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Convergence rate of extremes from Maxwell sample

Abstract

For the partial maximum from a sequence of independent and identically distributed random variables with Maxwell distribution, we establish the uniform convergence rate of its distribution to the extreme value distribution.

MSC:62E20, 60E05, 60F15, 60G15.

1 Introduction

One interesting problem in extreme value theory is to consider the convergence rate of some extremes. For the uniform convergence rate of extremes under the second-order regular variation conditions, see Falk [1], Balkema and de Haan [2], de Haan and Resnick [3] and Cheng and Jiang [4]. For the extreme value distributions and their associated uniform convergence rates for given distributions, see Hall and Wellner [5], Hall [6], Peng et al. [7], Lin and Peng [8] and Lin et al. [9].

In this note, we discuss the uniform convergence rate of extremes from a sequence of independent and identically distributed (iid) random variables with Maxwell distribution (MD). The probability density function of MD is given by

f(x)= 2 π x 2 σ 3 exp ( x 2 2 σ 2 ) ,x>0.
(1.1)

The MD and the convergence rate of extremes from Maxwell sample have been widely used in the field of physics. We establish the uniform convergence rate of its distribution to the extreme value distribution and give an improved proof for the pointwise convergence rate of MD.

Throughout this paper, let ( ξ n ,n1) be a sequence of iid random variables with common distribution F(x)= 0 x f(t)dt with a probability density function f(x) given by (1.1), and let M n = max 1 k n ξ k be the partial maximum. Liu and Fu [10] proved that

lim n P ( α n 1 ( M n β n ) x ) =exp ( exp ( x ) ) :=Λ(x)

with the normalizing constants α n and β n given by

α n = σ ( 2 log n ) 1 2 , β n = ( 2 σ 2 log n ) 1 2 + σ log ( 2 log n ) + σ log 2 π 2 ( 2 log n ) 1 2 .
(1.2)

By arguments similar to those of Hall [6], Peng et al. [7] and Lin et al. [9], the appropriate normalizing constants a n and b n can be given by the following equations:

a n = σ 2 b n 1
(1.3)

and

π 2 b n σ exp ( b n 2 2 σ 2 ) =n.
(1.4)

By arguments similar to those of Example 2 of Resnick [11], we have

b n = ( 2 σ 2 log n ) 1 2 + σ log ( 2 log n ) + σ log 2 π 2 ( 2 log n ) 1 2 +o ( ( log n ) 1 / 2 ) .

Hence

α n / a n 1,( β n b n )/ a n 0,

implying

lim n P( M n a n x+ b n )= lim n F n ( a n x+ b n )=Λ(x),

cf. Leadbetter et al. [12] or Resnick [11].

This paper is organized as follows. Section 2 gives some auxiliary results. In Section 3, we present the main result. Related proofs are given in Section 4.

2 Auxiliary results

To establish the uniform convergence of F n ( a n x+ b n ) to its extreme value distribution Λ(x), we need some auxiliary results. The first result is the decomposition of F(x), which is the following result.

Lemma 1 Let F(x) be the Maxwell distribution function. Then, for x>0, we have

1F(x)= 2 π x σ ( 1 + σ 2 x 2 ) exp ( x 2 2 σ 2 ) r(x)
(2.1)

with

0<r(x)= 2 π x σ y 2 exp ( y 2 2 σ 2 ) dy< 2 π σ 3 x 3 exp ( x 2 2 σ 2 ) .
(2.2)

For simplicity, throughout this paper, let C be a generic positive constant whose value may change from line to line, and let C i , C i j (iN, jN) be absolute positive constants.

For the normalizing constants a n , b n defined by (1.3) and (1.4), respectively, let

a n = a n r n , b n = b n + a n δ n ,
(2.3)

where r n 1, δ n 0, n. So, a n / a n 1, ( b n b n )/ a n 0, implying F n ( a n x+ b n )Λ(x). For large n, we have the following result.

