chapter 11 - covariance and correlationkwng/fall2010/phy335/lecture/lecture 14.pdf3. the binomial...
TRANSCRIPT
Topics Part 1 – Single measurement
1. Basic stuff (Chapter 1 and 2)2. Propagation of uncertainties (Chapter 3)
Part 2 – Multiple measurements as independent results
1. Mean and standard deviation (Chapter 4)2. Basic on probability distribution function (not in text explicitly)3. The Binomial distribution (Chapter 10)4. The Poisson distribution (Chapter 11)5. Normal distribution (first half of Chapter 5)6. χ2 test – how well does the data fit the distribution model? (Chapter 12)
Part 3 – Multiple measurements as one sample
1. Central limit theorem (not in text explicitly)2. Normal distribution (second half of Chapter 5)3. Propagation of error (Chapter 3)4. Rejection of data (Chapter 6)5. Merging two sets of data together (Chapter 7)
Part 4 - Dependent variablesPart 4 Dependent variables
1. Curve fitting (Chapter 8)2. Covariance and correlation (Chapter 9)
Basic Idea Given two random variables x and y, how to show they are independent of each other (or correlated in the opposite sense)?sense)?
0 8
1
Independent random variables x and yy
0 2
0.4
0.6
0.8
‐0.4
‐0.2
0
0.2
‐1 ‐0.8 ‐0.6 ‐0.4 ‐0.2 0 0.2 0.4 0.6 0.8 1x
‐1
‐0.8
‐0.6
0.4
Each dot is a pair of independent random numbers (x, y) in the range of [-1, 1]. Total of 100 dots.
Depending on the distribution the scattering may not center about the origin but it always
Basic Idea Depending on the distribution, the scattering may not center about the origin, but it always centers about the mean (x,y).
Consider the product of (x-x) and (y-y). It will be positive in quadrants I and III, and negative in quadrants II and IV If x and y are independent of each other the scatter
Independent random variables x and y
negative in quadrants II and IV. If x and y are independent of each other, the scatter randomly in all quadrants and the sum of their products should tend to 0.
y
(x,y) x
. variablesrandomt independen arey and x if 0 )yy()xx( ii
N
1i→−−∑
=
CovarianceCovariance
Covariance between two random variables x and y is defined as
)yy()xx(N1 ii
N
xy −−= ∑σN 1i=
If the random variables x and y are independent of each other, | | 0 If d l t d ill b f f 0 ( b|σxy|→ 0. If x and y are correlated, σxy will be far from 0 (can be positive or negative) but its magnitude will not exceed σxσy(Schwartz inequality).
ExampleI h t d 15 d d 15 d i d d tl
x y (x‐x) (y‐y) (x‐x)(y‐y)0 07253 0 964829 0 39924 0 397462 0 15868
I have generated 15 random x and 15 random y independently.
0.07253 0.964829 ‐0.39924 0.397462 ‐0.158680.711853 0.539325 0.240082 ‐0.02804 ‐0.006730.744859 0.077395 0.273088 ‐0.48997 ‐0.133810.118883 0.887473 ‐0.35289 0.320106 ‐0.112960.770617 0.527768 0.298846 ‐0.0396 ‐0.011830.08248 0.765561 ‐0.38929 0.198194 ‐0.077160.099049 0.992887 ‐0.37272 0.42552 ‐0.15860 833047 0 166098 0 361276 ‐0 40127 ‐0 144970.5
1
1.5
0.833047 0.166098 0.361276 0.40127 0.144970.322289 0.230947 ‐0.14948 ‐0.33642 0.0502890.864182 0.20435 0.392411 ‐0.36302 ‐0.142450.77726 0.415119 0.305489 ‐0.15225 ‐0.04651
0
0 0.2 0.4 0.6 0.8 1
0.423413 0.853539 ‐0.04836 0.286172 ‐0.013840.718309 0.671949 0.246538 0.104582 0.0257830.083665 0.946898 ‐0.38811 0.379531 ‐0.14730.454136 0.266361 ‐0.01763 ‐0.30101 0.005308
Sum 7.076572 8.5105 ‐1.07346
Mean 0.471771 0.567367 ‐0.07156←σxy
ExampleNow I generate y randomly about the value of x as x changes from 0 g y y gto 1.5 with an inclement of 0.1.
