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The limit theorems on extreme order statistics and partial sums of i.i.d. random variables

  • Gaoyu Li , Chengxiu Ling and Zhongquan Tan EMAIL logo
Published/Copyright: August 17, 2024

Abstract

This article proves several weak limit theorems for the joint version of extreme order statistics and partial sums of independently and identically distributed random variables. The results are also extended to almost sure limit version.

MSC 2010: 60G70; 60G55

1 Introduction

Since the maxima and partial sums of a sequence of random variables played a very important role in theoretical and applied probability, many authors studied their joint asymptotic properties. Chow and Teugels [1] first investigated asymptotic relation between the maxima and partial sums of a sequence of independently and identically distributed (i.i.d.) random variables { X n , n 1 } . Let M n = max { X 1 , X 2 , , X n } and S n = i = 1 n X i denote the maximum and partial sum of { X 1 , X 2 , , X n } , respectively. Suppose that there exist constants a n > 0 , b n R , n 1 and a non-degenerate distribution G ( x ) such that

lim n P ( a n ( M n b n ) x ) = G ( x ) ,

where G must be one of the following three types of extreme value distributions (see, e.g., Leadbetter et al. [2], Embrechts et al. [3], and de Haan and Ferreira [4]):

Gumbel: Λ ( x ) = exp ( e x ) , < x < ; Fréchet: Φ α ( x ) = 0 , x 0 , exp ( x α ) , x > 0 , α > 0 ; Weibull: Ψ α ( x ) = exp ( ( x ) α ) , x 0 , 1 , x > 0 , α > 0 .

If { X n , n 1 } has finite variance (without loss of generality, let E ( X 1 ) = 0 , Var ( X 1 ) = 1 ), Chow and Teugels [1] showed that

(1) lim n P a n ( M n b n ) x , S n n y = G ( x ) Φ ( y ) ,

which indicates that the maximum and the partial sum are asymptotically independent. Hereafter, Φ ( ) stands for the standard normal distribution function. If { X n , n 1 } has infinite variance, Chow and Teugels [1] showed that the maximum and the partial sum can be asymptotically independent or dependent.

The result of (1) has many extensions. A natural extension is to consider the dependent case. Anderson and Turkman [5,6] extended Chow and Teugels’s result to strong mixing case under a technical condition, and then, Hsing [7] completed their results by deleting the technical condition. Ho and Hsing [8], Ho and McCormick [9], McCormick and Qi [10], and Peng and Nadarajah [11] dealt with this problem for some dependent Gaussian cases, and James et al. [12] considered the multivariate stationary Gaussian cases. We refer to Chen and Tan [13], Tan and Tang [14], and Tan [15] for this problem for continuous times Gaussian processes and Gaussian random fields.

In applied fields, it is important to study the asymptotic relation between the extreme order statistics and the partial sum. For example[1], in the context of reliability theory, let { X n , n 1 } be the lifetime of the components of a system and M n ( n ) M n ( n 1 ) M n ( 1 ) be the n th order statistics of { X 1 , X 2 , , X n } , then the statistic M n ( k ) represents the lifetime of an k -out-of- n system and, furthermore, the lifetime of a coherent system is equal with the k th maximum of the sample. The random variable S n also determines the lifetime of a standby system. The asymptotic relation between the lifetimes of parallel systems in comparison with standby systems can be derived from the model of joint asymptotic between the extreme order statistics and the partial sum. Peng [16] and Hu et al. [17] derived the joint asymptotic distribution of the extreme order statistics and partial sum for stationary Gaussian sequences. Let { X n , n 1 } be a stationary Gaussian sequence with covariance functions r ( n ) = E ( X 1 X n + 1 ) . Suppose that the well-known Berman’s condition was satisfied, i.e.,

r ( n ) log n 0 , as n ,

then for any x , y R and fixed k ,

(2) lim n P c n ( M n ( k ) d n ) x , S n Var ( S n ) y = exp ( e x ) i = 0 k 1 ( e x ) i i ! Φ ( y ) ,

where c n > 0 , d n R , n 1 are the real sequences. For more studies in this direction, we refer to Tan and Peng [18], Peng et al. [19], and Tan and Yang [20].

Another interesting extension is to consider the almost sure limit theorem for the maxima and partial sums. Based on the result (1), Peng et al. [21] studied the almost sure limit theorem for the maxima and partial sums for i.i.d. random variables and showed that

(3) lim N 1 log N n = 1 N 1 n I a n ( M n b n ) x , S n n y = G ( x ) Φ ( y ) a.s.

for any x , y R , where I ( ) denotes the indicator function. For more studies on almost sure limit theorem for the maxima and partial sums for dependent cases, we refer to Dudziński [22,23], Zang et al. [24], Zhao et al. [25], Tan and Wang [26], Zang [27], Wu [28], and Wu and Jiang [29].

Most of the aforementioned studies focus on the Gaussian cases, especially regarding the joint version between extreme order statistics and partial sums. We even do not know the asymptotic relation between extreme order statistics and partial sums for i.i.d. cases. In this article, we continue to study the joint asymptotic properties of extreme order statistics and partial sums for i.i.d. random sequence and have two goals. The first one is to derive the joint asymptotic distribution of extreme order statistics and partial sums, which will extend the classic result (1). The second one is to prove the almost sure limit theorem for the joint version of extreme order statistics and partial sums, which will extend the result (3). Some related limit results on the joint asymptotic of the k largest maxima and partial sums are also given.

2 Weak limit theorems

The point process of exceedances plays a very important role in extreme value theory, since the asymptotic distributions of extreme order statistics can be derived with its aid. For more details about the theory of the point process of exceedances, we refer to Chapter 5 of Leadbetter et al. [2]. In this section, we will study the asymptotic relation between the extreme order statistics and the partial sum by virtue of the point process of exceedances. We first give the definition of the point process of exceedances.