Lemma 2 Let a n , b n be defined by (2.3). For fixed xR and sufficiently large n, we have

F n ( a n x + b n ) Λ ( x ) = Λ ( x ) e x ( ( x 2 2 x 1 ) a n b n 1 + ( r n 1 ) x + δ n + O [ ( a n b n 1 ) 2 + ( r n 1 ) 2 + δ n 2 ] ) .
(2.4)

Proof Note that b n σ ( 2 log n ) 1 2 , which means

a n b n 1 1 2 log n 0.

For large n, we have

2 π a n x + b n σ exp ( ( a n x + b n ) 2 2 σ 2 ) = 2 π b n σ ( 1 + a n b n 1 ( r n x + δ n ) ) exp ( b n 2 2 σ 2 ) × exp ( a n 2 ( r n 2 x 2 + δ n 2 + 2 r n δ n x ) 2 σ 2 ( r n 1 ) x x δ n ) = n 1 e x ( 1 ( x 2 2 x ) a n b n 1 ( r n 1 ) x δ n + O [ ( a n b n 1 ) 2 + ( r n 1 ) 2 + δ n 2 ] ) .

Since

σ 2 ( a n x + b n ) 2 = a n b n 1 2x ( a n b n 1 ) 2 +O ( ( a n b n 1 ) 3 ) ,

we have

σ 4 ( a n x + b n ) 4 = ( a n b n 1 ) 2 +O ( ( a n b n 1 ) 3 ) .

Similarly,

σ 5 ( a n x + b n ) 5 exp ( ( a n x + b n ) 2 2 σ 2 ) =O ( n 1 ( a n b n 1 ) 2 ) .

Hence,

1 F ( a n x + b n ) = n 1 e x ( 1 ( x 2 2 x 1 ) a n b n 1 ( r n 1 ) x δ n + O [ ( a n b n 1 ) 2 + ( r n 1 ) 2 + δ n 2 ] ) .
(2.5)

So,

F n ( a n x + b n ) Λ ( x ) = ( 1 n 1 e x ( 1 ( x 2 2 x 1 ) a n b n 1 ( r n 1 ) x δ n + O [ ( a n b n 1 ) 2 + ( r n 1 ) 2 + δ n 2 ] ) ) n Λ ( x ) = Λ ( x ) e x ( ( x 2 2 x 1 ) a n b n 1 + ( r n 1 ) x + δ n + O [ ( a n b n 1 ) 2 + ( r n 1 ) 2 + δ n 2 ] ) ,

which is the desired result. □

3 Main results

In this section we present the pointwise convergence rate and the uniform convergence rate of F n () to its extreme value distribution under different normalizing constants. The first result is the pointwise convergence of extremes under the normalizing constants given by (1.2).

Theorem 1 Let { ξ n ,n1} be a sequence of independent identically distributed random variables with common distribution MD. Then

F n ( α n x+ β n )Λ(x)Λ(x) e x ( log ( 2 log n ) ) 2 16 log n ,
(3.1)

for large n, where α n , β n are defined in (1.2).

Recently Liu and Fu [10] proved the result, we present an improved proof for the pointwise convergence rate in Section 4.

The following is the uniform convergence rate of extremes under the appropriate normalizing constants a n and b n given by (1.3) and (1.4), which shows that the optimal convergence rate is proportional to 1/logn.

Theorem 2 Let ( ξ n ,n1) be a sequence of independent identically distributed random variables with common distribution MD. For large n, there exist absolute constants 0< d 1 < d 2 such that

d 1 log n < sup x R | F n ( a n x + b n ) Λ ( x ) | < d 2 log n ,
(3.2)

where a n and b n are defined by (1.3) and (1.4), respectively.

