Statistics & Probability Letters 65 (2003) 103–109

Limit points of independent copies of sample maximaGeorge Mathew

Department of Mathematics, Southwest Missouri State University, 901 South National Avenue,Spring�eld, MO 65804-0094, USA

Received February 2003; received in revised form March 2003

Abstract

Let {X (i)n ; n¿ 1}, i=1; 2 be two independent copies of a sequence of independent and identically distributed

random variables. Let Min = max16j6n X

(i)j . For suitable normalizing constants an and bn, we investigate the

set of almost sure limit points of the vector sequence

Zn =(M 1

n − bnan

;M 2

n − bnan

); n¿ 1

and further investigate the limit set for vector subsequences Znk .c© 2003 Elsevier B.V. All rights reserved.

Keywords: Sample maxima; Set of limit points; Independent copies

1. Introduction

Let {X (i)n ; n¿ 1}, i=1; 2 be two independent copies of a sequence of independent and identically

distributed random variables. Let Min = max16j6n X

(i)j . Hebbar (1980) has investigated the set of

almost sure limit points of vector sequences with independent components based on the maxima ofa stationary Gaussian sequence. He showed that the limit set was a triangular region thus providingfor the maxima another example of the phenomenon that when considering limit points of vectorsequences with independent components, the resulting limit set is not the product set of the set oflimit points for each component. For sums, the asymptotic behavior is discussed in a quite generalsetting in LePage (1973).Nayak (1985) addresses for general distributions the same problem as Hebbar but in the i.i.d.

setting and takes as his starting point the one-dimensional result the law of the iterated logarithmsprovided by De Haan and Hordijk (1972). Also, another look at the results of De Haan and Hordijk

E-mail address: gem057f@smsu.edu (G. Mathew).

0167-7152/$ - see front matter c© 2003 Elsevier B.V. All rights reserved.doi:10.1016/j.spl.2003.07.006

104 G. Mathew / Statistics & Probability Letters 65 (2003) 103–109

was taken by HAusler (1985). His point of view was to examine the rate of maximal growth of themaxima when sampled along a subsequence nk . More precisely, he elucidated the eBect of takingsubsequences on the magnitude of Climk→∞M ∗

nk where M ∗n is the maximum suitably normalized.

In this paper, we consider a problem similar to the one considered by HAusler but directed towardsvector subsequences. We show that if the subsequence {nk ; k¿ 1} satisEes limk→∞ (nk−1)=nk ¿ 0then the entire set of almost sure limit points of

{Znk ; k¿ 1}={(

M 1nk − bnkank

);(M 2

nk − bnkank

); k¿ 1

}(1.1)

coincides with the limit set for the whole sequence. However, for thin subsequences, that is, forsubsequences satisfying limk→∞ (nk−1)=nk =0, we show that the upper boundary of the limit set for{Znk ; k¿ 1} is contained in the interior of the limit set for {Z n; n¿ 1}.We present the results in the next section.

2. Results

Let F denote the common distribution function of the sequence of random variables {X (i)n ; n¿ 1}.

Set

f(x) =1− F(x)F ′(x)

; (2.1)

where it is assumed that F ′(x) exists and is positive for all suFciently large x, and further set

g(x) = f(x) log log[1− F(x)]−1: (2.2)

From the work of De Haan and Hordijk (1972), the asymptotic behavior of g(x) as x → ∞ playsan important role in determining sample path properties of max16j6n Xj. Of considerable use is thefollowing identity giving the rate of decrease of the tail of a distribution belonging to the class ofdistributions considered in this paper. Let

bn = F−1(1− 1

n

)and an = f(bn) log log x = g(bn); (2.3)

where f is deEned in (2.1). Suppose in (2.2) that g′ exists and

limt→∞ g′(t) = 0:

Then, by De Haan and Hordijk (1972, p. 1192), we have

1− F(bn + xan) =(log n)rn

n; (2.4)

where limn→∞ rn(x) = −x. In order to prove our results we need the following result (Pathak andQualls, 1973, Lemma 2.5).

Lemma 2.1 (Pathak and Qualls). Let {In; n¿ 1} be a sequence of indicator functions such that∑∞n=1 EIn =∞. If for some A, 06A¡∞, Var(∑n

1 Ik)6A(∑n

1 EIk)2 for all n su7ciently large,

then P(∑∞

1 Ik =∞)¿ 1=(1 + A).

G. Mathew / Statistics & Probability Letters 65 (2003) 103–109 105

To reduce the notational burden, we let Un = (M 1n − bn)=an and Vn = (M 2

n − bn)=an, where an andbn are deEned in (2.3).

Theorem 2.2. Suppose F is twice di8erentiable with F ′(x) positive for all real x, and let r ¡ 1. Iflimt→∞ g′(t) = 0, then we have∑

k

(log nk)−r ¡∞ implies P{Unk ¿x; Vnk ¿y; i:o:}= 0 (2.5)

for x and y such that x + y¿r, and∑k

(log nk)−r =∞ implies P{Unk ¿x; Vnk ¿y; i:o:}= 1 (2.6)

for x and y such that x + y¡r.