For any random sequence { X n , n 1 } , define the point process of exceedances of level u n formed by X i , 1 i n as

(4) N n X ( B ) = i = 1 n I ( X i > u n , i n B ) ,

for any Borel set B on ( 0 , 1 ] .

In the following part, let M n ( k ) be the k th maximum of { X 1 , X 2 , , X n } and S n = i = 1 n X i . Let u n = u n ( x ) = a n 1 x + b n , where a n > 0 , b n R , n 1 will be chosen in Theorem 2.1.

Now, we state our main results.

Theorem 2.1

Let { X n , n 1 } be a sequence of i.i.d. random variables with non-degenerate common distribution function F and E ( X 1 ) = 0 , E ( X 1 2 ) = 1 . Suppose that there exist constants a n > 0 and b n R , n 1 such that

(5) lim n P ( a n ( M n b n ) x ) = G ( x )

holds, where G is one of the extreme value distribution Λ , Φ α for some α > 2 and Ψ α for some α > 0 . Then N n X d N and N n X and S n are asymptotically independent, where N is a Poisson point process on ( 0 , 1 ] with intensity log G 1 ( x ) .

Remark 2.1

  1. The convergence of N n X has been proved in Theorem 5.2.1 of Leadbetter et al. [2].

  2. Note that G = Φ α for some α ( 0 , 2 ) implies that { X n , n 1 } has infinite variance.

Theorem 2.2

Under the conditions of Theorem 2.1, we have for any x , y R and fixed k 1 ,

(6) lim n P a n ( M n ( k ) b n ) x , S n n y = G ( x ) i = 0 k 1 ( log G ( x ) ) i i ! Φ ( y ) .

Note that the common continuous distribution functions such as uniform, exponential, normal satisfy the conditions of Theorem 2.1. We give uniform as an example and several figures to illustrate Theorem 2.2.

Example 2.1

Let { X n , n 1 } be a sequence of independent uniform U ( 3 , 3 ) random variables. We have for any x R , y R and fixed k 1 ,

(7) lim n P a n ( M n ( k ) b n ) x , S n n y = e x i = 0 k 1 ( x ) i i ! Φ ( y ) G ( x , y , k ) ,

where

(8) a n = 3 n 6 and b n = 3 .

Note that (7) is a two-dimensional distribution convergence, which is difficult to illustrate clearly by figures. Thus, in the following figures, we fix y = 0 or fix x = 5 . To calculate the probability, we repeated the experiment 10,000 times.

Figure 1 illustrates (7) for k = 1 . Figure 1(a) shows that the rate of convergence of the first maximum after filtering by the sum is very faster, while Figure 1(b) shows that the convergence of the sum after filtering by the first maximum is not stable for uniform random variables, since too many variables are filtered of by the first maximum.

Figure 1 
               Plot 
                     
                        
                        
                           P
                                                         
                              
                                 
                                    
                                       a
                                    
                                    
                                       n
                                    
                                 
                                 
                                    (
                                    
                                       
                                          
                                             M
                                          
                                          
                                             n
                                          
                                          
                                             
                                                (
                                                
                                                   1
                                                
                                                )
                                             
                                          
                                       
                                       ‒
                                       
                                          
                                             b
                                          
                                          
                                             n
                                          
                                       
                                    
                                    )
                                 
                                 ≤
                                 x
                                 ,
                                 
                                    
                                       
                                          
                                             S
                                          
                                          
                                             n
                                          
                                       
                                    
                                    
                                       
                                          
                                             n
                                          
                                       
                                    
                                 
                                 ≤
                                 y
                              
                           
                        
                        P\left(\phantom{\rule[-0.75em]{}{0ex}},{a}_{n}\left({M}_{n}^{\left(1)}&#x2012;{b}_{n})\le x,\frac{{S}_{n}}{\sqrt{n}}\le y\right)
                     
                   for 
                     
                        
                        
                           n
                           =
                           1
                           
                              
                                 0
                              
                              
                                 2
                              
                           
                        
                        n=1{0}^{2}
                     
                   (+) and for 
                     
                        
                        
                           n
                           =
                           1
                           
                              
                                 0
                              
                              
                                 3
                              
                           
                        
                        n=1{0}^{3}
                     
                   (
                     
                        
                        
                           
                              ∘
                           
                        
                        \circ 
                     
                  ) and 
                     
                        
                        
                           G
                           
                              (
                              
                                 x
                                 ,
                                 y
                                 ,
                                 1
                              
                              )
                           
                        
                        G\left(x,y,1)
                     
                   (·), where 
                     
                        
                        
                           y
                           =
                           0
                        
                        y=0
                     
                   for (a) and 
                     
                        
                        
                           x
                           =
                           ‒
                           5
                        
                        x=&#x2012;5
                     
                   for (b).
Figure 1

Plot P a n ( M n ( 1 ) b n ) x , S n n y for n = 1 0 2 (+) and for n = 1 0 3 ( ) and G ( x , y , 1 ) (·), where y = 0 for (a) and x = 5 for (b).

Figure 2 illustrates (7) for k = 5 . It can be seen from Figure 2 that the rate of convergence of the fifth maximum and the sum after filtering by each other is almost the same for uniform random variables.