4 Proofs

Proof of Theorem 1 Firstly, we derive the following asymptotic expansions of b n defined by (1.4)

b n = β n +o ( ( log n ) 1 2 )
(4.1)

and

b n = β n σ ( log ( 2 log n ) + log 2 π ) 2 16 2 ( log n ) 3 2 + ( σ log 4 log n + log ( 2 log n ) + log 2 π 4 log n ) / ( 2 log n ) 1 2 + O ( ( log ( 2 log n ) ) 2 ( log n ) 5 2 ) .
(4.2)

Setting b n = β n + θ n and substituting into (1.4), we obtain by taking logarithms that

log π 2 +logσlog( β n + θ n )+ β n 2 2 σ 2 + β n θ n σ 2 + θ n 2 σ 2 =logn.

So,

( log ( 2 log n ) + log 2 π ) 2 16 log n log 4 log n + log ( 2 log n ) + log 2 π 4 log n + β n θ n σ 2 + θ n 2 σ 2 log ( 1 + θ n β n ) = 0 ,
(4.3)

therefore

β n θ n σ 2 ( log ( 2 log n ) + log 2 π ) 2 16 log n +log 4 log n + log ( 2 log n ) + log 2 π 4 log n ,
(4.4)

which implies

θ n σ ( log ( 2 log n ) + log 2 π ) 2 16 2 ( log n ) 3 2 + ( σ log 4 log n + log ( 2 log n ) + log 2 π 4 log n ) / ( 2 log n ) 1 2 .
(4.5)

Once again let

θ n = σ ( log ( 2 log n ) + log 2 π ) 2 16 2 ( log n ) 3 2 + ( σ log 4 log n + log ( 2 log n ) + log 2 π 4 log n ) / ( 2 log n ) 1 2 + ν n ,

where ν n =o( ( log ( 2 log n ) ) 2 ( log n ) 3 2 ). By similar arguments, we can obtain (4.2).

Note that a n = σ 2 b n , we have

r n 1= α n a n 1 log ( 2 log n ) 4 log n , δ n = β n b n a n ( log ( 2 log n ) ) 2 16 log n .

Noting a n b n 1 1 2 log n , by Lemma 2, we have

F n ( α n x+ β n )Λ(x)Λ(x) e x ( log ( 2 log n ) ) 2 16 log n .

The proof is complete. □

Proof of Theorem 2 Letting r n =1, δ n =0 in (2.3) and noting a n b n 1 1 2 log n , and by Lemma 2, there exists an absolute constant d 1 >0 such that

sup x R | F n ( a n x + b n ) Λ ( x ) | > d 1 log n .
(4.6)

Thus, in order to obtain the upper bound, we need to prove

(a) sup c n x < 0 | F n ( a n x + b n ) Λ ( x ) | < D 1 a n b n 1 ,
(4.7)
(b) sup 0 x d n | F n ( a n x + b n ) Λ ( x ) | < D 2 a n b n 1 ,
(4.8)
(c) sup d n x < | F n ( a n x + b n ) Λ ( x ) | < D 3 a n b n 1 ,
(4.9)
(d) sup < x c n | F n ( a n x + b n ) Λ ( x ) | < D 4 a n b n 1 ,
(4.10)

where D i >0 (i=1,2,3,4), and

c n =:loglog b n 2 σ 2 >0, d n =:loglog b n 2 b n 2 σ 2 >0.

Obviously,

σ ( 2 log n ) 1 2 < b n <σ ( 2 log n ) 1 2 (1+ C 0 )

and

b n a n c n = b n ( 1 σ 2 b n 2 c n ) = b n ( 1 σ 2 b n 2 log log b n 2 σ 2 ) >0.

Define Ψ n (x)=1F( a n x+ b n ), then

n log ( 1 F ( a n x + b n ) ) = n Ψ n ( x ) + n Ψ n ( x ) + n log ( 1 Ψ n ( x ) ) = n Ψ n ( x ) R n ( x ) .
(4.11)

By the following inequality

x x 2 2 ( 1 x ) <log(1x)<x(0<x<1),

we have

0< R n (x)= ( n Ψ n ( x ) + n log ( 1 Ψ n ( x ) ) ) < n Ψ n 2 ( x ) 2 ( 1 Ψ n ( x ) ) .

