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 [1] - [6] . Among the above many results, we are interested in Fristedt [4] and Vasudeva and Divanji [3] whose results are the following limit theorems on the increments of stable subordinators.

Theorem 1 ( [4] )

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

Theorem 2 ( [3] ) 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 ) ) .

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 [3] or [7] ) 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 [8] , 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 1_k 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.

In this paper, we developed some limit theorems on increments of stable subordinators. We reformulated the result obtained by Vasudeva and Divanji [3] , and established our result by using λ β ( t ) .

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