=1 119877119905(120573) 119889997888rarr 119873(0 1205902) where 1205902 = (14)1198641205972 log1198911205761205971205722 + (120597 log119891120576120597120572)2
Proof By the ergodic theorem the first conclusion holdsNow let F119905 = 1205901199090 1199091 119909119905 119905 ge 1 and F0 is a sigma
field Note that 119876119899(120573) = sum119899119905=1 119877119905(120573) It is easy to see that[119877119899(120573)|F119899minus1] = 0 Then we have
119864 [119876119899 (120573) | F119899minus1] = 119864 [119876119899minus1 (120573) + 119877119899 (120573) | F119899minus1]= 119876119899minus1 (120573) (B10)
Thus 119876119899F119899 119899 ge 0 is a martingale We have shownsup119905119864|119877119905(120573)|2+120578 lt infin which means 119864[1198772119905 (120573)] is uniformlyintegrable ByTheorem 11 of Billingsley [22] we have that as119899 997888rarr infin1119899 119899sum119905=11198772119905 (120573) 119886119904997888997888rarr 119864[1198772119899 (120573) | F119899minus1] = 1205902 (B11)
Hence using the following version of Martingale Cen-tral Limit Theorem from Hall and Heyde [23] we have(1radic119899)sum119899119905=1 119877119905(120573) 119889997888rarr 119873(0 1205902) The proof of Lemma B1 isthus completed ◻
Similarly we can verify that 119878119899 = sum119899119905=1(120597log119891120576120597120573) is amartingale By ergodic and stationary properties we obtainthat as 119899 997888rarr infin
1119899 119899sum119905=11205972 log119891120576120597120573120597120573119879 119886119904997888997888rarr 119864[120597 log119891120576120597120573 (120597 log119891120576120597120573 )119879] = Σ (B12)
Therefore (1radic119899)119878119899 119889997888rarr 119873(0 Σ)
10 Mathematical Problems in Engineering
In the same way for any vector 119888 = (1198881 1198882 1198883)119879 isin 1198773 (0 0 0) we have1radic119899119888119879(
119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) = 1radic119899sdot 119899sum119905=1
(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 ) 119889997888rarr 119873(0 119864(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 )2)
(B13)
By the Cramer-Wold device we obtain
1radic119899 (119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) 119889997888rarr 119873[(00) (1205902 119882119879119882 Σ )] (B14)
where119882 = (1198821111988212)119879 Then we can make the conclusionthat 1radic119899 119899sum119905=1119877119905 () 119889997888rarr 119873(0 1205962) (B15)
where 1205962 = 1205902 minus 119882119879Σminus1119882 The proof of the theorem iscompleted
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that they have no conflicts of interest
Acknowledgments
This work is supported by National Natural Science Foun-dation of China (Nos 11871028 11731015 11571051 and11501241) Natural Science Foundation of Jilin Province (Nos20180101216JC 20170101057JC and 20150520053JH) Pro-gram for Changbaishan Scholars of Jilin Province (2015010)and Science and Technology Program of Jilin EducationalDepartment during the ldquo13th Five-Yearrdquo Plan Period (No2016316)
References
[1] A Hoque ldquoFinite sample analysis of the first order autoregres-sive modelrdquoCalcutta Statistical Association Bulletin vol 34 no1-2 pp 51ndash63 1985
[2] T Ogawa H Sonoda S Ishiwa and Y Shigeta ldquoAn applicationof autoregressive model to pattern discrimination of brainelectrical activity mappingrdquo Brain Topography vol 6 no 1 pp3ndash11 1992
[3] A Subasi A Alkan E Koklukaya and M K Kiymik ldquoWaveletneural network classification of EEG signals by using ARmodelwith MLE preprocessingrdquo Journal of the International NeuralNetwork Society vol 18 pp 985ndash997 2005
[4] M Maleki and A R Nematollahi ldquoAutoregressive models withmixture of scale mixtures of gaussian innovationsrdquo IranianJournal of Science amp Technology Transactions A Science vol 41no 4 pp 1099ndash1107 2017
[5] J Conlisk ldquoStability in a random coefficient modelrdquo Interna-tional Economic Review vol 15 no 2 pp 529ndash533 1974
[6] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models Irdquo Journal of Time SeriesAnalysis vol 1 no 1 pp 37ndash46 1980
[7] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models IIrdquo Journal of Time SeriesAnalysis vol 2 no 3 pp 185ndash203 1981
[8] D F Nicholls and B G Quinn Random Coefficient Autore-gressive Models An Introduction Lecture Notes in StatisticsSpringer-Verlag New York NY USA 1982
[9] A Brandt ldquoThe stochastic equation Y119899+1 = A119899Y119899 + B119899 withstationary coefficientsrdquo Advances in Applied Probability vol 18no 1 pp 211ndash220 1986