x y (x‐x) (y‐y) (x‐x)(y‐y)0 0.929658 ‐0.75 0.021211 ‐0.01591
0.1 0.17865 ‐0.65 ‐0.7298 0.4743680.2 ‐0.64521 ‐0.55 ‐1.55366 0.8545110.3 1.074947 ‐0.45 0.1665 ‐0.074920 4 0 455536 ‐0 35 ‐0 45291 0 1585192.5 0.4 0.455536 0.35 0.45291 0.1585190.5 1.031123 ‐0.25 0.122676 ‐0.030670.6 1.585773 ‐0.15 0.677326 ‐0.10160.7 0.032196 ‐0.05 ‐0.87625 0.043813
0 5
1
1.5
2
0.8 0.261895 0.05 ‐0.64655 ‐0.032330.9 0.3087 0.15 ‐0.59975 ‐0.089961 0.830238 0.25 ‐0.07821 ‐0.01955
1.1 1.807077 0.35 0.89863 0.314521‐1
‐0.5
0
0.5
0 0.5 1 1.5 2
1.2 1.543898 0.45 0.635451 0.2859531.3 2.193795 0.55 1.285348 0.7069411.4 0.932722 0.65 0.024275 0.0157791 5 2 014146 0 75 1 105699 0 8292741.5 2.014146 0.75 1.105699 0.829274
Sum 12 14.53515 3.318736
Mean 0.75 0.908447 0.207421←σxy
Schwarz Inequality
[ ] ≥−+−=
≤
∑ 2ii
yxxy
0 )yy(t)xx(N1 A(t)function Construct
Proof.
σσσ
( )2jj2 0)xx()yy(
)xx(10A ≥⎥⎤
⎢⎡ −−
−−⇒≥ ∑∑[ ]
[ ]−−
−=⇒
=−+−−⇒
=−+−−=
∑∑
∑∑
2ii
iii
iiimin
)()xx()yy(
t
0 )yy(t)xx()yy(
0 )yy(t)xx()yy(N2
dtdA :Adetermint To
N
[ ] ( )2
jj2
j2
i
2jj
2j
2i
2j
imin
)xx()yy()yy()xx(
)xx()yy(N1 )yy()xx(
N1
0)yy(
)xx(N
0A
⎟⎟⎞
⎜⎜⎛ −−
≥⎥⎤
⎢⎡ −−
⇒
−−≥−−⇒
≥⎥⎥⎦⎢
⎢⎣ −
⇒≥
∑∑∑
∑∑∑
∑∑
⎥⎤
⎢⎡
⎟⎟⎞
⎜⎜⎛ −−
−−−−
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛
−
−−−−−=∴
−
∑∑∑
∑∑∑
∑
jj2
2
2j
jjiimin
2i
)xx()yy()yy()xx(2)xx(
)yy()xx()yy(
)yy()xx(N1 A
)yy(
jj2
j2
i
2
N)xx()yy(
N
)yy(N
)xx(
NNN
≥⇒
−−≥
⎥⎥⎦
⎤
⎢⎢⎣
⎡ −−⇒
⎟⎟⎠
⎜⎜⎝
≥⎥⎥⎦⎢
⎢⎣
⇒
∑∑∑
( )⎥⎤
⎢⎡
⎟⎞
⎜⎛ −−−−
⎥⎥⎥⎥⎥⎥
⎦⎢⎢⎢⎢⎢⎢
⎣⎟⎟⎠
⎞⎜⎜⎝
⎛
−
−−−+
⎟⎟⎠
⎜⎜⎝ −
=
∑∑∑∑
∑∑∑
∑∑∑
2
jj22
jj2
2
2j
jj2i
2j
iii
)xx()yy()(
)xx()yy(2)(1
)yy()xx()yy(
)yy(
)yy()yy()xx(2)xx(
N1
xyyx σσσ ≥⇒
( )
( ) ( )
( ) ⎤⎡
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−−−
+−
−−−−=
⎥⎥
⎦⎢⎢
⎣⎟⎟⎠
⎜⎜⎝ −
−+−
−−=
∑∑
∑∑
∑∑
∑∑∑∑
∑∑
2
2j
2jj
2j
2jj2
i
2j
jj2i2
j
jj2i
)()(
)yy()xx()yy(
)yy()xx()yy(
2)xx(N1
)yy()()yy(
)yy()yy(
)()yy(2)xx(
N1
( )⎥⎥⎦
⎤
⎢⎢⎣
⎡
−−−
−−=∑
∑∑ 2j
jj2i )yy(
)xx()yy()xx(
N1
Covariance in Another Form
−−= ∑1
)yy()xx(N1 iixyσ
+−−=
+−−=
∑∑∑∑
∑
yx1x1yy1xyx1
)y xyxyxyx(N1
iiii
iiii
+−−= ∑
∑∑∑∑
y xx yy xyxN1
yNN
yyN
yN
ii
iiii
=
−= ∑yx)xy(
y xyxN1 ii
>><<><−=∴
−=
yx-xyor yx)xy(
yx)xy(
xyσ yyy)y(xy
Error propagation
For simplicity, we assume q is a function of two variables x and y. We have learned two equations of error propagation:
Sum)ial(Different qq :equationFirst yxq σσσ ∂+
∂=
Sum) re(Quadratui yq
xq :equation Second
)(yx
q
2y
22
x
22
q
yxq
σσσ ⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+⎟⎠⎞
⎜⎝⎛∂∂
=
∂∂
yx yq ⎟⎠
⎜⎝ ∂⎠⎝ ∂
Is there any relation between these two equations?
