Let be a stable subordinator defined on a probability space and let at for t>0 be a non-negative valued function. In this paper, it is shown that under varying conditions on at, there exists a function such that where , , and .

Increments Stable Subordinators Iterated Logarithm Laws
1. Introduction

Let { X ( t ) , t ≥ 0 } be a stable ordinator with exponent α with 0 < α < 1 , defined on a probability space ( Ω , F , A ) . Let a t for t > 0 be a non-negative valued function and Y ( t ) = X ( t + a t ) − X ( t ) , t > 0 . Define

λ β ( t ) = θ α a t 1 α ( log t a t + β log log t + ( 1 − β ) log log a t ) α − 1 α ,

where 0 ≤ β ≤ 1 ,

θ α = ( B ( α ) ) 1 − α α and B ( α ) = ( 1 − α ) α α 1 − α ( cos ( π α 2 ) ) 1 α − 1 .

For any value of t, the characteristic function of X ( t ) is of the form

E ( e i u X ( t ) ) = exp ( − t | u | α ( 1 − u i | u | tan ( π α 2 ) ) ) ,   0 < α < 1.

Limit theorems on the increments of stable subordinators have been investigated in various directions by many authors  -  . Among the above many results, we are interested in Fristedt  and Vasudeva and Divanji  whose results are the following limit theorems on the increments of stable subordinators.

Theorem 1 (  )

lim inf t → ∞ θ α t − 1 α ( log log t ) 1 − α α X ( t ) = 1     almost   surely     ( a . s ) .

Theorem 2 (  ) Let 0 < a t for t > 0 , be a non-decreasing function of t such that

(i) 0 < a t ≤ t for t > 0 ,

(ii) a t → ∞ as t → ∞ , and

(iii) a t / t is non-increasing. Then

lim inf t → ∞ ( X ( t + a t ) − X ( t ) ) ξ ( t ) = 1   a . s . , (1)

where ξ ( t ) = θ α a t 1 α ( log t a t + log log t ) α − 1 α .

In this paper, our aim is to investigate Liminf behaviors of the increments of Y. We establish that, under certain conditions on a t ,

lim inf t → ∞ Y ( t ) λ β ( t ) = 1   a . s . ,     where     Y ( t ) = X ( t + a t ) − X ( t ) . (2)

Throughout the paper c and k (integer), with or without suffix, stand for positive constants. i.o. means infinitely often. We shall define for each u ≥ 0 the functions log u = log ( max ( u , 1 ) ) and log log u = log log ( max ( u , 3 ) ) .

2. Main Result

In this section, we reformulate the result obtained in Theorem 2 and establish our main result using λ β ( t ) with 0 ≤ β ≤ 1 instead of ξ ( t ) .

Theorem 3 Let a t , t > 0 , be a non-decreasing function of t such that

(i) 0 < a t ≤ t for t > 0 ,

(ii) a t → ∞ as t → ∞ , and

(iii) a t / t is non-increasing. Then

lim inf t → ∞ Y ( t ) λ β ( t ) = 1   a . s .

Remark 1 Let us mention some particular cases

1. For a t = t we obtain Fristedt’s iterated logarithm laws (see Thorem 1).

2. If β = 1 we have Vasudeva and Divanji theorem (see Theorem 2).

3. If β = 0 under assumptions (i), (ii) and (iii) of Theorem 3 we also have

lim inf t → ∞ Y ( t ) λ 0 ( t ) = 1   a . s .

In order to prove Theorem 3, we need the following Lemma

Lemma 1 (see  or  ) Let X 1 be a positive stable random variable with characteristic function

E ( exp { i u X 1 } ) = exp { − | u | α ( 1 − i u | u | tan ( π α 2 ) ) } ,   0 < α < 1.

Then, as x → 0 ,

P ( X 1 ≤ x ) ≃ x α 2 ( 1 − α ) 2 π α B ( α ) exp { − B ( α ) x α α − 1 }

where

B ( α ) = ( 1 − α ) α α − 1 α ( cos ( π α 2 ) ) 1 α − 1 .

Proof of Theorem 3. Firstly, we show that for any given ε > 0 , as t → ∞ ,

P ( Y ( t ) ≤ ( 1 + ε ) λ β ( t )     i . o ) = 1. (3)

Let u 1 be a number such that a u 1 > 1 . Define a sequence ( u k ) through u k + 1 = u k + a u k , for k = 1 , 2 , ⋯ . Now we show that

P ( Y ( u k ) ≤ ( 1 + ε ) λ β ( u k )     i . o ) = 1.