[10] D Wang and S K Ghosh ldquoBayesian estimation and unitroot tests for random coefficient autoregressive modelsrdquoModelAssisted Statistics and Applications vol 3 no 4 pp 281ndash2952008
[11] DWang S K Ghosh and S G Pantula ldquoMaximum likelihoodestimation and unit root test for first order random coefficientautoregressive modelsrdquo Journal of Statistical Theory and Prac-tice vol 4 no 2 pp 261ndash278 2010
[12] T V Ramanathan and M B Rajarshi ldquoRank tests for testingthe randomness of autoregressive coefficientsrdquo Statistics ampProbability Letters vol 21 no 2 pp 115ndash120 1994
[13] S Lee J Ha O Na and S Na ldquoThe cusum test for parameterchange in time seriesmodelsrdquo Scandinavian Journal of Statisticsvol 30 no 4 pp 781ndash796 2003
[14] M Moreno and J Romo ldquoRobust unit root tests with autore-gressive errorsrdquo Communications in StatisticsmdashTheory andMethods vol 45 no 20 pp 5997ndash6021 2016
[15] V K Rohatgi A K Md Ehsanes Sleh R Ahluwalia and P JildquoNull distribution of locally most powerful tests for the twosample problemwhen the combined sample is type II censoredrdquoCommunications in Statistics - Theory and Methods vol 19 pp2337ndash2355 1992
[16] M S Chikkagoudar and B S Biradar ldquoLocally most powerfulrank tests for comparison of two failure rates based on multipletype-ii censored datardquoCommunications in Statistics Theory andMethods vol 41 no 23 pp 4315ndash4331 2012
[17] A Manik N Balakrishna and T V Ramanathan ldquoTesting theconstancy of the thinning parameter in a random coefficientinteger autoregressive modelrdquo Statistical Papers pp 1ndash25 2017
[18] P J Huber Robust Statistics JohnWiley amp Sons New York NYUSA 1981
[19] T P Hettmansperger Statistical Inference Based on Ranks JohnWiley amp Sons Inc New York NY USA 1984
[20] Z-W Zhao D-H Wang and C-X Peng ldquoCoefficient con-stancy test in generalized random coefficient autoregressivemodelrdquoApplied Mathematics and Computation vol 219 no 20pp 10283ndash10292 2013
[21] H C Thode Testing for Normality CRC Press New York NYUSA 2002
Mathematical Problems in Engineering 11
[22] P Billingsley ldquoThe lindeberg-Lvy theorem for martingalesrdquoProceedings of the American Mathematical Society vol 12 no1 pp 788ndash792 1961
[23] P Hall and C C Heyde Martingale Limit Theory and ItsApplication Academic Press New York NY USA 1980
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8 Mathematical Problems in Engineering
Appendix
A Process of Establish the Test Statistic
In this Appendix A we present the detailed steps to obtainthe test statistics
Proof Therefore we can get the likelihood function
1198711198671 (1199091 119909119905) = 1198911198830 (1199090) 119899prod119905=1
119891119883119905|119883119905minus1 (119909119905 | 119909119905minus1)= 1198911198830 (1199090) 119899prod
119905=1
int119891120576 (119909119905 minus 120572119905119909119905minus1) 119889119865120572 (A1)
Now the Taylor series expansion of 1198711198671 around 120572 which isthe mean of 120572119905 gives that
1198711198671 (1199091 119909119905) = 1198911198830 (1199090) 119899prod119905=1
int119891120576 (119909119905minus 120572119905119909119905minus1) 119889119865120572
= 1198911198830 (1199090) 119899prod119905=1
int[119891120576 (119909119905 minus 120572119909119905minus1)
+ (120572119905 minus 120572) 1198911015840120576 (119909119905 minus 120572119909119905minus1)+ (120572119905 minus 120572)22 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)+ (120572119905 minus 120572)33 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) + sdot sdot sdot] 119889119865120572
= 1198911198830 (1199090) 119899prod119905=1
[119891120576 (119909119905 minus 120572119909119905minus1) + 12120590212057211989110158401015840120576 (119909119905minus 120572119909119905minus1) + 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) int (120572119905 minus 120572)33 119889119865120572+ sdot sdot sdot]
(A2)
Obviously we can conclude that 119875(120572119905 minus 120572 = 0) = 1 if 1205902120572 = 0That is to say the distribution function of random variable 120572119905is degenerated Furthermore we derive that
int (120572119905 minus 120572)119896119896 119889119865120572 = 0 119896 = 1 2 3 sdot sdot sdot (A3)
By taking the derivative of log 1198711198671 with respect to120573 at1205902120572 = 0we have the following equation