Figure 2 
               Plot 
                     
                        
                        
                           P
                                                         
                              
                                 
                                    
                                       a
                                    
                                    
                                       n
                                    
                                 
                                 
                                    (
                                    
                                       
                                          
                                             M
                                          
                                          
                                             n
                                          
                                          
                                             
                                                (
                                                
                                                   5
                                                
                                                )
                                             
                                          
                                       
                                       ‒
                                       
                                          
                                             b
                                          
                                          
                                             n
                                          
                                       
                                    
                                    )
                                 
                                 ≤
                                 x
                                 ,
                                 
                                    
                                       
                                          
                                             S
                                          
                                          
                                             n
                                          
                                       
                                    
                                    
                                       
                                          
                                             n
                                          
                                       
                                    
                                 
                                 ≤
                                 y
                              
                           
                        
                        P\left(\phantom{\rule[-0.75em]{}{0ex}},{a}_{n}\left({M}_{n}^{\left(5)}&#x2012;{b}_{n})\le x,\frac{{S}_{n}}{\sqrt{n}}\le y\right)
                     
                   for 
                     
                        
                        
                           n
                           =
                           1
                           
                              
                                 0
                              
                              
                                 2
                              
                           
                        
                        n=1{0}^{2}
                     
                   (+) and for 
                     
                        
                        
                           n
                           =
                           1
                           
                              
                                 0
                              
                              
                                 3
                              
                           
                        
                        n=1{0}^{3}
                     
                   (
                     
                        
                        
                           
                              ∘
                           
                        
                        \circ 
                     
                  ) and 
                     
                        
                        
                           G
                           
                              (
                              
                                 x
                                 ,
                                 y
                                 ,
                                 5
                              
                              )
                           
                        
                        G\left(x,y,5)
                     
                   (·), where 
                     
                        
                        
                           y
                           =
                           0
                        
                        y=0
                     
                   for (a) and 
                     
                        
                        
                           x
                           =
                           ‒
                           5
                        
                        x=&#x2012;5
                     
                   for (b).
Figure 2

Plot P a n ( M n ( 5 ) b n ) x , S n n y for n = 1 0 2 (+) and for n = 1 0 3 ( ) and G ( x , y , 5 ) (·), where y = 0 for (a) and x = 5 for (b).

Figure 3 illustrates (7) for k = 9 . Figure 3(a) shows that the rate of convergence of the ninth maximum after filtering by the sum is smaller than that of the sum after filtering by the ninth maximum for uniform random variables.

Figure 3 
               Plot 
                     
                        
                        
                           P
                                                         
                              
                                 
                                    
                                       a
                                    
                                    
                                       n
                                    
                                 
                                 
                                    (
                                    
                                       
                                          
                                             M
                                          
                                          
                                             n
                                          
                                          
                                             
                                                (
                                                
                                                   9
                                                
                                                )
                                             
                                          
                                       
                                       ‒
                                       
                                          
                                             b
                                          
                                          
                                             n
                                          
                                       
                                    
                                    )
                                 
                                 ≤
                                 x
                                 ,
                                 
                                    
                                       
                                          
                                             S
                                          
                                          
                                             n
                                          
                                       
                                    
                                    
                                       
                                          
                                             n
                                          
                                       
                                    
                                 
                                 ≤
                                 y
                              
                           
                        
                        P\left(\phantom{\rule[-0.75em]{}{0ex}},{a}_{n}\left({M}_{n}^{\left(9)}&#x2012;{b}_{n})\le x,\frac{{S}_{n}}{\sqrt{n}}\le y\right)
                     
                   for 
                     
                        
                        
                           n
                           =
                           1
                           
                              
                                 0
                              
                              
                                 2
                              
                           
                        
                        n=1{0}^{2}
                     
                   (+) and for 
                     
                        
                        
                           n
                           =
                           1
                           
                              
                                 0
                              
                              
                                 3
                              
                           
                        
                        n=1{0}^{3}
                     
                   (
                     
                        
                        
                           
                              ∘
                           
                        
                        \circ 
                     
                  ) and 
                     
                        
                        
                           G
                           
                              (
                              
                                 x
                                 ,
                                 y
                                 ,
                                 9
                              
                              )
                           
                        
                        G\left(x,y,9)
                     
                   (·), where 
                     
                        
                        
                           y
                           =
                           0
                        
                        y=0
                     
                   for (a) and 
                     
                        
                        
                           x
                           =
                           ‒
                           5
                        
                        x=&#x2012;5
                     
                   for (b).
Figure 3

Plot P a n ( M n ( 9 ) b n ) x , S n n y for n = 1 0 2 (+) and for n = 1 0 3 ( ) and G ( x , y , 9 ) (·), where y = 0 for (a) and x = 5 for (b).

The aforementioned results can be extended to the point process of exceedances of multiple levels. Let u n ( 1 ) u n ( 2 ) u n ( s ) be s levels, where u n ( j ) = u n ( x j ) = a n 1 x j + b n , x j R , j = 1 , 2 , , s . Define the point process N ˜ n of exceedances of levels u n ( 1 ) , u n ( 2 ) , , u n ( s ) as

(9) N ˜ n X ( B ) = j = 1 s N n ( j ) ( B ) = j = 1 s i = 1 n I ( X i > u n ( j ) , ( i n , j ) B ) ,

for any Borel set B on ( 0 , 1 ] × Z .

Construct a point process N ˜ on ( 0 , 1 ] × R such that N ˜ ( ) = j = 1 s N ( j ) ( ) , where for each j = 1 , 2 , , s 1 , N ( j ) is a Poisson process independent thinning of the Poisson process N ( j + 1 ) with deleting probability 1 log G 1 ( x j ) log G 1 ( x j + 1 ) , the initial Poisson process N ( s ) has intensity log G 1 ( x s ) . It is known that N ( j ) , j = 1 , 2 , , s are the Poisson processes with intensity log G 1 ( x j ) , respectively, and are independent on disjoint intervals on the plane (cf. Section 5.5 in Leadbetter et al. [2]).

Theorem 2.3

Let { X n , n 1 } be a sequence of standard i.i.d. random variables with non-degenerate common distribution function F and E ( X 1 ) = 0 , E ( X 1 2 ) = 1 . Suppose that there exist constants a n > 0 , b n R , n 1 , and a non-degenerate distribution G ( x ) such that (5) holds, where G is one of the extreme value distribution Λ , Φ α for some α > 2 , and Ψ α for some α > 0 . Then, N ˜ n X d N ˜ and N ˜ n X and S n are asymptotically independent.

