Abdelkader Bahram^1,2, Bader Almohaimeed^2
1Department of Mathematics, Djillali Liabes University, Sidi-Bel-Abbes, Algeria.
²Department of Mathematics, Faculty of Science, Qassim University, Saudi Arabia.
DOI: 10.4236/am.2017.85053   PDF    HTML   XML   1,528 Downloads   2,775 Views   Citations

Abstract

Let be a stable subordinator defined on a probability space 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 such that

where , , and .

Keywords

Increments, Stable Subordinators, Iterated Logarithm Laws

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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.

Conflicts of Interest

The authors declare no conflicts of interest.

References

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