120597 log 11987111986711205971205902120572 1003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 120597sum119899119905=1 log [119891120576 (119909119905 minus 120572119909119905minus1) + (12) 120590212057211989110158401015840120576 (119909119905 minus 120572119909119905minus1) + 119891101584010158401015840120576 (119909119905 minus 120572119909119905minus1) int ((120572119905 minus 120572)3 3) 119889119865120572 + sdot sdot sdot ]1205971205902120572 100381610038161003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 120597sum119899119905=1 log [119891120576 (119909119905 minus 120572119909119905minus1) + (12) 120590212057211989110158401015840120576 (119909119905 minus 120572119909119905minus1)]1205971205902120572 100381610038161003816100381610038161003816100381610038161003816100381610038161205902120572=0
= 12 119899sum119905=111989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) (A4)
Thus the LMP test statistic has the form119876119899 (120573) = 119899sum119905=1
12 11989110158401015840120576 (119909119905 minus 120572119909119905minus1)119891120576 (119909119905 minus 120572119909119905minus1) (A5)
This completes the proof
B Proof of Theorem 1
Proof of Theorem 1 In the following we can write down thelog-likelihood function under the null hypothesis
log 119871 = 119899sum119905=1
log119891120576 (119909119905 minus 120572119909119905minus1) (B1)
According to the maximum likelihood principle the maxi-mum likelihood estimate can be obtained by maximizing
the log-likelihood function log 119871 with respect to 120573 Applyingthe Taylor series expansion to 120597 log119891120576120597120573 gives
0 = 120597log119891120576120597120573 = 120597log119891120576120597120573 + ( minus 120573) 1205972log119891120576 (120573lowast)1205971205731205971205731015840 (B2)
where 120573lowast is on the line segment between 120573 and so 120573lowast119901997888rarr
120573(119899 997888rarr infin) Under the null hypothesis we can write
minus 120573 = minus 119899sum119905=1
1205972 log1198911205761205971205731205971205731015840 + 119900119901 (119899)minus1 119899sum119905=1120597 log119891120576120597120573 (B3)
Mathematical Problems in Engineering 9
Using the similar method as in the above expanding(1radic119899)sum119899119905=1 119877119905() around 120573 gives that1radic119899 119899sum119905=1119877119905 () = 1radic119899 119899sum119905=1119877119905 (120573)+ ( minus 120573)119879 1radic119899 119899sum119905=1120597119877119905 (120573)120597120573 + 119900119901 (1) (B4)
with 120597119877119905(120573)120597120573 = (120597119877119905120597120572 1205971198771199051205971205902)119879 The remainder termis 119900119901(1) since 119901997888rarr 120573Under (C3) and (C4) we conclude that1205972119877119905(120573)1205971205731205971205731015840 = 119874119901(119899) Therefore
( minus 120573)119879 ( 1radic119899) 119899sum119905=1120597119877119905 (120573)120597120573= minus1119899 119899sum119905=1[120597119877119905 (120573)120597120573 ]119879times 1119899 119899sum119905=11205972 log1198911205761205971205731205971205731015840 + 119900119901 (1)minus1times 1radic119899 119899sum119905=1120597 log119891120576120597120573
(B5)
Now the first term of (B5) converges to1119899 119899sum119905=1120597119877119905 (120573)120597120573 = 1119899 119899sum119905=1 120597120597120573 120597 log1198911205761205971205902120572 1205902120572=0
= (1119899 119899sum119905=11205972 log11989112057612059712057212059712059021205721119899 119899sum119905=11205972 log11989112057612059712059021205761205971205902120572 )1205902120572=0
997888rarr 119882as 119899 997888rarr infin
(B6)
with119882 = (1198821111988212)119879This is because
11988211 = 119864(120597 log1198911205761205971205721205971205902120572 ) = 119864(minus120597 log119891120576120597120572 120597 log1198911205761205971205902120572 )= minus12119864[120597 log119891120576120597120572 1205972 log1198911205761205971205722 + (120597 log119891120576120597120572 )2]
at 1205902120572 = 011988212 = 119864(120597 log1198911205761205971205902120576 1205971205902120572) = 119864(minus120597 log1198911205761205971205902120576 120597 log1198911205761205971205902120572 )= minus12119864[120597 log1198911205761205971205902120576 1205972 log1198911205761205971205722 + (120597 log119891120576120597120572 )2]
at 1205902120572 = 0
(B7)
Next we can obtain the elements of Fisher informationmatrix
Σ = 1119899 119899sum119905=1119864[120597 log119891120576120597120573 (120597 log119891120576120597120573 )119879 | F119910119905minus1]= [[[[[
minus119864[1205972 log1198911205761205971205722 ] minus119864[1205972 log1198911205761205971205721205971205902120576 ]minus119864[1205972 log1198911205761205971205721205971205902120576 ] minus119864[1205972 log11989112057612059712059021205761205971205902120576 ]]]]]]
(B8)
Thus (B5) is asymptotically equivalent to
minus119882Σminus1 1radic119899 119899sum119905=1120597 log119891120576120597120573 (B9)