Remark 2.2

The convergence of N ˜ n X has been proved in Theorem 5.5.1 of Leadbetter et al. [2].

It is easy to see from Theorem 2.3 that the l largest maxima and the partial sum are asymptotically independent.

Theorem 2.4

Under the conditions of Theorem 2.3, we have for any fixed l 1 and any x 1 , , x l , y R ,

(10) lim n P a n ( M n ( 1 ) b n ) x 1 , , a n ( M n ( l ) b n ) x l , S n n y = H ( x 1 , x 2 , , x l ) Φ ( y ) ,

where H ( x 1 , x 2 , , x l ) is the joint limit distribution of ( a n ( M n ( 1 ) b n ) , , a n ( M n ( l ) b n ) ) .

The distribution function H ( x 1 , x 2 , , x l ) has a very complicated representation. When the distribution function of X 1 is absolutely continuous with density satisfying some regularity conditions, the density function of H has been given in Definition 4.2.7 in Embrechts et al. [3]. For more details, we refer to Section 4.2 of Embrechts et al. [3] and Section 2.1 of de Haan and Ferreira [4]. The following corollary states a two-dimensional case.

Corollary 2.1

Under the conditions of Theorem 2.3, we have for any fixed l > k 1 and any x 1 , x 2 , y R ,

(11) lim n P a n ( M n ( k ) b n ) x 1 , a n ( M n ( l ) b n ) x 2 , S n n y = i = 0 k 1 j = i l 1 G ( x 2 ) ( log G 1 ( x 1 ) ) i ( log G 1 ( x 2 ) log G 1 ( x 1 ) ) j i i ! ( j i ) ! Φ ( y ) .

We need the following lemmas to prove Theorems 2.1–2.4.

Lemma 2.1

Let { X n , n 1 } be a sequence of i.i.d. random variables with non-degenerate common continuous distribution function F and E ( X 1 ) = 0 , E ( X 1 2 ) = 1 . Let { Y n , n 1 } be an independent copy of { X n , n 1 } . If (5) holds, then for any 0 < a < b 1 ,

i = [ n a ] + 1 [ n b ] P X i > u n , S n n > y P Y i > u n , S n n > y 0 ,

as n .

Proof

Note that G defined in (5) is either a Gumbel distribution Λ ( x ) or a Fréchet distribution Φ α ( x ) with α > 2 or a Weibull distribution Ψ α with α > 0 . Define x F = sup { x : F ( x ) < 1 } and S n i = S n X i , i = 1 , 2 , , n . Integration by parts shows that

P X i > u n , S n n > y = u n x F P S n i + x n > y d F ( x ) = ( 1 F ( u n ) ) P S n i + u n n > y + ( 1 F ( u n ) ) u n x F 1 F ( x ) 1 F ( u n ) d P S n i + x n > y .

Thus, the sum in Lemma 2.1 is bounded above by

i = [ n a ] + 1 [ n b ] ( 1 F ( u n ) ) P S n i + u n n > y ( 1 F ( u n ) ) P S n n > y + i = [ n a ] + 1 [ n b ] ( 1 F ( u n ) ) u n x F 1 F ( x ) 1 F ( u n ) d P S n i + x n > y = ( [ n b ] [ n a ] ) ( 1 F ( u n ) ) P S n 1 + u n n > y P S n n > y + ( [ n b ] [ n a ] ) ( 1 F ( u n ) ) u n x F 1 F ( x ) 1 F ( u n ) d P S n 1 + x n > y ( [ n b ] [ n a ] ) ( 1 F ( u n ) ) [ T n 1 + T n 2 ] ,

where we used the stationarity of { X n , n 1 } in the second step. Taking into account (5), we obtain from Theorem 1.5.1 of Leadbetter et al. [2] that

(12) n ( 1 F ( u n ) ) log G 1 ( x ) ,

as n . Thus, to prove the lemma, it suffices to prove T n i 0 , i = 1 , 2 , as n . If G = Λ , then u n is slowly varying and so u n n 0 , as n (see, e.g., Anderson and Turkman [5]). If G = Φ α with α > 2 , then u n is regularly varying with index 1 α < 1 2 and so again u n n 0 , as n (see, e.g., Anderson and Turkman [5]). If G = Ψ α with α > 0 then u n x F < for all n 1 , and obviously, u n n 0 , as n . Hence, by the central limit theorem for i.i.d. random variables, T n 1 0 , as n . It follows from the aforementioned arguments that we can also choose a constant ε > 0 such that ( u n ) 1 + ε n 0 , as n . For the term T n 2 , we will split it into two cases: x F = and x F < . For the case x F = , G must be either Λ or Φ α . Note that

T n 2 = 1 u n 1 F ( t u n ) 1 F ( u n ) d P S n 1 + t u n n > y .

For X in the domain of attraction of Φ α , 1 F ( t u n ) 1 F ( u n ) t α as u n . For X in the domain of attraction of Λ , 1 F ( t u n ) 1 F ( u n ) t α for large enough u n whatever α > 0 . We thus have for α > 2 ,

T n 2 1 u n t α d P S n 1 + t u n n > y = u n x u n α d P S n 1 + x n > y = u n u n 1 + ε x u n α d P S n 1 + x n > y + u n 1 + ε x u n α d P S n 1 + x n > y u n u n 1 + ε 1 d P S n 1 + x n > y + u n 1 + ε ( u n ) ε α d P S n 1 + x n > y P S n 1 + u n 1 + ε n > y P S n 1 + u n n > y + ( u n ) ε α ,

which tends to 0, by central limit theorem for i.i.d. random variables again, since ( u n ) 1 + ε n 0 and u n , as n . For the case x F < , G must be either Λ or Ψ α . We have

T n 2 = u n x F 1 F ( x ) 1 F ( u n ) d P S n 1 + x n > y P S n 1 + x F n > y P S n 1 + u n n > y 0 ,

as n , by central limit theorem for i.i.d. random variables again, since u n x F < for all n 1 . The proof of the lemma is complete.□

Lemma 2.2

Let { X n , n 1 } be a sequence of i.i.d. random variables with non-degenerate common continuous distribution function F and E ( X 1 ) = 0 , E ( X 1 2 ) = 1 . Let M n X ( a , b ] = max ( X i , i n ( a , b ] ) for any 0 < a < b 1 . If (5) holds, then for any 0 < a 1 < b 1 a 2 < b 2 a k < b k 1 ,

P i = 1 k { M n X ( a i , b i ] u n } , S n n y i = 1 k [ G ( x ) ] ( b i a i ) Φ ( y ) ,

as n .