Proof. Observe that for r ¡x + y we have

P{Unk ¿x; Vnk ¿y} = [1− Fnk (bnk + xank )][1− Fnk (bnk + yank )]

= (1 + o(1))(log nk)rnk (x)+rnk (y)

6 (log nk)−r for all k suFciently large; (2.7)

where we have used relation (2.4). Thus by (2.7), it is clear that (2.5) holds by the Borel–CantelliLemma.Suppose the condition in (2.6) holds where r ¡ 1 and x+ y¡r, for some x; y. Clearly, to show

P{Unk ¿x; Vnk ¿y; i:o:}=1, it suFces to establish P{Unk ¿x; Vnk ¿y; i:o:}=1 for any subsequence{tk ; k¿ 1}. In particular, consider the subsequence deEned by

ntk = [exp(kr−1)]: (2.8)

Let Ik = I{Untk¿ x; Vntk

¿y}. Then as in the Erst part of the proof, we have for all suFcientlylarge k

EIk = P{Untk¿ x; Vntk

¿y}¿ (log ntk )

−�r for some 0¡�¡ 1 such that x + y¡�r

¿ k−�; (2.9)

where the last inequality follows from (2.8). Therefore by (2.9), it follows that∑k

EIk =∞: (2.10)

Next observe that as noted by De Haan and Hordijk (1972, p. 1192), bn + xan is ultimately non-decreasing in n so that it follows that for j¡k suFciently large, and with n′j = ntj

P{Un′j ¿ x; Vn′j ¿y;Un′k ¿ x; Vn′k ¿y}= p1p2; (2.11)

106 G. Mathew / Statistics & Probability Letters 65 (2003) 103–109

where

p1 =P{M 1n′j¿ bn′k + xan′k}

+P{(bn′j + xan′j ¡M 1n′j6 bn′k + xan′k ) ∩ (M 1

n′j ;n′k¿bn′k + xan′k )} (2.12)

with Mn′j ;n′k= maxn′j¡i6n′k Xi. Also, p2 is similarly deEned with M 2

n in place of M1n and y in place

of x.Letting un = un(x) = bn + xan, we can write (2.12) as

p1 = 1− Fn′j(un′j)− Fn′k (un′k ) + Fn′j(un′j)Fn′k−n′j(un′k ): (2.13)

Similarly, we have an expression for p2 with un(y) in place of un(x).Also, in the same way one Ends that

P{Un′j ¿ x; Vn′j ¿y}P{Un′k ¿ x; Vn′k ¿y}= q1q2; (2.14)

where

q1 = [1− Fn′j(un′j)][1− Fn′k (un′k )] (2.15)

and where again q2 is likewise deEned with un(y) in place of un(x).Let di = pi − qi, i = 1; 2 and observe by deEnitions (2.13) and (2.15)

d1 = Fn′j(un′j)[Fn′k−n′j(un′k )− Fn′k (un′k )] (2.16)

with a similar expression for d2.Then by (2.11) and (2.14), we have

Cov(Ij; Ik) = p1p2 − q1q2 = q1d2 + q2d1 + d1d2: (2.17)

Further, it can be checked that for suFciently large j; k

q16 (log n′j)−x=2(log n′k)

−x=2 (2.18)

and

d16 (n′j=n′k)(log n

′k)

−x=2 (2.19)

with similar expressions for q2 and d2. Therefore in view of the bounds in (2.18) and (2.19), itfollows that Cov(Ij; Ik) is at most 3(n′j=n′k).Now, observe that

Var

(m∑1

Ij

)=

m∑j=1

Var(Ij) + 2m−1∑j=1

Cov(Ij; Ij+1) + 2m−2∑j=1

m∑k=j+2

Cov(Ij; Ik)

6 5m∑1

EIj + 6m−2∑j=1

m∑k=j+2

(n′j=n′k): (2.20)

The above inequality holds since for indicator random variables Ij, Var(Ij)6EIj,∑m−1

j=1 Cov(Ij; Ij+1)6 2

∑mj=1 EIj, and by the above stated upper bound for Cov(Ij; Ik). Let

’k = exp(kr−1): (2.21)

G. Mathew / Statistics & Probability Letters 65 (2003) 103–109 107

Let sm = [m%], where 0¡%¡ 1− �. Then,m−2∑j=1

m∑k=j+2

(n′j=n′k)6

m−2∑j=1

m−1∑k=j+1

(’j=’k)

6sm∑j=1

m∑k=j+1

(’j=’k) +m−1∑

j=sm+1

m∑k=j+1

(’j=’k)

6 smm∑

k=1

(’1=’k+1) + mm∑

k=1

(’sm=’sm+k); (2.22)

where the last inequality follows since

max{(’j=’k): k − j = t and j¿ j0}= (’j0=’j0+t):

Now as∑∞

1 ’1=’k+1¡∞ and ’sm=’sm+k ¡ exp{−r−1sr−1−1m k}, the Enal expression in (2.22) isbounded above by

csm + mm∑

k=1

exp{−r−1sr−1−1m k}6 c′

m∑1

EIk ; (2.23)

where c=∑m

k=1 (’1=’k+1) and c′ a suitable constant. The last inequality follows from the deEnitionof sm and (2.9).Therefore by (2.10), (2.22), and (2.23) we have that for any '¿ 0 and m suFciently large

Var

(m∑1

Ij

)6 '

(m∑1

EIj

)2:

Therefore by Lemma 2.1, it follows that P{Untk¿ x; Vntk

¿y; i:o:}= 1, which completes the proofof Theorem 2.2.