We present the following lemma recommended by Billingsley[22] to establish the asymptotic normality of the test statistic
Lemma B1Under assumption of (A1) we have as 119899 997888rarr infin(i) lim119899997888rarrinfin(1119899)sum119899119905=1 119864(1198772119905 (120573) | F119877119905minus1) = 1205902 119886119904 (ii)(1radic119899)sum119899119905=1 119877119905(120573) 119889997888rarr 119873(0 1205902) where 1205902 = (14)1198641205972 log1198911205761205971205722 + (120597 log119891120576120597120572)2
Proof By the ergodic theorem the first conclusion holdsNow let F119905 = 1205901199090 1199091 119909119905 119905 ge 1 and F0 is a sigma
field Note that 119876119899(120573) = sum119899119905=1 119877119905(120573) It is easy to see that[119877119899(120573)|F119899minus1] = 0 Then we have
119864 [119876119899 (120573) | F119899minus1] = 119864 [119876119899minus1 (120573) + 119877119899 (120573) | F119899minus1]= 119876119899minus1 (120573) (B10)
Thus 119876119899F119899 119899 ge 0 is a martingale We have shownsup119905119864|119877119905(120573)|2+120578 lt infin which means 119864[1198772119905 (120573)] is uniformlyintegrable ByTheorem 11 of Billingsley [22] we have that as119899 997888rarr infin1119899 119899sum119905=11198772119905 (120573) 119886119904997888997888rarr 119864[1198772119899 (120573) | F119899minus1] = 1205902 (B11)
Hence using the following version of Martingale Cen-tral Limit Theorem from Hall and Heyde [23] we have(1radic119899)sum119899119905=1 119877119905(120573) 119889997888rarr 119873(0 1205902) The proof of Lemma B1 isthus completed ◻
Similarly we can verify that 119878119899 = sum119899119905=1(120597log119891120576120597120573) is amartingale By ergodic and stationary properties we obtainthat as 119899 997888rarr infin
1119899 119899sum119905=11205972 log119891120576120597120573120597120573119879 119886119904997888997888rarr 119864[120597 log119891120576120597120573 (120597 log119891120576120597120573 )119879] = Σ (B12)
Therefore (1radic119899)119878119899 119889997888rarr 119873(0 Σ)
10 Mathematical Problems in Engineering
In the same way for any vector 119888 = (1198881 1198882 1198883)119879 isin 1198773 (0 0 0) we have1radic119899119888119879(
119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) = 1radic119899sdot 119899sum119905=1
(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 ) 119889997888rarr 119873(0 119864(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 )2)
(B13)
By the Cramer-Wold device we obtain
1radic119899 (119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) 119889997888rarr 119873[(00) (1205902 119882119879119882 Σ )] (B14)
where119882 = (1198821111988212)119879 Then we can make the conclusionthat 1radic119899 119899sum119905=1119877119905 () 119889997888rarr 119873(0 1205962) (B15)
where 1205962 = 1205902 minus 119882119879Σminus1119882 The proof of the theorem iscompleted
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that they have no conflicts of interest
Acknowledgments
This work is supported by National Natural Science Foun-dation of China (Nos 11871028 11731015 11571051 and11501241) Natural Science Foundation of Jilin Province (Nos20180101216JC 20170101057JC and 20150520053JH) Pro-gram for Changbaishan Scholars of Jilin Province (2015010)and Science and Technology Program of Jilin EducationalDepartment during the ldquo13th Five-Yearrdquo Plan Period (No2016316)
References
[1] A Hoque ldquoFinite sample analysis of the first order autoregres-sive modelrdquoCalcutta Statistical Association Bulletin vol 34 no1-2 pp 51ndash63 1985
[2] T Ogawa H Sonoda S Ishiwa and Y Shigeta ldquoAn applicationof autoregressive model to pattern discrimination of brainelectrical activity mappingrdquo Brain Topography vol 6 no 1 pp3ndash11 1992
[3] A Subasi A Alkan E Koklukaya and M K Kiymik ldquoWaveletneural network classification of EEG signals by using ARmodelwith MLE preprocessingrdquo Journal of the International NeuralNetwork Society vol 18 pp 985ndash997 2005
[4] M Maleki and A R Nematollahi ldquoAutoregressive models withmixture of scale mixtures of gaussian innovationsrdquo IranianJournal of Science amp Technology Transactions A Science vol 41no 4 pp 1099ndash1107 2017
[5] J Conlisk ldquoStability in a random coefficient modelrdquo Interna-tional Economic Review vol 15 no 2 pp 529ndash533 1974
[6] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models Irdquo Journal of Time SeriesAnalysis vol 1 no 1 pp 37ndash46 1980
[7] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models IIrdquo Journal of Time SeriesAnalysis vol 2 no 3 pp 185ndash203 1981
[8] D F Nicholls and B G Quinn Random Coefficient Autore-gressive Models An Introduction Lecture Notes in StatisticsSpringer-Verlag New York NY USA 1982