Proof

Since { X n , n 1 } is a sequence of i.i.d. random variables, by Theorem 5.4.5 of Leadbetter et al. [2], we have, as n

P i = 1 k { M n X ( a i , b i ] u n } i = 1 k [ G ( x ) ] ( b i a i ) .

The asymptotic independence between the maxima and the sum can be proved by the same way as the roof of Theorem 1 of Anderson and Turkman [5].□

Lemma 2.3

Let { Y n , n 1 } be an independent copy of { X n , n 1 } . For any fixed y R , constructing a new sequence as

Z i = X i I S n n y + Y i I S n n > y , 1 i n ,

and define

(13) N n Z ( B ) = i = 1 n I ( Z i > u n , i n B ) ,

for any Borel set B on ( 0 , 1 ] . Then, N n Z d N , where N is a Poisson point process on ( 0 , 1 ] with intensity log G 1 ( x ) .

Proof

By Theorem A.1 in Leadbetter et al. [2], it is sufficient to show that

( i ) E N n Z ( ( a , b ] ) E N ( ( a , b ] ) = ( b a ) log G 1 ( x ) , 0 < a < b 1 ; ( i i ) P i = 1 k { N n Z ( ( a i , b i ] ) = 0 } P i = 1 k { N ( ( a i , b i ] ) = 0 } = exp i = 1 k ( b i a i ) log G 1 ( x ) ,

where 0 < a 1 < b 1 a 2 < b 2 a k < b k 1 .

For (i), we have

E N n Z ( ( a , b ] ) = E i n ( a , b ] I ( Z i > u n ) = i = [ n a ] + 1 [ n b ] P Z i > u n , S n n y + P Z i > u n , S n n > y = i = [ n a ] + 1 [ n b ] P X i > u n , S n n y + P Y i > u n , S n n > y = i = [ n a ] + 1 [ n b ] P ( X i > u n ) + P Y i > u n , S n n > y P X i > u n , S n n > y = ( [ n b ] [ n a ] ) ( 1 F ( u n ) ) + i = [ n a ] + 1 [ n b ] P Y i > u n , S n n > y P X i > u n , S n n > y .

Thus, by Lemma 2.1 and the fact (12), we have

E N n Z ( ( a , b ] ) ( b a ) log G 1 ( x ) = E N ( ( a , b ] ) ,

as n .

Next, for (ii), it follows that

(14) P i = 1 k { N n Z ( ( a i , b i ] ) = 0 } = P i = 1 k { M n X ( a i , b i ] u n } , S n n y + P i = 1 k { M n Y ( a i , b i ] u n } P S n n > y .

By Lemma 2.2, we have

(15) P i = 1 k { M n X ( a i , b i ] u n } , S n n y i = 1 k exp ( ( b i a i ) log G 1 ( x ) ) Φ ( y ) ,

as n . Recalling that { Y n , n 1 } is a sequence of i.i.d. random variables and noting the fact (12) again, we obtain

(16) P i = 1 k { M n Y ( a i , b i ] u n } i = 1 k exp ( ( b i a i ) log G 1 ( x ) ) ,

as n . Since { X n , n 1 } is also a sequence of i.i.d. random variables with E ( X 1 ) = 0 , E ( X 1 2 ) = 1 , we have

(17) P S n n > y 1 Φ ( y ) ,

as n . Thus, substituting (15)–(17) into (14), we obtain

P i = 1 k { N n Z ( ( a i , b i ] ) = 0 } exp i = 1 k ( b i a i ) log G 1 ( x ) .

This completes the proof of Lemma 2.3.□

Proof of Theorem 2.1

As stating in Remark 2.1, the convergence of N n X has been proved by Theorem 5.2.1 in Leadbetter et al. [2]. Hence, we only need to show the asymptotic independence between N n X and S n . First, noting that for disjoint Borel sets B 1 , B 2 , , B k on ( 0 , 1 ] with m ( B i ) = 0 , i = 1 , 2 , , k , we have

P i = 1 k { N n X ( B i ) = l i } , S n n y = P i = 1 k { N n Z ( B i ) = l i } , S n n y = P i = 1 k { N n Z ( B i ) = l i } P i = 1 k { N n Z ( B i ) = l i } , S n n > y = P i = 1 k { N n Z ( B i ) = l i } P i = 1 k { N n Y ( B i ) = l i } , S n n > y = P i = 1 k { N n Z ( B i ) = l i } P i = 1 k { N n Y ( B i ) = l i } P S n n > y ,

where l 1 , l 2 , , l k are non-negative integer numbers. Since { Y n , n 1 } is an independent copy of { X n , n 1 } , N n Y converges in distribution to a Poisson process on (0,1] with parameter log G 1 ( x ) . Thus, we have

P i = 1 k { N n Y ( B i ) = l i } i = 1 k ( m ( B i ) log G 1 ( x ) ) l i l i ! e m ( B i ) log G 1 ( x ) ,

as n . By Lemma 2.3,

P i = 1 k { N n Z ( B i ) = l i } i = 1 k ( m ( B i ) log G 1 ( x ) ) l i l i ! e m ( B i ) log G 1 ( x ) ,

as n . Thus,

P i = 1 k { N n X ( B i ) = l i } , S n n y Φ ( y ) i = 1 k ( m ( B i ) log G 1 ( x ) ) l i l i ! e m ( B i ) log G 1 ( x ) ,

as n , which proves the asymptotic independence between N n X and S n . The proof of Theorem 2.1 is complete.□