Error propagation and Covariance Iiii -(1)--- )y-y(
yq )x-x(
xq )y,xq( q
∂∂
+∂∂
+≈
ii
ii
)y-y(1q)x-x(1q)y,xq(1
)y-y(yq )x-x(
xq )y,xq(
N1 q
yx
∂+
∂+=
⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+=∴
∂∂
∑∑∑∑∑
∑
ii
)yy(q)xx(q)yxq(q
:(1) intoback (2) Substitute-(2)--- )y,xq(
)yy(Ny
)xx(Nx
)y,xq(N
∂+
∂+≈
=∂
+∂
+ ∑∑∑∑∑
ii
iii
q,in deviation standardtheasqin y uncertaintheconsider t weIf
)y-y(yq )x-x(
xq q
)y-y(y
)x-x(x
)y,xq( q
∂∂
+∂∂
+=
∂+
∂+≈
2
ii2
q
2i
2q
q)y-y(yq )x-x(
xq q
N1
)qq(N1
qqy
σ
σ
⎟⎟⎠
⎞⎜⎜⎝
⎛−
∂∂
+∂∂
+=∴
−=
∑
∑
22
22
2
ii
)()(1qq2)(1q)(1q
)y-y(yq )x-x(
xq
N1
yxN
⎟⎟⎞
⎜⎜⎛ ∂⎟⎞
⎜⎛ ∂+⎟⎟
⎞⎜⎜⎛ ∂
+⎟⎞
⎜⎛ ∂
⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
=
⎠⎝ ∂∂
∑∑∑
∑
xy2
y
22
x
2
iiii
yq
xq2
yq
xq
)y-y()x-x(Ny
qxq2 )y-y(
Nyq)x-x(
Nxq
σσσ ⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
⎟⎠⎞
⎜⎝⎛∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+⎟⎠⎞
⎜⎝⎛∂∂
=
⎟⎟⎠
⎜⎜⎝ ∂⎟⎠
⎜⎝ ∂
+⎟⎟⎠
⎜⎜⎝ ∂
+⎟⎠
⎜⎝ ∂
= ∑∑∑
Error propagation and Covariance II22 ⎞⎛⎞⎛⎞⎛⎞⎛
xy2
y
22
x
22
q yq
xq2
yq
xq σσσσ ⎟⎟
⎠
⎞⎜⎜⎝
⎛∂∂
⎟⎠⎞
⎜⎝⎛∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+⎟⎠⎞
⎜⎝⎛∂∂
= This is a general case.
Case I. x and y are independent variables. σxy = 0
Sum)e(Quadratur qq 2y
22
x
22
q σσσ ⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+⎟⎠⎞
⎜⎝⎛∂∂
= )(Qyx yxq ⎟⎠
⎜⎝ ∂
⎟⎠
⎜⎝ ∂
Case 2. x and y are dependent variables. σxy ≠ 0. When x and y are highly correlated, | σxy | is maximum and equal to σ σ (Schwarz inequality)equal to σxσy (Schwarz inequality).
yq
xq2
yq
xq yx
2y
22
x
22
q σσσσσ ⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
⎟⎠⎞
⎜⎝⎛∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+⎟⎠⎞
⎜⎝⎛∂∂
=
yq
xq
2
yx σσ ⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
=
Sum) ial(Different yq
xq yxq σσσ
∂∂
+∂∂
=∴
Error propagation and Covariance III
In Summary,
Sum)e(Quadraturqq 22
22
2 σσσ ⎟⎟⎞
⎜⎜⎛ ∂
+⎟⎞
⎜⎛ ∂= Sum)e(Quadratur
yx yxq σσσ ⎟⎟
⎠⎜⎜⎝ ∂
+⎟⎠
⎜⎝ ∂
qq2 ∂∂corresponds to the case when x and y are absolute independent of each other.