[9] A Brandt ldquoThe stochastic equation Y119899+1 = A119899Y119899 + B119899 withstationary coefficientsrdquo Advances in Applied Probability vol 18no 1 pp 211ndash220 1986
[10] D Wang and S K Ghosh ldquoBayesian estimation and unitroot tests for random coefficient autoregressive modelsrdquoModelAssisted Statistics and Applications vol 3 no 4 pp 281ndash2952008
[11] DWang S K Ghosh and S G Pantula ldquoMaximum likelihoodestimation and unit root test for first order random coefficientautoregressive modelsrdquo Journal of Statistical Theory and Prac-tice vol 4 no 2 pp 261ndash278 2010
[12] T V Ramanathan and M B Rajarshi ldquoRank tests for testingthe randomness of autoregressive coefficientsrdquo Statistics ampProbability Letters vol 21 no 2 pp 115ndash120 1994
[13] S Lee J Ha O Na and S Na ldquoThe cusum test for parameterchange in time seriesmodelsrdquo Scandinavian Journal of Statisticsvol 30 no 4 pp 781ndash796 2003
[14] M Moreno and J Romo ldquoRobust unit root tests with autore-gressive errorsrdquo Communications in StatisticsmdashTheory andMethods vol 45 no 20 pp 5997ndash6021 2016
[15] V K Rohatgi A K Md Ehsanes Sleh R Ahluwalia and P JildquoNull distribution of locally most powerful tests for the twosample problemwhen the combined sample is type II censoredrdquoCommunications in Statistics - Theory and Methods vol 19 pp2337ndash2355 1992
[16] M S Chikkagoudar and B S Biradar ldquoLocally most powerfulrank tests for comparison of two failure rates based on multipletype-ii censored datardquoCommunications in Statistics Theory andMethods vol 41 no 23 pp 4315ndash4331 2012
[17] A Manik N Balakrishna and T V Ramanathan ldquoTesting theconstancy of the thinning parameter in a random coefficientinteger autoregressive modelrdquo Statistical Papers pp 1ndash25 2017
[18] P J Huber Robust Statistics JohnWiley amp Sons New York NYUSA 1981
[19] T P Hettmansperger Statistical Inference Based on Ranks JohnWiley amp Sons Inc New York NY USA 1984
[20] Z-W Zhao D-H Wang and C-X Peng ldquoCoefficient con-stancy test in generalized random coefficient autoregressivemodelrdquoApplied Mathematics and Computation vol 219 no 20pp 10283ndash10292 2013
[21] H C Thode Testing for Normality CRC Press New York NYUSA 2002
Mathematical Problems in Engineering 11
[22] P Billingsley ldquoThe lindeberg-Lvy theorem for martingalesrdquoProceedings of the American Mathematical Society vol 12 no1 pp 788ndash792 1961
[23] P Hall and C C Heyde Martingale Limit Theory and ItsApplication Academic Press New York NY USA 1980
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
Mathematical Problems in Engineering 9
Using the similar method as in the above expanding(1radic119899)sum119899119905=1 119877119905() around 120573 gives that1radic119899 119899sum119905=1119877119905 () = 1radic119899 119899sum119905=1119877119905 (120573)+ ( minus 120573)119879 1radic119899 119899sum119905=1120597119877119905 (120573)120597120573 + 119900119901 (1) (B4)
with 120597119877119905(120573)120597120573 = (120597119877119905120597120572 1205971198771199051205971205902)119879 The remainder termis 119900119901(1) since 119901997888rarr 120573Under (C3) and (C4) we conclude that1205972119877119905(120573)1205971205731205971205731015840 = 119874119901(119899) Therefore
( minus 120573)119879 ( 1radic119899) 119899sum119905=1120597119877119905 (120573)120597120573= minus1119899 119899sum119905=1[120597119877119905 (120573)120597120573 ]119879times 1119899 119899sum119905=11205972 log1198911205761205971205731205971205731015840 + 119900119901 (1)minus1times 1radic119899 119899sum119905=1120597 log119891120576120597120573
(B5)
Now the first term of (B5) converges to1119899 119899sum119905=1120597119877119905 (120573)120597120573 = 1119899 119899sum119905=1 120597120597120573 120597 log1198911205761205971205902120572 1205902120572=0
= (1119899 119899sum119905=11205972 log11989112057612059712057212059712059021205721119899 119899sum119905=11205972 log11989112057612059712059021205761205971205902120572 )1205902120572=0
997888rarr 119882as 119899 997888rarr infin
(B6)
with119882 = (1198821111988212)119879This is because
11988211 = 119864(120597 log1198911205761205971205721205971205902120572 ) = 119864(minus120597 log119891120576120597120572 120597 log1198911205761205971205902120572 )= minus12119864[120597 log119891120576120597120572 1205972 log1198911205761205971205722 + (120597 log119891120576120597120572 )2]