Proof of Theorem 2.2

It is easy to see that

P a n ( M n ( k ) b n ) x , S n n y = P N n X ( ( 0 , 1 ] ) k 1 , S n n y = l = 0 k 1 P N n X ( ( 0 , 1 ] ) = l , S n n y ,

and the result follows by Theorem 2.1.□

Proof of Theorem 2.3

The proof of Theorem 2.3 is essentially the same as that of Theorem 2.1, and the details are omitted here.□

Proof of Theorem 2.4

By Theorem 2.3, we have as n ,

P a n ( M n ( 1 ) b n ) x 1 , , a n ( M n ( k ) b n ) x k , S n n y = P N n ( 1 ) ( ( 0 , 1 ] ) 0 , , N n ( k ) ( ( 0 , 1 ] ) k 1 , S n n y P ( N ( 1 ) ( ( 0 , 1 ] ) 0 , , N ( k ) ( ( 0 , 1 ] ) k 1 ) Φ ( y ) ,

which shows the asymptotic independence of the k largest maxima and the partial sum.□

Proof of Corollary 2.1

By Theorem 2.3 again, we have

P a n ( M n ( k ) b n ) x 1 , a n ( M n ( l ) b n ) x 2 , S n n y = P N n ( k ) ( ( 0 , 1 ] ) k 1 , N n ( l ) ( ( 0 , 1 ] ) l 1 , S n n y P ( N ( k ) ( ( 0 , 1 ] ) k 1 , N ( l ) ( ( 0 , 1 ] ) l 1 ) Φ ( y ) ,

as n . Note that

P ( N ( k ) ( ( 0 , 1 ] ) k 1 , N ( l ) ( ( 0 , 1 ] ) l 1 ) = i = 0 k 1 j = i l 1 P ( N ( k ) ( ( 0 , 1 ] ) = i , N ( l ) ( ( 0 , 1 ] ) = j ) = i = 0 k 1 j = i l 1 P ( N ( l ) ( ( 0 , 1 ] ) = j ) P ( N ( k ) ( ( 0 , 1 ] ) = i N ( l ) ( ( 0 , 1 ] ) = j ) = i = 0 k 1 j = i l 1 ( log G 1 ( x 2 ) ) j j ! e log G 1 ( x 2 ) j ! i ! ( j i ) ! log G 1 ( x 1 ) log G 1 ( x 2 ) i 1 log G 1 ( x 1 ) log G 1 ( x 2 ) j i = i = 0 k 1 j = i l 1 G ( x 2 ) ( log G 1 ( x 1 ) ) i ( log G 1 ( x 2 ) log G 1 ( x 1 ) ) j i i ! ( j i ) ! ,

which completes the proof.□

3 Almost sure limit theorem

In this section, we extend Theorems 2.2 and 2.4 to the almost sure version.

Theorem 3.1

Let { X n , n 1 } be a sequence of standard i.i.d. random variables with non-degenerate common distribution function F and E ( X 1 ) = 0 , E ( X 1 2 ) = 1 . Suppose that there exist constants a n > 0 , b n R , n 1 , and a non-degenerate distribution G ( x ) such that (5) holds, where G is one of the extreme value distribution Λ , Φ α for some α > 2 and Ψ α for some α > 0 . Then, for any x , y R and fixed k 1 ,

(18) lim N 1 log N n = 1 N 1 n I a n ( M n ( k ) b n ) x , S n n y = G ( x ) i = 0 k 1 ( log G ( x ) ) i i ! Φ ( y ) a.s. ,

and for any x 1 , , x k , y R and fixed k 1 ,

(19) lim N 1 log N n = 1 N 1 n I a n ( M n ( 1 ) b n ) x 1 , , a n ( M n ( k ) b n ) x k , S n n y = H ( x 1 , x 2 , , x k ) Φ ( y ) a.s.

We give an example and several figures to illustrate Theorem 3.1.

Example 3.1

Let { X n , n 1 } be a sequence of independent uniform U ( 3 , 3 ) random variables. We have for any x R , y R and fixed k 1 ,

(20) lim N 1 log N n = 1 N 1 n I a n ( M n ( k ) b n ) x , S n n y = G ( x , y , k ) a.s. ,

where G ( x , y , k ) is defined in (7) and a n and b n are defined in (8).

Figures 4, 5, and 6 illustrate (20) for k = 2 , 6, and 10, respectively. In these figures, we fix y = 0 for (a) and fix x = 2 for (b). Part (a) of these figures illustrates the a.s. convergence of the extreme order statistics after filtering by the sum for uniform random variables, while part (b) illustrates the a.s. convergence of the sum after filtering by the extreme order statistics for uniform random variables.

Figure 4 
               Plot 
                     
                        
                        
                           
                              
                                 1
                              
                              
                                 log
                                 N
                              
                           
                           
                              
                                 
                                    ∑
                                 
                              
                              
                                 n
                                 =
                                 1
                              
                              
                                 N
                              
                           
                           
                              
                                 1
                              
                              
                                 n
                              
                           
                           I
                                                         
                              
                                 
                                    
                                       a
                                    
                                    
                                       n
                                    
                                 
                                 
                                    (
                                    
                                       
                                          
                                             M
                                          
                                          
                                             n
                                          
                                          
                                             
                                                (
                                                
                                                   2
                                                
                                                )
                                             
                                          
                                       
                                       ‒
                                       
                                          
                                             b
                                          
                                          
                                             n
                                          
                                       
                                    
                                    )
                                 
                                 ≤
                                 x
                                 ,
                                 
                                    
                                       
                                          
                                             S
                                          
                                          
                                             n
                                          
                                       
                                    
                                    
                                       
                                          
                                             n
                                          
                                       
                                    
                                 
                                 ≤
                                 y
                              
                           
                        
                        \frac{1}{\log N}{\sum }_{n=1}^{N}\frac{1}{n}I\left(\phantom{\rule[-0.75em]{}{0ex}},{a}_{n}\left({M}_{n}^{\left(2)}&#x2012;{b}_{n})\le x,\frac{{S}_{n}}{\sqrt{n}}\le y\right)
                     
                   for 
                     
                        
                        
                           N
                           =
                           1
                           
                              
                                 0
                              
                              
                                 3
                              
                           
                        
                        N=1{0}^{3}
                     
                   (+) and for 
                     
                        
                        
                           N
                           =
                           1
                           
                              
                                 0
                              
                              
                                 4
                              
                           
                        
                        N=1{0}^{4}
                     
                   (
                     
                        
                        
                           
                              ∘
                           
                        
                        \circ 
                     
                  ) and 
                     
                        
                        
                           G
                           
                              (
                              
                                 x
                                 ,
                                 y
                                 ,
                                 2
                              
                              )
                           
                        
                        G\left(x,y,2)
                     
                   (·), where 
                     
                        
                        
                           y
                           =
                           0
                        
                        y=0
                     
                   for (a) and 
                     
                        
                        
                           x
                           =
                           ‒
                           2
                        
                        x=&#x2012;2
                     
                   for (b).
Figure 4

Plot 1 log N n = 1 N 1 n I a n ( M n ( 2 ) b n ) x , S n n y for N = 1 0 3 (+) and for N = 1 0 4 ( ) and G ( x , y , 2 ) (·), where y = 0 for (a) and x = 2 for (b).

Figure 5 
               Plot 
                     
                        
                        
                           
                              
                                 1
                              
                              
                                 log
                                 N
                              
                           
                           
                              
                                 
                                    ∑
                                 
                              
                              
                                 n
                                 =
                                 1
                              
                              
                                 N
                              
                           
                           
                              
                                 1
                              
                              
                                 n
                              
                           
                           I
                                                         
                              
                                 
                                    
                                       a
                                    
                                    
                                       n
                                    
                                 
                                 
                                    (
                                    
                                       
                                          
                                             M
                                          
                                          
                                             n
                                          
                                          
                                             
                                                (
                                                
                                                   6
                                                
                                                )
                                             
                                          
                                       
                                       ‒
                                       
                                          
                                             b
                                          
                                          
                                             n
                                          
                                       
                                    
                                    )
                                 
                                 ≤
                                 x
                                 ,
                                 
                                    
                                       
                                          
                                             S
                                          
                                          
                                             n
                                          
                                       
                                    
                                    
                                       
                                          
                                             n
                                          
                                       
                                    
                                 
                                 ≤
                                 y
                              
                           
                        
                        \frac{1}{\log N}{\sum }_{n=1}^{N}\frac{1}{n}I\left(\phantom{\rule[-0.75em]{}{0ex}},{a}_{n}\left({M}_{n}^{\left(6)}&#x2012;{b}_{n})\le x,\frac{{S}_{n}}{\sqrt{n}}\le y\right)
                     
                   for 
                     
                        
                        
                           N
                           =
                           1
                           
                              
                                 0
                              
                              
                                 3
                              
                           
                        
                        N=1{0}^{3}
                     
                   (+) and for 
                     
                        
                        
                           N
                           =
                           1
                           
                              
                                 0
                              
                              
                                 4
                              
                           
                        
                        N=1{0}^{4}
                     
                   (
                     
                        
                        
                           
                              ∘
                           
                        
                        \circ 
                     
                  ) and 
                     
                        
                        
                           G
                           
                              (
                              
                                 x
                                 ,
                                 y
                                 ,
                                 6
                              
                              )
                           
                        
                        G\left(x,y,6)
                     
                   (·), where 
                     
                        
                        
                           y
                           =
                           0
                        
                        y=0
                     
                   for (a) and 
                     
                        
                        
                           x
                           =
                           ‒
                           2
                        
                        x=&#x2012;2
                     
                   for (b).
Figure 5

Plot 1 log N n = 1 N 1 n I a n ( M n ( 6 ) b n ) x , S n n y for N = 1 0 3 (+) and for N = 1 0 4 ( ) and G ( x , y , 6 ) (·), where y = 0 for (a) and x = 2 for (b).

Figure 6 
               Plot 
                     
                        
                        
                           
                              
                                 1
                              
                              
                                 log
                                 N
                              
                           
                           
                              
                                 
                                    ∑
                                 
                              
                              
                                 n
                                 =
                                 1
                              
                              
                                 N
                              
                           
                           
                              
                                 1
                              
                              
                                 n
                              
                           
                           I
                                                         
                              
                                 
                                    
                                       a
                                    
                                    
                                       n
                                    
                                 
                                 
                                    (
                                    
                                       
                                          
                                             M
                                          
                                          
                                             n
                                          
                                          
                                             
                                                (
                                                
                                                   10
                                                
                                                )
                                             
                                          
                                       
                                       ‒
                                       
                                          
                                             b
                                          
                                          
                                             n
                                          
                                       
                                    
                                    )
                                 
                                 ≤
                                 x
                                 ,
                                 
                                    
                                       
                                          
                                             S
                                          
                                          
                                             n
                                          
                                       
                                    
                                    
                                       
                                          
                                             n
                                          
                                       
                                    
                                 
                                 ≤
                                 y
                              
                           
                        
                        \frac{1}{\log N}{\sum }_{n=1}^{N}\frac{1}{n}I\left(\phantom{\rule[-0.75em]{}{0ex}},{a}_{n}\left({M}_{n}^{\left(10)}&#x2012;{b}_{n})\le x,\frac{{S}_{n}}{\sqrt{n}}\le y\right)
                     
                   for 
                     
                        
                        
                           N
                           =
                           1
                           
                              
                                 0
                              
                              
                                 3
                              
                           
                        
                        N=1{0}^{3}
                     
                   (+) and for 
                     
                        
                        
                           N
                           =
                           1
                           
                              
                                 0
                              
                              
                                 4
                              
                           
                        
                        N=1{0}^{4}
                     
                   (
                     
                        
                        
                           
                              ∘
                           
                        
                        \circ 
                     
                  ) and 
                     
                        
                        
                           G
                           
                              (
                              
                                 x
                                 ,
                                 y
                                 ,
                                 10
                              
                              )
                           
                        
                        G\left(x,y,10)
                     
                   (·), where 
                     
                        
                        
                           y
                           =
                           0
                        
                        y=0
                     
                   for (a) and 
                     
                        
                        
                           x
                           =
                           ‒
                           2
                        
                        x=&#x2012;2
                     
                   for (b).
Figure 6

Plot 1 log N n = 1 N 1 n I a n ( M n ( 10 ) b n ) x , S n n y for N = 1 0 3 (+) and for N = 1 0 4 ( ) and G ( x , y , 10 ) (·), where y = 0 for (a) and x = 2 for (b).

Before giving the proof of Theorem 3.1, we state two lemmas, which will be used in the proofs of our main results. For n m k , let M m , n ( k ) be the k th maximum of { X m + 1 , , X n } and S m , n = l = m + 1 n X l . As usual, a n b n means limsup n a n b n < + .

Lemma 3.1

Let { ξ n , n 1 } be a sequence of uniformly bounded random variables, i.e., there exists some M ( 0 , ) such that ξ n M a.s. for all n N . If

Var n = 1 N 1 n ξ n ( log N ) 2 ( log log N ) ( 1 + ε )

for some ε > 0 , then

lim N 1 log N n = 1 N 1 n ( ξ n E ξ n ) = 0 a.s.

Proof

See Lemma 3.1 of Csáki and Gonchigdanzan [30].□

Lemma 3.2

Let { X n , n 1 } be a sequence of standard i.i.d. random variables. We have for any fixed k and m k

P ( M m ( k ) > M m , n ( k ) ) k m n .

Proof

See Lemma 1 of Peng et al. [31].□

Proof of Theorem 3.1

Let

η n = I M n ( k ) u n , S n n y P M n ( k ) u n , S n n y .

Note that ( η n ) n = 1 is a sequence of bounded random variables with Var ( η n ) 1 . We first show

(21) lim N 1 log N n = 1 N 1 n η n = 0 , a.s.

Using Lemma 3.1, we only need to show

(22) Var n = 1 N 1 n η n ( log N ) 2 ( log log N ) ( 1 + ε ) .

We have

Var n = 1 N 1 n η n = E n = 1 N 1 n η n 2 = n = 1 N E η n 2 n 2 + 2 1 m < n N E ( η m η n ) n m L N , 1 + 2 L N , 2 .

Clearly,

L N , 1 = n = 1 N 1 n 2 E η n 2 n = 1 N 1 n 2 = O ( 1 ) .

For L N , 2 , for n m + k , we have

E ( η m η n ) = Cov I M m ( k ) u m , S m m y , I M n ( k ) u n , S n n y R 1 + R 2 + R 3 ,

where

R 1 = Cov I M m ( k ) u m , S m m y , I M n ( k ) u n , S n n y I M m , n ( k ) u n , S n n y ,

R 2 = Cov I M m ( k ) u m , S m m y , I M m , n ( k ) u n , S n n y I M m , n ( k ) u n , S m , n n y ,

and

R 3 = Cov I M m ( k ) u m , S m m y , I M m , n ( k ) u n , S m , n n y .

For the first term R 1 , by Lemma 3.2, we have

R 1 E I M n ( k ) u n , S n n y I M m , n ( k ) u n , S n n y P ( M m , n ( k ) u n ) P ( M n ( k ) u n ) P ( M m ( k ) > M m , n ( k ) ) k m n .

For the second term R 2 , we have

R 2 E I M m , n ( k ) u n , S n n y I M m , n ( k ) u n , S m , n n y E S m n 1 n ( E ( S m 2 ) ) 1 2 = m n 1 2 .

By the independence of { X n , n 1 } , we have

R 3 = 0 .

Then, we conclude that

L N , 2 1 m < n N n m + k 1 m n m n + m n 1 2 + 1 m < n N n < m + k 1 m n n = 2 N 1 n 2 m = 1 n 1 1 + n = 2 N 1 n 3 2 m = 1 n 1 1 m 1 2 + m = 1 N 1 1 m n = m + 1 m + k 1 n n = 2 N 1 n + 2 n = 2 N 1 n + m = 1 N 1 1 m log N .

Thus, (22) holds. Note that Theorem 2.2 implies

(23) lim N 1 log N n = 1 N 1 n P M n ( k ) u n , S n n y = G ( x ) i = 0 k 1 ( log G ( x ) ) i i ! Φ ( y ) a.s. ,

and then, the first assertion of Theorem 3.1 follows from (21) and (23). The second assertion can be proved similarly.□

Acknowledgements

The authors are most grateful to the referees and the editor for the thorough reading and valuable suggestions, which greatly improve the original results of this manuscript.

  1. Funding information: This research was supported by Innovation of Jiaxing City: a program to support the talented persons and Project of new economy research center of Jiaxing City (No. WYZB202254).

  2. Author contributions: Li was a major contributor in writing the manuscript. Tan provided some helpful discussions in writing the manuscript. The examples and simulation studies are provided by Ling. All authors read and approved the manuscript.

  3. Conflict of interest: The authors state no conflicts of interest.

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Received: 2023-06-29
Revised: 2024-06-13
Accepted: 2024-07-25
Published Online: 2024-08-17

© 2024 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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