Sum)ial(Different yq
xq yx
2q σσσ
∂∂
+∂∂
=
corresponds to the case when x and y are completely correlated.
Anything is between these two extreme cases.
Coefficient of Linear CorrelationIn linear regression, we expect x and y follow a linear relationship y=A+Bx. In other words, for our least square fit to work, we expect a high
l ti b t d
Coefficient of Linear Correlation
correlation between x and y.
To measure the correlation between x and y, we define the coefficient of Linear Correlation (r) as the covariance normalized by σxσy.ea Co e a o ( ) as e co a a ce o a ed by σxσy
)yy()xx(r
ii
N
1ixy−−
==∑=
σ
)yy()xx(r
2i
N
1i
2i
N
1i
yx −−
==
∑∑==
σσ
Coefficient of Linear CorrelationSchwarz Inequality: σxy ≤ σxσy
1 r 1- r xy ≤≤⇒=∴σσ
σ
yxσσIf x and y are independent (uncorrelated) r=0 and we should not try to do linear regression in this case.
y y y
x x x
Positive Correlation (r > 0) Negative Correlation (r < 0) No Correlation (r = 0)
If All Points Falls Exactly on a Straight Line
)xB(x)yy(xBAy BxAy
ii
ii
−=−∴+=⇒+=
If all data points falls exactly on a straight line y=A+Bx,
)yy()xx(
)yy()xx( r
2i
N2
i
N
ii
N
1i
yx
xy
−−
−−==
∑∑
∑=
σσσ
)(B)(
)xx(B
22N
2N
2i
N
1i
1i1i
−=
∑∑
∑
∑∑
=
==
)xx(B
)xx(B)xx(
N
2i
N
1i
2i
2
1i
2i
1i
−=
−−
∑
∑∑
=
==
1
)xx(B 2i
N
1i
=
−∑=
Note: If you have only two data points r=1 Hence correlation has noNote: If you have only two data points, r=1. Hence correlation has no meaning if there are only two data points.
ExampleConsider the following data points and do a linear regression:
x y x2 xy [y‐(A+Bx)]2
‐0.04275 0.046483 0.001827 ‐0.00199 0.005989175
6 1 data points
g p g
0.021185 0.003933 0.000449 8.33E‐05 0.000671560.024486 ‐0.04226 0.0006 ‐0.00103 0.00566167‐0.03811 0.038747 0.001453 ‐0.00148 0.004257470.027062 0.002777 0.000732 7.51E‐05 0.001066320 04175 0 026556 0 001743 0 00111 0 00319422
3
4
5
‐0.04175 0.026556 0.001743 ‐0.00111 0.00319422‐0.0401 0.049289 0.001608 ‐0.00198 0.006033470.033305 ‐0.03339 0.001109 ‐0.00111 0.00558818‐0.01777 ‐0.02691 0.000316 0.000478 0.000389380.036418 ‐0.02956 0.001326 ‐0.00108 0.00545939
0
1
2
15 data points
0.027726 ‐0.00849 0.000769 ‐0.00024 0.00198476‐0.00766 0.035354 5.87E‐05 ‐0.00027 0.001083520.021831 0.017195 0.000477 0.000375 0.00017598‐0.04163 0.04469 0.001733 ‐0.00186 0.00555598‐0.00459 ‐0.02336 2.1E‐05 0.000107 0.00082487
‐1
0
‐1 0 1 2 3 4 5 6 7
5.4321 5.1762 29.50771 28.11764 1.973E‐05Sum 5.389757 5.27725 29.52193 28.10662 0.04795566
Δ= 443.3014 A= 0.009715 σy= 0.058527
B= 0.950285 σA= 0.006453‐0.02
0
0.02
0.04
0.06
‐0.05 0 0.05
σB= 0.011119
‐0.06
‐0.04
ExampleNow calculate the coefficient of linear correlation:
5
6 1 data points
Now calculate the coefficient of linear correlation:
x y (x‐x)2 (y‐y)2 (x‐x)(y‐y)
‐0.04275 0.046483 0.144101 0.080284 0.10755979
0 021185 0 003933 0 099651 0 106208 0 10287697
3
4
5 0.021185 0.003933 0.099651 0.106208 0.10287697
0.024486 ‐0.04226 0.097578 0.13845 0.11623079
‐0.03811 0.038747 0.140604 0.084728 0.10914701
0.027062 0.002777 0.095975 0.106962 0.10131992
‐0.04175 0.026556 0.143347 0.091974 0.11482235
0
1
2
0.04 75 0.0 6556 0. 43347 0.09 974 0. 48 35
‐0.0401 0.049289 0.142095 0.078702 0.10575072
0.033305 ‐0.03339 0.092146 0.131927 0.11025681
‐0.01777 ‐0.02691 0.125763 0.127259 0.12650871
0.036418 ‐0.02956 0.090265 0.129163 0.10797667
15 data points
‐1
0
‐1 0 1 2 3 4 5 6 70.027726 ‐0.00849 0.095564 0.114458 0.10458502
‐0.00766 0.035354 0.118693 0.086715 0.10145184
0.021831 0.017195 0.099243 0.097739 0.09848852
‐0.04163 0.04469 0.143257 0.081304 0.10792297
0 00459 0 02336 0 116586 0 124745 0 12059609
‐0.02
0
0.02
0.04
0.06
‐0.05 0 0.05
‐0.00459 ‐0.02336 0.116586 0.124745 0.12059609
5.4321 5.1762 25.96147 23.48732 24.6934285
Mean 0.33686 0.329828σx: σy: σxy:
1.31592 1.251697 1.64555767
r= 0 999043
‐0.06
‐0.04r= 0.999043
ExampleRecalculate the coefficient of linear correlation without the last data point:
5
6 1 data points
Recalculate the coefficient of linear correlation without the last data point:
x y (x‐x)2 (y‐y)2 (x‐x)(y‐y)
‐0.04275 0.046483 0.001594 0.00158 ‐0.0015869
0 021185 0 003933 0 000576 7 87E 06 6 732E 05
3
4
5 0.021185 0.003933 0.000576 7.87E‐06 ‐6.732E‐05
0.024486 ‐0.04226 0.000746 0.002401 ‐0.0013379
‐0.03811 0.038747 0.001246 0.001025 ‐0.0011297
0.027062 0.002777 0.000893 1.57E‐05 ‐0.0001183
0 04175 0 026556 0 001516 0 000393 0 0007716
0
1
2
‐0.04175 0.026556 0.001516 0.000393 ‐0.0007716
‐0.0401 0.049289 0.001389 0.001811 ‐0.0015861
0.033305 ‐0.03339 0.001305 0.00161 ‐0.0014496
‐0.01777 ‐0.02691 0.000224 0.001132 0.00050299
0 036418 0 02956 0 00154 0 001318 0 0014244
15 data points
‐1
0
‐1 0 1 2 3 4 5 6 7
0.036418 ‐0.02956 0.00154 0.001318 ‐0.0014244
0.027726 ‐0.00849 0.000933 0.000232 ‐0.0004651
‐0.00766 0.035354 2.34E‐05 0.000819 ‐0.0001385
0.021831 0.017195 0.000608 0.000109 0.0002578
0 04163 0 04469 0 001506 0 00144 0 0014731
‐0.02
0
0.02
0.04
0.06
‐0.05 0 0.05
‐0.04163 0.04469 0.001506 0.00144 ‐0.0014731
‐0.00459 ‐0.02336 3.12E‐06 0.000906 5.3169E‐05
Mean ‐0.00282 0.006737σx: σy: σxy:
0 030661 0 03141 0 0007156
‐0.06
‐0.040.030661 0.03141 ‐0.0007156
r= ‐0.74309
Quantitative Significance of r
Provided x and y are independent, we do not expect r to be 0. Quantitatively, we can calculate the probability of getting |r| greater than a
Q g
y p y g g | | gcertain value r0 (given x and y are independent variables): ProbN(|r|> r0)
You can find a table of this in the Appendix C .
Notes.
1. ProbN(|r|> 0) = 100% and ProbN(|r|> 1) = 0%
1. ProbN(|r|> r0) decreases as r0 increases, since we expect small r for independent x and y.
2. . ProbN(|r|> r0) falls as N increases, since larger N will make the measurements more “truthful”.
ExampleNow calculate the coefficient of linear correlation:
5
6 1 data points
Now calculate the coefficient of linear correlation:
x y (x‐x)2 (y‐y)2 (x‐x)(y‐y)
‐0.04275 0.046483 0.144101 0.080284 0.10755979
0 021185 0 003933 0 099651 0 106208 0 10287697
3
4
5 0.021185 0.003933 0.099651 0.106208 0.10287697
0.024486 ‐0.04226 0.097578 0.13845 0.11623079
‐0.03811 0.038747 0.140604 0.084728 0.10914701
0.027062 0.002777 0.095975 0.106962 0.10131992
‐0.04175 0.026556 0.143347 0.091974 0.11482235
0
1
2
0.04 75 0.0 6556 0. 43347 0.09 974 0. 48 35
‐0.0401 0.049289 0.142095 0.078702 0.10575072
0.033305 ‐0.03339 0.092146 0.131927 0.11025681
‐0.01777 ‐0.02691 0.125763 0.127259 0.12650871
0.036418 ‐0.02956 0.090265 0.129163 0.10797667
15 data points
‐1
0
‐1 0 1 2 3 4 5 6 70.027726 ‐0.00849 0.095564 0.114458 0.10458502
‐0.00766 0.035354 0.118693 0.086715 0.10145184
0.021831 0.017195 0.099243 0.097739 0.09848852
‐0.04163 0.04469 0.143257 0.081304 0.10792297
0 00459 0 02336 0 116586 0 124745 0 12059609
‐0.02
0
0.02
0.04
0.06
‐0.05 0 0.05
‐0.00459 ‐0.02336 0.116586 0.124745 0.12059609
5.4321 5.1762 25.96147 23.48732 24.6934285
Mean 0.33686 0.329828σx: σy: σxy:
1.31592 1.251697 1.64555767
r= 0 999043
‐0.06
‐0.04r= 0.999043
Prob16(|r|> 0.999)=0%
ExampleRecalculate the coefficient of linear correlation without the last data point:
5
6 1 data points
Recalculate the coefficient of linear correlation without the last data point:
x y (x‐x)2 (y‐y)2 (x‐x)(y‐y)
‐0.04275 0.046483 0.001594 0.00158 ‐0.0015869
0 021185 0 003933 0 000576 7 87E 06 6 732E 05
3
4
5 0.021185 0.003933 0.000576 7.87E‐06 ‐6.732E‐05
0.024486 ‐0.04226 0.000746 0.002401 ‐0.0013379
‐0.03811 0.038747 0.001246 0.001025 ‐0.0011297
0.027062 0.002777 0.000893 1.57E‐05 ‐0.0001183
0 04175 0 026556 0 001516 0 000393 0 0007716
0
1
2
‐0.04175 0.026556 0.001516 0.000393 ‐0.0007716
‐0.0401 0.049289 0.001389 0.001811 ‐0.0015861
0.033305 ‐0.03339 0.001305 0.00161 ‐0.0014496
‐0.01777 ‐0.02691 0.000224 0.001132 0.00050299
0 036418 0 02956 0 00154 0 001318 0 0014244
15 data points
‐1
0
‐1 0 1 2 3 4 5 6 7
0.036418 ‐0.02956 0.00154 0.001318 ‐0.0014244
0.027726 ‐0.00849 0.000933 0.000232 ‐0.0004651
‐0.00766 0.035354 2.34E‐05 0.000819 ‐0.0001385
0.021831 0.017195 0.000608 0.000109 0.0002578
0 04163 0 04469 0 001506 0 00144 0 0014731
‐0.02
0
0.02
0.04
0.06
‐0.05 0 0.05
‐0.04163 0.04469 0.001506 0.00144 ‐0.0014731
‐0.00459 ‐0.02336 3.12E‐06 0.000906 5.3169E‐05
Mean ‐0.00282 0.006737σx: σy: σxy:
0 030661 0 03141 0 0007156
‐0.06
‐0.040.030661 0.03141 ‐0.0007156
r= ‐0.74309
Prob15(|r|> 0.75) ~ 40%