at 1205902120572 = 011988212 = 119864(120597 log1198911205761205971205902120576 1205971205902120572) = 119864(minus120597 log1198911205761205971205902120576 120597 log1198911205761205971205902120572 )= minus12119864[120597 log1198911205761205971205902120576 1205972 log1198911205761205971205722 + (120597 log119891120576120597120572 )2]
at 1205902120572 = 0
(B7)
Next we can obtain the elements of Fisher informationmatrix
Σ = 1119899 119899sum119905=1119864[120597 log119891120576120597120573 (120597 log119891120576120597120573 )119879 | F119910119905minus1]= [[[[[
minus119864[1205972 log1198911205761205971205722 ] minus119864[1205972 log1198911205761205971205721205971205902120576 ]minus119864[1205972 log1198911205761205971205721205971205902120576 ] minus119864[1205972 log11989112057612059712059021205761205971205902120576 ]]]]]]
(B8)
Thus (B5) is asymptotically equivalent to
minus119882Σminus1 1radic119899 119899sum119905=1120597 log119891120576120597120573 (B9)
We present the following lemma recommended by Billingsley[22] to establish the asymptotic normality of the test statistic
Lemma B1Under assumption of (A1) we have as 119899 997888rarr infin(i) lim119899997888rarrinfin(1119899)sum119899119905=1 119864(1198772119905 (120573) | F119877119905minus1) = 1205902 119886119904 (ii)(1radic119899)sum119899119905=1 119877119905(120573) 119889997888rarr 119873(0 1205902) where 1205902 = (14)1198641205972 log1198911205761205971205722 + (120597 log119891120576120597120572)2
Proof By the ergodic theorem the first conclusion holdsNow let F119905 = 1205901199090 1199091 119909119905 119905 ge 1 and F0 is a sigma
field Note that 119876119899(120573) = sum119899119905=1 119877119905(120573) It is easy to see that[119877119899(120573)|F119899minus1] = 0 Then we have
119864 [119876119899 (120573) | F119899minus1] = 119864 [119876119899minus1 (120573) + 119877119899 (120573) | F119899minus1]= 119876119899minus1 (120573) (B10)
Thus 119876119899F119899 119899 ge 0 is a martingale We have shownsup119905119864|119877119905(120573)|2+120578 lt infin which means 119864[1198772119905 (120573)] is uniformlyintegrable ByTheorem 11 of Billingsley [22] we have that as119899 997888rarr infin1119899 119899sum119905=11198772119905 (120573) 119886119904997888997888rarr 119864[1198772119899 (120573) | F119899minus1] = 1205902 (B11)
Hence using the following version of Martingale Cen-tral Limit Theorem from Hall and Heyde [23] we have(1radic119899)sum119899119905=1 119877119905(120573) 119889997888rarr 119873(0 1205902) The proof of Lemma B1 isthus completed ◻
Similarly we can verify that 119878119899 = sum119899119905=1(120597log119891120576120597120573) is amartingale By ergodic and stationary properties we obtainthat as 119899 997888rarr infin
1119899 119899sum119905=11205972 log119891120576120597120573120597120573119879 119886119904997888997888rarr 119864[120597 log119891120576120597120573 (120597 log119891120576120597120573 )119879] = Σ (B12)
Therefore (1radic119899)119878119899 119889997888rarr 119873(0 Σ)
10 Mathematical Problems in Engineering
In the same way for any vector 119888 = (1198881 1198882 1198883)119879 isin 1198773 (0 0 0) we have1radic119899119888119879(
119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) = 1radic119899sdot 119899sum119905=1
(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 ) 119889997888rarr 119873(0 119864(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 )2)
(B13)
By the Cramer-Wold device we obtain
1radic119899 (119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) 119889997888rarr 119873[(00) (1205902 119882119879119882 Σ )] (B14)
where119882 = (1198821111988212)119879 Then we can make the conclusionthat 1radic119899 119899sum119905=1119877119905 () 119889997888rarr 119873(0 1205962) (B15)
where 1205962 = 1205902 minus 119882119879Σminus1119882 The proof of the theorem iscompleted
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that they have no conflicts of interest
Acknowledgments
This work is supported by National Natural Science Foun-dation of China (Nos 11871028 11731015 11571051 and11501241) Natural Science Foundation of Jilin Province (Nos20180101216JC 20170101057JC and 20150520053JH) Pro-gram for Changbaishan Scholars of Jilin Province (2015010)and Science and Technology Program of Jilin EducationalDepartment during the ldquo13th Five-Yearrdquo Plan Period (No2016316)
References
[1] A Hoque ldquoFinite sample analysis of the first order autoregres-sive modelrdquoCalcutta Statistical Association Bulletin vol 34 no1-2 pp 51ndash63 1985
[2] T Ogawa H Sonoda S Ishiwa and Y Shigeta ldquoAn applicationof autoregressive model to pattern discrimination of brainelectrical activity mappingrdquo Brain Topography vol 6 no 1 pp3ndash11 1992