First, suppose that x c n . By (2.1), we have

Ψ n ( x ) Ψ n ( c n ) = 1 F ( b n a n c n ) < 2 π b n a n c n σ ( 1 + σ 2 ( b n a n c n ) 2 ) exp ( ( b n a n c n ) 2 2 σ 2 ) < 2 2 π b n σ ( 1 a n b n 1 c n ) exp ( ( b n ) 2 2 σ 2 + c n a n b n 1 c n 2 2 ) < 2 n 1 e c n = 2 n 1 log b n 2 σ 2 < sup n n 0 2 log ( C 1 log n ) n < C 11 < 1

with C 1 =2 ( 1 + C 0 ) 2 , implying

inf x c n ( 1 Ψ n ( x ) ) >1 C 11 >0.

Therefore,

0 < R n ( x ) n Ψ n 2 ( x ) 2 ( 1 C 11 ) n Ψ n 2 ( c n ) 2 ( 1 C 11 ) < n 1 ( log ( C 1 log n ) ) 2 2 ( 1 C 11 ) = n 1 ( log ( C 1 log n ) ) 2 a n b n 1 2 ( 1 C 11 ) a n b n 1 < n 1 ( log ( C 1 log n ) ) 2 4 ( 1 C 11 ) log n a n b n 1 < C a n b n 1 .

By 1 e x <x, x>0, we have

| exp ( R n ( x ) ) 1 | < R n (x)<C a n b n 1 .

Setting A n (x)=exp(n Ψ n (x)+ e x ), B n (x)=exp( R n (x)), we obtain

| F n ( a n x + b n ) Λ ( x ) | = Λ ( x ) | A n ( x ) B n ( x ) 1 | = Λ ( x ) | A n ( x ) B n ( x ) B n ( x ) + B n ( x ) 1 | Λ ( x ) | A n ( x ) 1 | + | B n ( x ) 1 | < Λ ( x ) | A n ( x ) 1 | + C a n b n 1 .
(4.12)

By (2.1) and (2.2), we have

n Ψ n ( x ) + e x = n [ 2 π b n + a n x σ ( 1 + σ 2 ( b n + a n x ) 2 ) exp ( ( b n + a n x ) 2 2 σ 2 ) r ( a n x + b n ) ] + e x = ( 1 + a n b n 1 x ) e x C n ( x ) ,

where

C n (x)= ( 1 σ 2 ( b n + a n x ) 2 + σ 4 ( b n + a n x ) 4 δ n ( a n x + b n ) ) exp ( a n b n 1 x 2 2 ) + ( 1 + a n b n 1 x ) 1

with 0< δ n ( a n x+ b n )<1. To prove (4.7), we consider the case of c n x<0. By e x >1x, x>0, we have

C n ( x ) < ( 1 a n b n 1 x 2 2 ) ( ( 1 + σ 4 ( b n + a n x ) 4 ) δ n ( a n x + b n ) ) + ( 1 + a n b n 1 x ) 1 < ( 1 a n b n 1 x 2 2 ) { 1 + ( a n b n 1 ) 2 ( 1 + a n b n 1 x ) 4 } + ( 1 + a n b n 1 x ) 1 = ( ( 1 + a n b n 1 x ) 4 + x 2 2 x ( 1 + a n b n 1 x ) 1 ) a n b n 1
(4.13)

and

C n ( x ) > ( 1 σ 2 ( b n + a n x ) 2 ) exp ( a n b n 1 x 2 2 ) + ( 1 + a n b n 1 x ) 1 > ( 1 σ 2 ( b n + a n x ) 2 ) + ( 1 + a n b n 1 x ) 1 > ( ( 1 + a n b n 1 x ) 2 x ( 1 + a n b n 1 x ) 1 ) a n b n 1 > ( 1 + a n b n 1 x ) 2 .
(4.14)

Hence, for c n x<0, by combining (4.13) and (4.14) together, we have

| C n ( x ) | < ( ( 1 + a n b n 1 x ) 4 + x 2 2 x ( 1 + a n b n 1 x ) 1 + ( 1 + a n b n 1 x ) 2 ) a n b n 1 < ( ( 1 a n b n 1 c n ) 4 + c n 2 2 + c n ( 1 a n b n 1 c n ) 1 + ( 1 a n b n 1 c n ) 2 ) a n b n 1 < C 21 .

Furthermore, for c n x<0, we have

| n Ψ n ( x ) + e x | < ( 1 + a n b n 1 x ) e x | C n ( x ) | < ( ( 1 + a n b n 1 x ) 4 + x 2 2 x ( 1 + a n b n 1 x ) 1 + ( 1 + a n b n 1 x ) 2 ) e x a n b n 1 < ( ( 1 a n b n 1 c n ) 4 + c n 2 2 + c n ( 1 a n b n 1 c n ) 1 + ( 1 a n b n 1 c n ) 2 ) e c n a n b n 1 < C 22 .

Noting that 0<| e x 1|<|x|( e x +1), xR and e x >1x+ x 2 /2 for c n x<0, we have

Λ ( x ) | A n ( x ) 1 | = Λ ( x ) | exp ( n Ψ n ( x ) + e x ) 1 | < Λ ( x ) | n Ψ n ( x ) + e x | ( exp ( n Ψ n ( x ) + e x ) + 1 ) < ( e C 22 + 1 ) ( ( 1 + a n b n 1 x ) 4 + x 2 2 x ( 1 + a n b n 1 x ) 1 + ( 1 + a n b n 1 x ) 2 ) × a n b n 1 exp ( e x x ) < C 23 a n b n 1 .

Together with (4.12), we establish (4.7).

Second, we prove (4.8). Note that

C n ( x ) < ( 1 + σ 4 ( b n + a n x ) 4 δ n ( a n x + b n ) ) ( 1 a n b n 1 x 2 2 ) + ( 1 + a n b n 1 x ) 1 < ( a n b n 1 ) 2 ( 1 + a n b n 1 x ) 4 + x 2 2 a n b n 1 a n b n 1 x ( 1 + a n b n 1 x ) 1 < ( a n b n 1 ) 2 ( 1 + a n b n 1 x ) 4 + x 2 2 a n b n 1 < ( 1 + x 2 2 ) a n b n 1
(4.15)

and

C n (x)> ( ( 1 + a n b n 1 x ) 2 x ( 1 + a n b n 1 x ) 1 ) a n b n 1 .
(4.16)

By (4.15) and (4.16), for 0x< d n , we have

| C n ( x ) | < ( 1 + x 2 2 + ( 1 + a n b n 1 x ) 2 + x ( 1 + a n b n 1 x ) 1 ) a n b n 1 < ( 2 + x + x 2 2 ) a n b n 1 .

Hence,

| n Ψ n ( x ) + e x | < ( 1 + a n b n 1 x ) e x | C n ( x ) | < ( 1 + a n b n 1 x ) e x ( 2 + x + x 2 2 ) a n b n 1 < C 31 a n b n 1 < C 32 .

Therefore

Λ ( x ) | A n ( x ) 1 | < Λ ( x ) | n Ψ n ( x ) + e x | ( exp ( n Ψ n ( x ) + e x ) + 1 ) < C 31 ( e C 32 + 1 ) Λ ( d n ) a n b n 1 < C 33 a n b n 1 .
(4.17)

Combining (4.12) and (4.17) together, we can derive that

sup 0 x d n | F n ( a n x + b n ) Λ ( x ) | <( C 12 + C 33 ) a n b n 1 =: D 2 a n b n 1 .

Hence (4.8) is proved.

Third, for x d n , we have

sup x d n ( 1 Λ ( x ) ) 1Λ( d n )= a n b n 1 .
(4.18)

By 1 e x <x, xR, we have

1 F n ( a n d n + b n ) = 1 exp ( n log F ( a n d n + b n ) ) < n log F ( a n d n + b n ) = n Ψ n ( d n ) + R n ( d n ) .
(4.19)

By (2.1) and log(1+x)<x, 0<x<1, we have

n Ψ n ( d n ) = n ( 1 F ( a n d n + b n ) ) < ( 1 + a n b n 1 d n ) e d n ( 1 + a n b n 1 ( 1 + a n b n 1 d n ) 2 ) < 2 ( 1 + a n b n 1 d n ) log b n 2 b n 2 σ 2 < 2 ( d n + a n 1 b n ) σ 2 b n 2 σ 2 a n b n 1 = ( 1 + σ 2 b n 2 σ 2 σ 2 b n 2 σ 2 log log ( σ 2 b n 2 σ 2 ) ) a n b n 1 < C 41 a n b n 1 .
(4.20)

Noting that R n ( d n )< C 12 a n b n 1 , and combining (4.18), (4.19), (4.20) and (4.14) together, we have

sup x d n | F n ( a n x + b n ) Λ ( x ) | < sup x d n ( 1 F n ( a n x + b n ) ) + sup x d n ( 1 Λ ( x ) ) < n Ψ n ( d n ) + R n ( d n ) + a n b n 1 < ( C 41 + C 12 + 1 ) a n b n 1 = : D 3 a n b n 1 ,

which is (4.9).

Finally, consider the case of <x< c n . If a n x+ b n 0, then F n ( a n x+ b n )=0. By Λ(x)< 1 x , x>1, we have

sup x b n / a n | F n ( a n x + b n ) Λ ( x ) | = sup x b n / a n Λ(x)Λ ( b n 2 / σ 2 ) < σ 2 b n 2 = a n b n 1 .

So, we only need to consider the case of a n x+ b n >0. By using the monotonicity of Λ(x), we have

sup x c n Λ(x)Λ( c n )= a n b n 1 .
(4.21)

Noting log(1x)<x, 0<x<1 and e x >1x, xR, and combining (2.1) and (2.2) together, we have

sup x c n F n ( a n x + b n ) F n ( b n a n c n ) < ( 1 n 1 ( 1 a n b n 1 c n ) ( 1 ( a n b n 1 ) 2 ( 1 a n b n 1 c n ) 4 ) exp ( c n a n b n 1 c n 2 2 ) ) n < exp ( e c n ( 1 a n b n 1 c n ) ( 1 ( a n b n 1 ) 2 ( 1 a n b n 1 c n ) 4 ) exp ( a n b n 1 c n 2 2 ) ) < exp ( e c n ( 1 ( a n b n 1 c n + ( a n b n 1 ) 2 ( 1 a n b n 1 c n ) 3 + a n b n 1 c n 2 2 ) ) ) < exp ( e c n ) exp ( ( a n b n 1 c n + ( a n b n 1 ) 2 ( 1 a n b n 1 c n ) 3 + a n b n 1 c n 2 2 ) e c n ) < C 51 a n b n 1 .

Together with (4.21), we have

sup < x c n | F n ( a n x + b n ) Λ ( x ) | sup < x < c n F n ( a n x + b n ) + sup < x < c n Λ ( x ) F n ( b n a n c n ) + Λ ( c n ) < ( C 12 + 1 ) a n b n 1 = : D 4 a n b n 1 .

This is (4.10). The proof is complete. □

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Acknowledgements

The research was supported by the National Natural Science Foundation of China (11171275) and the Fundamental Research Funds for the Central Universities (XDJK2012C045).

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Liu, C., Liu, B. Convergence rate of extremes from Maxwell sample. J Inequal Appl 2013, 477 (2013). https://doi.org/10.1186/1029-242X-2013-477

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Keywords

  • extreme value distribution
  • maximum
  • Maxwell distribution
  • uniform convergence rate