From Mijhneer  , we have

P ( Y ( u k ) ≤ ( 1 + ε ) λ β ( u k ) ) = P ( X ( 1 ) ≤ ( 1 + ε ) λ β ( u k ) a u k 1 α ) . (4)

But

λ β ( u k ) a u k 1 α = θ α ( log u k a u k + β log log u k + ( 1 − β ) log log a u k ) α − 1 α .

Applying Lemma 1 with

x = ( 1 + ε ) θ α ( log u k a u k + β log log u k + ( 1 − β ) log log a u k ) α − 1 α ,

one can find a k 0 such that, for all k ≥ k 0 ,

P ( X ( 1 ) ≤ ( 1 + ε ) λ β ( u k ) a u k 1 α ) ≥ c 0 2 ( log ( u k ( log u k ) β ( log a u k ) 1 − β a u k ) ) 1 / 2     × exp { − ( 1 + ε ) α / ( α − 1 ) log ( u k ( log u k ) β ( log a u k ) 1 − β a u k ) } ,

where c 0 is some positive constant. Notice that

( 1 + ε ) α α − 1 = ( 1 − ε 1 ) < 1       for   some     ε 1 > 0.

Hence

P ( X ( 1 ) ≤ ( 1 + ε ) λ β ( u k ) a u k 1 α ) ≥ c 0 2 ( log ( u k ( log u k ) β ( log a u k ) 1 − β a u k ) ) 1 / 2 ( a u k u k )         × ( u k a u k ) ε 1 1 ( ( log u k ) β ( log a u k ) 1 − β ) ( 1 − ε 1 ) = c 0 2 ( log ( u k ( log u k ) β ( log a u k ) 1 − β a u k ) ) 1 / 2 ( u k + 1 − u k u k )         × ( u k a u k ) ε 1 1 ( ( log u k ) β ( log a u k ) 1 − β ) ( 1 − ε 1 ) .

Let 1 k = u k / a u k and m k = ( log u k ) β ( log a u k ) 1 − β . Note that 1k is non-decreasing and m k → ∞ as k → ∞ . In turn one finds a k 1 ≥ k 0 , such that

1 k ε 1 m k ε 1 ( l o g 1 k m k ) 1 / 2 ≥ 1,     whenever     k ≥ k 1 .

Therefore, for all k ≥ k 1 , we have

P ( X ( 1 ) ≤ ( 1 + ε ) λ β ( u k ) a u k 1 α ) ≥ c 0 ( u k + 1 − u k ) 2 u k ( log u k ) β ( log a u k ) 1 − β = c 0 ( u k + 1 − u k ) 2 u k ( log a u k log u k ) β 1 log a u k ≥ c 0 ( u k + 1 − u k ) 2 u k ( log a u k log u k ) 1 log a u k = c 0 ( u k + 1 − u k ) 2 u k log u k . (5)

Observe that

∫ k 1 ∞ d t t log t ≤ ∑ k = k 1 ∞ ( u k + 1 − u k ) u k log u k . (6)

From the fact that ∫ k 1 ∞ d t t log t = ∞ and from (4), (5), and (6) one gets

∑ k = 1 ∞     P ( Y ( u k ) ≤ ( 1 + ε ) λ β ( u k ) ) = ∞ .

Observe that ( Y ( u k ) ) is a sequence of mutually independent random variables (for, u k + 1 = u k + a u k ) and by applying Borel-Cantelli lemma, we get

P ( Y ( u k ) ≤ ( 1 + ε ) λ β ( u k )     i . o ) = 1

which establishes (3).

Now we complete the proof by showing that, for any ε ∈ ( 0,1 ) ,

P ( Y ( t ) ≤ ( 1 − ε ) λ β ( t k )   i . o ) = 0. (7)

Define a subsequence ( t k ) , such that

a t k = ( t k + 1 − t k ) / ( log t k ) ( 1 − β ) ( 1 + ε ) ,   k = 1 , 2 , ⋯ (8)

and the events A t and B k as

A t = { Y ( t ) ≤ ( 1 − ε ) λ β ( t ) }

and

B k = { inf t k ≤ t ≤ t k + 1 Y ( t ) ≤ ( 1 − ε ) λ β ( t k + 1 ) } ,   k = 1 , 2 , ⋯ .

Note that

( A t   i . o . , t → ∞ ) ⊂ ( B k   i . o . , k → ∞ ) .

Further, define

C k = { X ( t k + a t k ) − X ( t k + 1 ) ≤ ( 1 − ε ) λ β ( t k + 1 ) }

and observe that

( B k   i . o . , k → ∞ ) ⊂ ( C k   i . o . , k → ∞ ) .

Hence in order to prove (7) it is enough to show that

P ( C k   i . o . ) = 0. (9)

We have

P ( X ( t k + a t k ) − X ( t k + 1 ) ≤ ( 1 − ε ) λ β ( t k + 1 ) ) = P ( X ( 1 ) ≤ ( 1 − ε ) λ β ( t k + 1 ) ( a t k + t k − t k + 1 ) 1 / α )

and

( 1 − ε ) λ β ( t k + 1 ) ( a t k + t k − t k + 1 ) 1 / α ≃ ( 1 − ε ) θ α ( a t k + 1 a t k ) 1 / α ( log ( t k + 1 ( log t k + 1 ) β ( log a t k ) 1 − β a t k ) ) ( α − 1 ) / α .

The fact that a t / t is non-increasing as t → ∞ implies that

a t k + 1 t k + 1 ≤ a t k t k     or     a t k + 1 a t k ≤ t k + 1 t k .

Hence for a given ε 1 > 0 satisfying ( 1 − ε ) ( 1 + ε 1 ) 1 / α < 1 , there exists a k 2 such that

a t k + 1 / a t k ≤ ( 1 + ε 1 ) ,       for   all     k ≥ k 2 .

Let ( 1 − ε ) ) ( 1 + ε 1 ) 1 / α = ( 1 − ε 2 ) . Then, for k ≥ k 2 ,

P ( C k ) ≤ P ( X ( 1 ) ≤ ( 1 − ε 2 ) θ α ( log t k + 1 a t k + 1 ( log t k + 1 ) β ( log a t k + 1 ) 1 − β ) ( α − 1 ) / α ) .

From lemma 1, we can find a k 3 ( ≥ k 2 ) such that for all k ≥ k 3 ,

P ( C k ) ≤ c 1 ( log t k + 1 a t k + 1 ( log t k + 1 ) β ( log a t k + 1 ) 1 − β ) − 1 2     × exp { ( 1 − ε 2 ) α / ( α − 1 ) ( log t k + 1 a t k ( log t k + 1 ) β ( log a t k + 1 ) 1 − β ) } ,

where c 1 is a positive constant.

Let ( 1 − ε 2 ) α / ( α − 1 ) = ( 1 + ε 3 ) , ε 3 > 0. Then, for all k ≥ k 3 ,

P ( C k ) ≤ c 1 ( log t k + 1 a t k ( log t k + 1 ) β ( log a t k + 1 ) 1 − β ) − 1 / 2 ( a t k + 1 t k ) ( 1 + ε 3 )                         ( ( log t k + 1 ) β ( log a t k + 1 ) 1 − β ) − ( 1 + ε 3 ) .

Since

( a t k + 1 / t k + 1 ) ( 1 + ε 3 ) ≤ ( a t k / t k ) ( 1 + ε 3 ) ≤ a t k / t k ,

then from (8) and for all k ≥ k 3 , we have

P ( C k ) ≤ c 1 ( l o g t k a t k ( l o g t k ) β ( l o g a t k ) 1 − β ) − 1 / 2 ( a t k t k ) ( ( l o g t k ) β ( l o g a t k ) 1 − β ) − ( 1 + ε 3 ) .

P ( C k ) ≤ c 1 ( log t k a t k ( log t k ) β ( log a t k ) 1 − β ) − 1 / 2 ( t k + 1 − t k t k )     × 1 ( log t k ) 1 + ε 3 1 ( log a t k + 1 ) ( 1 − β ) ( 1 + ε 3 ) ≤ c 1 ( t k + 1 − t k t k ) 1 ( log t k ) ( 1 + ε 3 ) .

Observe that

∫ k 4 ∞ d t t ( log t ) ( 1 + ε 3 ) ≥ ∑ k = k 4 ∞ ( t k + 1 − t k ) t k + 1 ( log t k + 1 ) ( 1 + ε 3 ) ,

and

( t k + 1 − t k ) t k + 1 ( log t k + 1 ) ( 1 + ε 3 ) ≃ ( t k + 1 − t k ) t k ( log t k ) ( 1 + ε 3 ) .

Hence

∑ k = k 4 ∞ ( t k + 1 − t k ) t k ( log t k ) ( 1 + ε 3 ) < ∞ .

Now we get ∑ k = k 4 ∞ P ( C k ) < ∞ , which in turn establishes (9) by applying to the Borel-Cantelli lemma. The proof of Theorem 3 is complete.

3. Conclusion

In this paper, we developed some limit theorems on increments of stable subordinators. We reformulated the result obtained by Vasudeva and Divanji  , and established our result by using λ β ( t ) .

Acknowledgments

Our thanks to the experts who have contributed towards development of our paper.

Cite this paper

Bahram, A. and Almohaimeed, B. (2017) On the Increments of Stable Subordinators. Applied Mathematics, 8, 663-670. https://doi.org/10.4236/am.2017.85053

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