[3] A Subasi A Alkan E Koklukaya and M K Kiymik ldquoWaveletneural network classification of EEG signals by using ARmodelwith MLE preprocessingrdquo Journal of the International NeuralNetwork Society vol 18 pp 985ndash997 2005
[4] M Maleki and A R Nematollahi ldquoAutoregressive models withmixture of scale mixtures of gaussian innovationsrdquo IranianJournal of Science amp Technology Transactions A Science vol 41no 4 pp 1099ndash1107 2017
[5] J Conlisk ldquoStability in a random coefficient modelrdquo Interna-tional Economic Review vol 15 no 2 pp 529ndash533 1974
[6] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models Irdquo Journal of Time SeriesAnalysis vol 1 no 1 pp 37ndash46 1980
[7] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models IIrdquo Journal of Time SeriesAnalysis vol 2 no 3 pp 185ndash203 1981
[8] D F Nicholls and B G Quinn Random Coefficient Autore-gressive Models An Introduction Lecture Notes in StatisticsSpringer-Verlag New York NY USA 1982
[9] A Brandt ldquoThe stochastic equation Y119899+1 = A119899Y119899 + B119899 withstationary coefficientsrdquo Advances in Applied Probability vol 18no 1 pp 211ndash220 1986
[10] D Wang and S K Ghosh ldquoBayesian estimation and unitroot tests for random coefficient autoregressive modelsrdquoModelAssisted Statistics and Applications vol 3 no 4 pp 281ndash2952008
[11] DWang S K Ghosh and S G Pantula ldquoMaximum likelihoodestimation and unit root test for first order random coefficientautoregressive modelsrdquo Journal of Statistical Theory and Prac-tice vol 4 no 2 pp 261ndash278 2010
[12] T V Ramanathan and M B Rajarshi ldquoRank tests for testingthe randomness of autoregressive coefficientsrdquo Statistics ampProbability Letters vol 21 no 2 pp 115ndash120 1994
[13] S Lee J Ha O Na and S Na ldquoThe cusum test for parameterchange in time seriesmodelsrdquo Scandinavian Journal of Statisticsvol 30 no 4 pp 781ndash796 2003
[14] M Moreno and J Romo ldquoRobust unit root tests with autore-gressive errorsrdquo Communications in StatisticsmdashTheory andMethods vol 45 no 20 pp 5997ndash6021 2016
[15] V K Rohatgi A K Md Ehsanes Sleh R Ahluwalia and P JildquoNull distribution of locally most powerful tests for the twosample problemwhen the combined sample is type II censoredrdquoCommunications in Statistics - Theory and Methods vol 19 pp2337ndash2355 1992
[16] M S Chikkagoudar and B S Biradar ldquoLocally most powerfulrank tests for comparison of two failure rates based on multipletype-ii censored datardquoCommunications in Statistics Theory andMethods vol 41 no 23 pp 4315ndash4331 2012
[17] A Manik N Balakrishna and T V Ramanathan ldquoTesting theconstancy of the thinning parameter in a random coefficientinteger autoregressive modelrdquo Statistical Papers pp 1ndash25 2017
[18] P J Huber Robust Statistics JohnWiley amp Sons New York NYUSA 1981
[19] T P Hettmansperger Statistical Inference Based on Ranks JohnWiley amp Sons Inc New York NY USA 1984
[20] Z-W Zhao D-H Wang and C-X Peng ldquoCoefficient con-stancy test in generalized random coefficient autoregressivemodelrdquoApplied Mathematics and Computation vol 219 no 20pp 10283ndash10292 2013
[21] H C Thode Testing for Normality CRC Press New York NYUSA 2002
Mathematical Problems in Engineering 11
[22] P Billingsley ldquoThe lindeberg-Lvy theorem for martingalesrdquoProceedings of the American Mathematical Society vol 12 no1 pp 788ndash792 1961
[23] P Hall and C C Heyde Martingale Limit Theory and ItsApplication Academic Press New York NY USA 1980
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
10 Mathematical Problems in Engineering
In the same way for any vector 119888 = (1198881 1198882 1198883)119879 isin 1198773 (0 0 0) we have1radic119899119888119879(
119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) = 1radic119899sdot 119899sum119905=1
(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 ) 119889997888rarr 119873(0 119864(1198881119877119905 (120573) + 1198882 120597 log119891120576120597120572 + 1198883 120597 log1198911205761205971205902120576 )2)
(B13)
By the Cramer-Wold device we obtain
1radic119899 (119899sum119905=1
119877119905 (120573)119899sum119905=1
120597 log119891120576120597120573 ) 119889997888rarr 119873[(00) (1205902 119882119879119882 Σ )] (B14)
where119882 = (1198821111988212)119879 Then we can make the conclusionthat 1radic119899 119899sum119905=1119877119905 () 119889997888rarr 119873(0 1205962) (B15)
where 1205962 = 1205902 minus 119882119879Σminus1119882 The proof of the theorem iscompleted
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The authors declare that they have no conflicts of interest
Acknowledgments
