ABSTRACT

Let $\left\{X\left(t\right),t\ge 0\right\}$ be a stable subordinator defined on a probability space $\left(\Omega \mathrm{,}\mathcal{F}\mathrm{,}\mathcal{A}\right)$ and let $a_t$ for $t>0$ be a non-negative valued function. In this paper, it is shown that under varying conditions on $a_t$ , there exists a function ${\lambda }_{\beta }\left(t\right)$ such that

$\underset{t\to \infty }{\lim \inf }\frac{\left(X\left(t+a_T\right)-X\left(t\right)\right)}{{\lambda }_{\beta }\left(t\right)}=1a.s.,$

where ${\lambda }_{\beta }\left(t\right)={\theta }_{\alpha }{a}_t^{\frac{1}{\alpha }}{\left(\log \frac{t}{a_t}+\beta \log \log t+\left(1-\beta \right)\log \log a_t\right)}^{\frac{\alpha -1}{\alpha }}$ , $0\le \beta \le 1$ , ${\theta }_{\alpha }={\left(B\left(\alpha \right)\right)}^{\frac{1-\alpha }{\alpha }}$ and $B\left(\alpha \right)=\left(1-\alpha \right){\alpha }^{\frac{\alpha }{1-\alpha }}{\left(\cos \left(\frac{\text{π}\alpha }{2}\right)\right)}^{\frac{1}{\alpha -1}}.$

Keywords:

Increments, Stable Subordinators, Iterated Logarithm Laws

1. Introduction

Let $\left\{X\left(t\right)\mathrm{,}t\ge 0\right\}$ be a stable ordinator with exponent $\alpha$ with $0<\alpha <1$ , defined on a probability space $\left(\Omega \mathrm{,}\mathcal{F}\mathrm{,}\mathcal{A}\right)$ . Let $a_t$ for $t>0$ be a non-negative valued function and $Y\left(t\right)=X\left(t+a_t\right)-X\left(t\right)$ , $t>0$ . Define

${\lambda }_{\beta }\left(t\right)={\theta }_{\alpha }{a}_t^{\frac{1}{\alpha }}{\left(\log \frac{t}{a_t}+\beta \log \log t+\left(1-\beta \right)\log \log a_t\right)}^{\frac{\alpha -1}{\alpha }}$ ,

where $0\le \beta \le 1$ ,

${\theta }_{\alpha }={\left(B\left(\alpha \right)\right)}^{\frac{1-\alpha }{\alpha }}$ and $B\left(\alpha \right)=\left(1-\alpha \right){\alpha }^{\frac{\alpha }{1-\alpha }}{\left(\cos \left(\frac{\text{π}\alpha }{2}\right)\right)}^{\frac{1}{\alpha -1}}.$

For any value of t, the characteristic function of $X\left(t\right)$ is of the form

$E\left({\text{e}}^{iuX\left(t\right)}\right)=\exp \left(-t{\left|u\right|}^{\alpha }\left(1-\frac{ui}{\left|u\right|}\tan \left(\frac{\text{π}\alpha }{2}\right)\right)\right),0<\alpha <1.$

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] )

$\underset{t\to \infty }{\lim \inf }{\theta }_{\alpha }{t}^{-\frac{1}{\alpha }}{\left(\log \log t\right)}^{\frac{1-\alpha }{\alpha }}X\left(t\right)=1\text{almost}\text{surely}\left(a.s\right).$

Theorem 2 ( [3] ) Let $0<a_t$ for $t>0$ , be a non-decreasing function of $t$ such that

(i) $0<a_t\le t$ for $t>0$ ,

(ii) $a_t\to \infty$ as $t\to \infty$ , and

(iii) $a_t/t$ is non-increasing. Then

$\underset{t\to \infty }{\lim \inf }\frac{\left(X\left(t+a_t\right)-X\left(t\right)\right)}{\xi \left(t\right)}=1a.s.,$ (1)

where $\xi \left(t\right)={\theta }_{\alpha }{a}_t^{\frac{1}{\alpha }}{\left(\log \frac{t}{a_t}+\log \log t\right)}^{\frac{\alpha -1}{\alpha }}.$

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

$\begin{array}{l}\underset{t\to \infty }{\lim \inf }\frac{Y\left(t\right)}{{\lambda }_{\beta }\left(t\right)}=1a.s.,\\ \text{where}Y\left(t\right)=X\left(t+a_t\right)-X\left(t\right).\end{array}$ (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 $u\ge 0$ the functions $\log u=\log \left(\max \left(u,1\right)\right)$ and $\log \log u=\log \log \left(\max \left(u,3\right)\right)$ .

2. Main Result

In this section, we reformulate the result obtained in Theorem 2 and establish our main result using ${\lambda }_{\beta }\left(t\right)$ with $0\le \beta \le 1$ instead of $\xi \left(t\right)$ .

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

(i) $0<a_t\le t$ for $t>0$ ,

(ii) $a_t\to \infty$ as $t\to \infty$ , and

(iii) $a_t/t$ is non-increasing. Then

$\lim {\inf }_{t\to \infty }\frac{Y\left(t\right)}{{\lambda }_{\beta }\left(t\right)}=1a.s.$

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 $\beta =1$ we have Vasudeva and Divanji theorem (see Theorem 2).

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

$\underset{t\to \infty }{\lim \inf }\frac{Y\left(t\right)}{{\lambda }_0\left(t\right)}=1a.s.$

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

$E\left(\exp \left\{iu{X}_1\right\}\right)=\exp \left\{-{\left|u\right|}^{\alpha }\left(1-\frac{iu}{\left|u\right|}\tan \left(\frac{\text{π}\alpha }{2}\right)\right)\right\},0<\alpha <1.$

Then, as $x\to 0,$

$P\left(X_1\le x\right)\simeq \frac{x^{\frac{\alpha }{2\left(1-\alpha \right)}}}{\sqrt{2\text{π}\alpha B\left(\alpha \right)}}\exp \left\{-B\left(\alpha \right)x^{\frac{\alpha }{\alpha -1}}\right\}$

where

$B\left(\alpha \right)=\left(1-\alpha \right){\alpha }^{\frac{\alpha -1}{\alpha }}{\left(\cos \left(\frac{\text{π}\alpha }{2}\right)\right)}^{\frac{1}{\alpha -1}}.$

Proof of Theorem 3. Firstly, we show that for any given $\epsilon >0$ , as $t\to \infty \mathrm{,}$

$P\left(Y\left(t\right)\le \left(1+\epsilon \right){\lambda }_{\beta }\left(t\right)i.o\right)=1.$ (3)

Let $u_1$ be a number such that $a_{u_1}>1$ . Define a sequence $\left(u_k\right)$ through $u_{k+1}=u_k+a_{u_k}$ , for $k=1,2,\cdots .$ Now we show that

$P\left(Y\left(u_k\right)\le \left(1+\epsilon \right){\lambda }_{\beta }\left(u_k\right)i.o\right)=1.$

From Mijhneer [8] , we have

$P\left(Y\left(u_k\right)\le \left(1+\epsilon \right){\lambda }_{\beta }\left(u_k\right)\right)=P\left(X\left(1\right)\le \frac{\left(1+\epsilon \right){\lambda }_{\beta }\left(u_k\right)}{a_{u_k}^{\frac{1}{\alpha }}}\right).$ (4)

But

$\frac{{\lambda }_{\beta }\left(u_k\right)}{a_{u_k}^{\frac{1}{\alpha }}}={\theta }_{\alpha }{\left(\log \frac{u_k}{a_{u_k}}+\beta \log \log u_k+\left(1-\beta \right)\log \log a_{u_k}\right)}^{\frac{\alpha -1}{\alpha }}.$

Applying Lemma 1 with

$x=\left(1+\epsilon \right){\theta }_{\alpha }{\left(\log \frac{u_k}{a_{u_k}}+\beta \log \log u_k+\left(1-\beta \right)\log \log a_{u_k}\right)}^{\frac{\alpha -1}{\alpha }},$

one can find a $k_0$ such that, for all $k\ge k_0$ ,

$\begin{array}{l}P\left(X\left(1\right)\le \frac{\left(1+\epsilon \right){\lambda }_{\beta }\left(u_k\right)}{a_{u_k}^{\frac{1}{\alpha }}}\right)\\ \ge \frac{c_0}{2{\left(\log \left(\frac{u_k{\left(\log u_k\right)}^{\beta }{\left(\log a_{u_k}\right)}^{1-\beta }}{a_{u_k}}\right)\right)}^{1/2}}\\ \times \exp \left\{-{\left(1+\epsilon \right)}^{\alpha /\left(\alpha -1\right)}\log \left(\frac{u_k{\left(\log u_k\right)}^{\beta }{\left(\log a_{u_k}\right)}^{1-\beta }}{a_{u_k}}\right)\right\},\end{array}$

where $c_0$ is some positive constant. Notice that

${\left(1+\epsilon \right)}^{\frac{\alpha }{\alpha -1}}=\left(1-{\epsilon }_1\right)<1\text{for}\text{some}{\epsilon }_1>0.$

Hence

$\begin{array}{l}P\left(X\left(1\right)\le \frac{\left(1+\epsilon \right){\lambda }_{\beta }\left(u_k\right)}{a_{u_k}^{\frac{1}{\alpha }}}\right)\\ \ge \frac{c_0}{2{\left(\log \left(\frac{u_k{\left(\log u_k\right)}^{\beta }{\left(\log a_{u_k}\right)}^{1-\beta }}{a_{u_k}}\right)\right)}^{1/2}}\left(\frac{a_{u_k}}{u_k}\right)\\ \times {\left(\frac{u_k}{a_{u_k}}\right)}^{{\epsilon }_1}\frac{1}{{\left({\left(\log u_k\right)}^{\beta }{\left(\log a_{u_k}\right)}^{1-\beta }\right)}^{\left(1-{\epsilon }_1\right)}}\\ =\frac{c_0}{2{\left(\log \left(\frac{u_k{\left(\log u_k\right)}^{\beta }{\left(\log a_{u_k}\right)}^{1-\beta }}{a_{u_k}}\right)\right)}^{1/2}}\left(\frac{u_{k+1}-u_k}{u_k}\right)\\ \times {\left(\frac{u_k}{a_{u_k}}\right)}^{{\epsilon }_1}\frac{1}{{\left({\left(\log u_k\right)}^{\beta }{\left(\log a_{u_k}\right)}^{1-\beta }\right)}^{\left(1-{\epsilon }_1\right)}}.\end{array}$

Let $1_k=u_k/a_{u_k}$ and $m_k={\left(\log u_k\right)}^{\beta }{\left(\log a_{u_k}\right)}^{1-\beta }$ . Note that 1_k is non-decreasing and $m_k\to \infty$ as $k\to \infty$ . In turn one finds a $k_1\ge k_0\mathrm{,}$ such that

$\frac{1_k^{{\epsilon }_1}{m}_k^{{\epsilon }_1}}{{\left(log{1}_k{m}_k\right)}^{1/2}}\ge \mathrm{1,}\text{whenever}k\ge k_1\mathrm{.}$

Therefore, for all $k\ge k_1$ , we have

$\begin{array}{l}P\left(X\left(1\right)\le \frac{\left(1+\epsilon \right){\lambda }_{\beta }\left(u_k\right)}{a_{u_k}^{\frac{1}{\alpha }}}\right)\\ \ge c_0\frac{\left(u_{k+1}-u_k\right)}{2u_k{\left(\log u_k\right)}^{\beta }{\left(\log a_{u_k}\right)}^{1-\beta }}=c_0\frac{\left(u_{k+1}-u_k\right)}{2u_k}{\left(\frac{\log a_{u_k}}{\log u_k}\right)}^{\beta }\frac{1}{\log a_{u_k}}\\ \ge c_0\frac{\left(u_{k+1}-u_k\right)}{2u_k}\left(\frac{\log a_{u_k}}{\log u_k}\right)\frac{1}{\log a_{u_k}}=c_0\frac{\left(u_{k+1}-u_k\right)}{2u_k\log u_k}.\end{array}$ (5)

Observe that

${\displaystyle {\int }_{k_1}^{\infty }}\frac{\text{d}t}{t\log t}\le {\displaystyle \underset{k=k_1}{\overset{\infty }{\sum }}}\frac{\left(u_{k+1}-u_k\right)}{u_k\log u_k}.$ (6)

From the fact that ${\displaystyle {\int }_{k_1}^{\infty }}\frac{\text{d}t}{t\log t}=\infty$ and from (4), (5), and (6) one gets

${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}P\left(Y\left(u_k\right)\le \left(1+\epsilon \right){\lambda }_{\beta }\left(u_k\right)\right)=\infty .$

Observe that $\left(Y\left(u_k\right)\right)$ 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

$P\left(Y\left(u_k\right)\le \left(1+\epsilon \right){\lambda }_{\beta }\left(u_k\right)i.o\right)=1$

which establishes (3).

Now we complete the proof by showing that, for any $\epsilon \in \left(\mathrm{0,1}\right)$ ,

$P\left(Y\left(t\right)\le \left(1-\epsilon \right){\lambda }_{\beta }\left(t_k\right)i.o\right)=0.$ (7)

Define a subsequence $\left(t_k\right)$ , such that

$a_{t_k}=\left(t_{k+1}-t_k\right)/{\left(\log t_k\right)}^{\left(1-\beta \right)\left(1+\epsilon \right)},k=1,2,\cdots$ (8)

and the events $A_t$ and $B_k$ as

$A_t=\left\{Y\left(t\right)\le \left(1-\epsilon \right){\lambda }_{\beta }\left(t\right)\right\}$

and

$B_k=\left\{\underset{t_k\le t\le t_{k+1}}{\inf }Y\left(t\right)\le \left(1-\epsilon \right){\lambda }_{\beta }\left(t_{k+1}\right)\right\},k=1,2,\cdots .$

Note that

$\left(A_{t}i.o.,t\to \infty \right)\subset \left(B_{k}i.o.,k\to \infty \right).$

Further, define

$C_k=\left\{X\left(t_k+a_{t_k}\right)-X\left(t_{k+1}\right)\le \left(1-\epsilon \right){\lambda }_{\beta }\left(t_{k+1}\right)\right\}$

and observe that

$\left(B_{k}i.o.,k\to \infty \right)\subset \left(C_{k}i.o.,k\to \infty \right).$

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

$P\left(C_{k}i.o.\right)=0.$ (9)

We have

$P\left(X\left(t_k+a_{t_k}\right)-X\left(t_{k+1}\right)\le \left(1-\epsilon \right){\lambda }_{\beta }\left(t_{k+1}\right)\right)=P\left(X\left(1\right)\le \frac{\left(1-\epsilon \right){\lambda }_{\beta }\left(t_{k+1}\right)}{{\left(a_{t_k}+t_k-t_{k+1}\right)}^{1/\alpha }}\right)$

and

$\begin{array}{l}\frac{\left(1-\epsilon \right){\lambda }_{\beta }\left(t_{k+1}\right)}{{\left(a_{t_k}+t_k-t_{k+1}\right)}^{1/\alpha }}\\ \simeq \left(1-\epsilon \right){\theta }_{\alpha }{\left(\frac{a_{t_{k+1}}}{a_{t_k}}\right)}^{1/\alpha }{\left(\log \left(\frac{t_{k+1}{\left(\log t_{k+1}\right)}^{\beta }{\left(\log a_{t_k}\right)}^{1-\beta }}{a_{t_k}}\right)\right)}^{\left(\alpha -1\right)/\alpha }.\end{array}$

The fact that $a_t/t$ is non-increasing as $t\to \infty$ implies that

$\frac{a_{t_{k+1}}}{t_{k+1}}\le \frac{a_{t_k}}{t_k}\text{or}\frac{a_{t_{k+1}}}{a_{t_k}}\le \frac{t_{k+1}}{t_k}\mathrm{.}$

Hence for a given ${\epsilon }_1>0$ satisfying $\left(1-\epsilon \right){\left(1+{\epsilon }_1\right)}^{1/\alpha }<1,$ there exists a $k_2$ such that

$a_{t_{k+1}}/a_{t_k}\le \left(1+{\epsilon }_1\right),\text{for}\text{all}k\ge k_2.$

Let $\left(1-\epsilon \right)){\left(1+{\epsilon }_1\right)}^{1/\alpha }=\left(1-{\epsilon }_2\right)$ . Then, for $k\ge k_2$ ,

$P\left(C_k\right)\le P\left(X\left(1\right)\le \left(1-{\epsilon }_2\right){\theta }_{\alpha }{\left(\log \frac{t_{k+1}}{a_{t_{k+1}}}{\left(\log t_{k+1}\right)}^{\beta }{\left(\log a_{t_{k+1}}\right)}^{1-\beta }\right)}^{\left(\alpha -1\right)/\alpha }\right).$

From lemma 1, we can find a $k_3\left(\ge k_2\right)$ such that for all $k\ge k_3$ ,

$\begin{array}{c}P\left(C_k\right)\le c_1{\left(\log \frac{t_{k+1}}{a_{t_{k+1}}}{\left(\log t_{k+1}\right)}^{\beta }{\left(\log a_{t_{k+1}}\right)}^{1-\beta }\right)}^{-\frac{1}{2}}\\ \times \exp \left\{{\left(1-{\epsilon }_2\right)}^{\alpha /\left(\alpha -1\right)}\left(\log \frac{t_{k+1}}{a_{t_k}}{\left(\log t_{k+1}\right)}^{\beta }{\left(\log a_{t_{k+1}}\right)}^{1-\beta }\right)\right\},\end{array}$

where $c_1$ is a positive constant.

Let ${\left(1-{\epsilon }_2\right)}^{\alpha /\left(\alpha -1\right)}=\left(1+{\epsilon }_3\right)$ , ${\epsilon }_3>0.$ Then, for all $k\ge k_3$ ,

$\begin{array}{l}P\left(C_k\right)\le c_1{\left(\log \frac{t_{k+1}}{a_{t_k}}{\left(\log t_{k+1}\right)}^{\beta }{\left(\log a_{t_{k+1}}\right)}^{1-\beta }\right)}^{-1/2}{\left(\frac{a_{t_{k+1}}}{t_k}\right)}^{\left(1+{\epsilon }_3\right)}\\ {\left({\left(\log t_{k+1}\right)}^{\beta }{\left(\log a_{t_{k+1}}\right)}^{1-\beta }\right)}^{-\left(1+{\epsilon }_3\right)}.\end{array}$

Since

${\left(a_{t_{k+1}}/t_{k+1}\right)}^{\left(1+{\epsilon }_3\right)}\le {\left(a_{t_k}/t_k\right)}^{\left(1+{\epsilon }_3\right)}\le a_{t_k}/t_k\mathrm{,}$

then from (8) and for all $k\ge k_3$ , we have

$P\left(C_k\right)\le c_1{\left(log\frac{t_k}{a_{t_k}}{\left(log{t}_k\right)}^{\beta }{\left(log{a}_{t_k}\right)}^{1-\beta }\right)}^{-1/2}\left(\frac{a_{t_k}}{t_k}\right){\left({\left(log{t}_k\right)}^{\beta }{\left(log{a}_{t_k}\right)}^{1-\beta }\right)}^{-\left(1+{\epsilon }_3\right)}\mathrm{.}$

$\begin{array}{c}P\left(C_k\right)\le c_1{\left(\log \frac{t_k}{a_{t_k}}{\left(\log t_k\right)}^{\beta }{\left(\log a_{t_k}\right)}^{1-\beta }\right)}^{-1/2}\left(\frac{t_{k+1}-t_k}{t_k}\right)\\ \times \frac{1}{{\left(\log t_k\right)}^{1+{\epsilon }_3}}\frac{1}{{\left(\log a_{t_{k+1}}\right)}^{\left(1-\beta \right)\left(1+{\epsilon }_3\right)}}\\ \le c_1\left(\frac{t_{k+1}-t_k}{t_k}\right)\frac{1}{{\left(\log t_k\right)}^{\left(1+{\epsilon }_3\right)}}.\end{array}$

Observe that

${\displaystyle {\int }_{k_4}^{\infty }}\frac{\text{d}t}{t{\left(\log t\right)}^{\left(1+{\epsilon }_3\right)}}\ge {\displaystyle \underset{k=k_4}{\overset{\infty }{\sum }}}\frac{\left(t_{k+1}-t_k\right)}{t_{k+1}{\left(\log t_{k+1}\right)}^{\left(1+{\epsilon }_3\right)}},$

and

$\frac{\left(t_{k+1}-t_k\right)}{t_{k+1}{\left(\log t_{k+1}\right)}^{\left(1+{\epsilon }_3\right)}}\simeq \frac{\left(t_{k+1}-t_k\right)}{t_k{\left(\log t_k\right)}^{\left(1+{\epsilon }_3\right)}}.$

Hence

${\displaystyle \underset{k=k_4}{\overset{\infty }{\sum }}}\frac{\left(t_{k+1}-t_k\right)}{t_k{\left(\log t_k\right)}^{\left(1+{\epsilon }_3\right)}}<\infty .$

Now we get ${\displaystyle {\sum }_{k=k_4}^{\infty }}P\left(C_k\right)<\infty$ , 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 [3] , and established our result by using ${\lambda }_{\beta }\left(t\right)$ .

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

References