Now we present the result for subsequences.

Theorem 2.3. If the subsequence {nk ; k¿ 1} satis�eslimk→∞

(nk−1=nk)¿ 0; (2.24)

then the set of almost sure limit points of {(Unk ; Vnk ); k¿ 1},lim set (Unk ; Vnk ) = {(x; y): 06 x; y and x + y6 1}: (2.25)

On the other hand if the subsequence {nk ; k¿ 1} satis�es the conditionslimk→∞

(nk−1=nk) = 0 (2.26)

∑k

(log nk)a ¡∞ if a¡− r and∑k

(log nk)a =∞ if a¿− r (2.27)

108 G. Mathew / Statistics & Probability Letters 65 (2003) 103–109

for some r ¡ 1, then the set of points {(x; r − x): 06 x6 r} forms the upper boundary of thelimit set; that is, for any x′; y′ such that x′ + y′¿r we have

P{Unk ¿x′; Vnk ¿y′; i:o:}= 0; (2.28)

and for any x′; y′ such that x′ + y′¡r

P{Unk ¿x′; Vnk ¿y′; i:o:}= 1: (2.29)

Proof. Assume condition (2.24) holds and suppose for some 0¡x; y that r = x + y¡ 1. To showthat (x; y) is a limit point of the sequence (Unk ; Vnk ) it suFces to establish for some subsequence(n′k ; k¿ 1)

P{Un′k ¿ x′; Vn′k ¿y′; i:o:}= 0 for any x′; y′ with x′ + y′¿x + y (2.30)

and

P{Un′k ¿ x′; Vn′k ¿y′; i:o:}= 1 for any x′; y′ with x′ + y′¡x + y: (2.31)

We choose as our subsequence the subsequence deEned in (2.8). Observe that under assumption(2.24) it follows that for all k suFciently large, ntk ¡ntk+1 so that for any r′¿r∑

k

(log ntk+1)−r′ 6

∑k

(log’k)−r′

=∑k

k−(r′=r)

¡∞:

Therefore, by Theorem 2.2 we have (2.30) holds. Similarly we End for any r′¡r that∑k

(log ntk )−r′ ¿

∑k

(log’k)−r′

=∑k

k−(r′=r)

=∞:

Hence, an appeal to Theorem 2.2 yields (2.31).Thus we have shown that

lim set (Unk ; Vnk ) ⊃ {(x; y): 0¡x; y and x + y¡ 1}:Therefore, since the limit set for the entire sequence is {(x; y): 06 x; y and x+ y6 1} as given inTheorem 3.2 of Nayak (1985), it follows that (2.25) holds.Finally, since condition (2.27) implying (2.28) and (2.29) is just a restatement of Theorem 2.2,

the proof of Theorem 2.3 is complete.

Remark. Let {X (i)n ; n¿ 1}, i=1; 2; : : : ; ‘ be ‘ independent copies of a sequence of independent and

identically distributed random variables, and for ‘¿ 2, let {Z nk ; k¿ 1} be deEned similarly as in(1.1). Then, an extension of Theorem 2.3 to the case of ‘¿ 2 can be obtained just by making the

G. Mathew / Statistics & Probability Letters 65 (2003) 103–109 109

following two changes: the limit set in (2.25) becomes {(x1; x2; : : : ; x‘): 06 xi and∑‘

1 xi6 1}; andif (2.27) holds, then the upper boundary becomes {(x1; x2; : : : ; x‘): 06 xi and

∑‘1 xi = r}.

Acknowledgements

The author is grateful to Dr. W.P. McCormick for valuable comments and suggestions.

References

De Haan, L., Hordijk, A., 1972. The rate of growth of sample maxima. Ann. Math. Statist. 43, 1185–1196.Hebbar, H.V., 1980. Almost sure limit points of maxima of stationary Gaussian sequences. Ann. Probab. 8, 393–399.HAusler, J., 1985. The growth rate of partial maxima on subsequences. Statist. Probab. Lett. 3, 245–249.LePage, R.D., 1973. Log log law for Gaussian processes. Z. Wahrsch. Verw. Gebiete 25, 103–108.Nayak, S.S., 1985. Almost sure limit points of independent copies of sample maxima. Stochastic Process. Appl. 20,353–360.

Pathak, P.K., Qualls, C., 1973. A law of iterated logarithm for stationary Gaussian processes. Trans. Amer. Math. Soc.181, 185–193.