This work is supported by National Natural Science Foun-dation of China (Nos 11871028 11731015 11571051 and11501241) Natural Science Foundation of Jilin Province (Nos20180101216JC 20170101057JC and 20150520053JH) Pro-gram for Changbaishan Scholars of Jilin Province (2015010)and Science and Technology Program of Jilin EducationalDepartment during the ldquo13th Five-Yearrdquo Plan Period (No2016316)
References
[1] A Hoque ldquoFinite sample analysis of the first order autoregres-sive modelrdquoCalcutta Statistical Association Bulletin vol 34 no1-2 pp 51ndash63 1985
[2] T Ogawa H Sonoda S Ishiwa and Y Shigeta ldquoAn applicationof autoregressive model to pattern discrimination of brainelectrical activity mappingrdquo Brain Topography vol 6 no 1 pp3ndash11 1992
[3] A Subasi A Alkan E Koklukaya and M K Kiymik ldquoWaveletneural network classification of EEG signals by using ARmodelwith MLE preprocessingrdquo Journal of the International NeuralNetwork Society vol 18 pp 985ndash997 2005
[4] M Maleki and A R Nematollahi ldquoAutoregressive models withmixture of scale mixtures of gaussian innovationsrdquo IranianJournal of Science amp Technology Transactions A Science vol 41no 4 pp 1099ndash1107 2017
[5] J Conlisk ldquoStability in a random coefficient modelrdquo Interna-tional Economic Review vol 15 no 2 pp 529ndash533 1974
[6] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models Irdquo Journal of Time SeriesAnalysis vol 1 no 1 pp 37ndash46 1980
[7] D F Nicholls and B G Quinn ldquoThe estimation of randomcoefficient autoregressive models IIrdquo Journal of Time SeriesAnalysis vol 2 no 3 pp 185ndash203 1981
[8] D F Nicholls and B G Quinn Random Coefficient Autore-gressive Models An Introduction Lecture Notes in StatisticsSpringer-Verlag New York NY USA 1982
[9] A Brandt ldquoThe stochastic equation Y119899+1 = A119899Y119899 + B119899 withstationary coefficientsrdquo Advances in Applied Probability vol 18no 1 pp 211ndash220 1986
[10] D Wang and S K Ghosh ldquoBayesian estimation and unitroot tests for random coefficient autoregressive modelsrdquoModelAssisted Statistics and Applications vol 3 no 4 pp 281ndash2952008
[11] DWang S K Ghosh and S G Pantula ldquoMaximum likelihoodestimation and unit root test for first order random coefficientautoregressive modelsrdquo Journal of Statistical Theory and Prac-tice vol 4 no 2 pp 261ndash278 2010
[12] T V Ramanathan and M B Rajarshi ldquoRank tests for testingthe randomness of autoregressive coefficientsrdquo Statistics ampProbability Letters vol 21 no 2 pp 115ndash120 1994
[13] S Lee J Ha O Na and S Na ldquoThe cusum test for parameterchange in time seriesmodelsrdquo Scandinavian Journal of Statisticsvol 30 no 4 pp 781ndash796 2003
[14] M Moreno and J Romo ldquoRobust unit root tests with autore-gressive errorsrdquo Communications in StatisticsmdashTheory andMethods vol 45 no 20 pp 5997ndash6021 2016
[15] V K Rohatgi A K Md Ehsanes Sleh R Ahluwalia and P JildquoNull distribution of locally most powerful tests for the twosample problemwhen the combined sample is type II censoredrdquoCommunications in Statistics - Theory and Methods vol 19 pp2337ndash2355 1992
[16] M S Chikkagoudar and B S Biradar ldquoLocally most powerfulrank tests for comparison of two failure rates based on multipletype-ii censored datardquoCommunications in Statistics Theory andMethods vol 41 no 23 pp 4315ndash4331 2012
[17] A Manik N Balakrishna and T V Ramanathan ldquoTesting theconstancy of the thinning parameter in a random coefficientinteger autoregressive modelrdquo Statistical Papers pp 1ndash25 2017
[18] P J Huber Robust Statistics JohnWiley amp Sons New York NYUSA 1981
[19] T P Hettmansperger Statistical Inference Based on Ranks JohnWiley amp Sons Inc New York NY USA 1984
[20] Z-W Zhao D-H Wang and C-X Peng ldquoCoefficient con-stancy test in generalized random coefficient autoregressivemodelrdquoApplied Mathematics and Computation vol 219 no 20pp 10283ndash10292 2013
[21] H C Thode Testing for Normality CRC Press New York NYUSA 2002
Mathematical Problems in Engineering 11
[22] P Billingsley ldquoThe lindeberg-Lvy theorem for martingalesrdquoProceedings of the American Mathematical Society vol 12 no1 pp 788ndash792 1961
[23] P Hall and C C Heyde Martingale Limit Theory and ItsApplication Academic Press New York NY USA 1980
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
Mathematical Problems in Engineering 11
[22] P Billingsley ldquoThe lindeberg-Lvy theorem for martingalesrdquoProceedings of the American Mathematical Society vol 12 no1 pp 788ndash792 1961
[23] P Hall and C C Heyde Martingale Limit Theory and ItsApplication Academic Press New York NY USA 1980
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom