On the Increments of Stable Subordinators

Abstract

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 .

Keywords

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

Keywords: 1. Introduction

Let $\left\{X\left(t\right),t\ge 0\right\}$ be a stable ordinator with exponent $\alpha$ with $0<\alpha <1$ , defined on a probability space $\left(\Omega ,\mathcal{F},\mathcal{A}\right)$ . 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

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

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

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

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

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

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 (  )

$\underset{t\to \infty }{\mathrm{lim}\mathrm{inf}}{\theta }_{\alpha }{t}^{-\frac{1}{\alpha }}{\left(\mathrm{log}\mathrm{log}t\right)}^{\frac{1-\alpha }{\alpha }}X\left(t\right)=1\text{ }\text{ }\text{almost}\text{\hspace{0.17em}}\text{surely}\text{ }\text{ }\left(a.s\right).$

Theorem 2 (  ) 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

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

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

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

$\begin{array}{l}\underset{t\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{Y\left(t\right)}{{\lambda }_{\beta }\left(t\right)}=1\text{ }a.s.,\text{ }\text{ }\\ \text{where}\text{ }\text{\hspace{0.17em}}Y\left(t\right)=X\left(t+{a}_{t}\right)-X\left(t\right).\end{array}$ (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 $\mathrm{log}u=\mathrm{log}\left(\mathrm{max}\left(u,1\right)\right)$ and $\mathrm{log}\mathrm{log}u=\mathrm{log}\mathrm{log}\left(\mathrm{max}\left(u,3\right)\right)$ .

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

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

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 $\beta =1$ 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

$\underset{t\to \infty }{\mathrm{lim}\mathrm{inf}}\frac{Y\left(t\right)}{{\lambda }_{0}\left(t\right)}=1\text{ }a.s.$

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

$E\left(\mathrm{exp}\left\{iu{X}_{1}\right\}\right)=\mathrm{exp}\left\{-{|u|}^{\alpha }\left(1-\frac{iu}{|u|}\mathrm{tan}\left(\frac{\text{π}\alpha }{2}\right)\right)\right\},\text{\hspace{0.17em}}0<\alpha <1.$

Then, as $x\to 0,$

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

where

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

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

$P\left(Y\left(t\right)\le \left(1+\epsilon \right){\lambda }_{\beta }\left(t\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}i.o\right)=1.$ (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

$P\left(Y\left({u}_{k}\right)\le \left(1+\epsilon \right){\lambda }_{\beta }\left({u}_{k}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}i.o\right)=1.$

From Mijhneer  , 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

$\frac{{\lambda }_{\beta }\left({u}_{k}\right)}{{a}_{{u}_{k}}^{\frac{1}{\alpha }}}={\theta }_{\alpha }{\left(\mathrm{log}\frac{{u}_{k}}{{a}_{{u}_{k}}}+\beta \mathrm{log}\mathrm{log}{u}_{k}+\left(1-\beta \right)\mathrm{log}\mathrm{log}{a}_{{u}_{k}}\right)}^{\frac{\alpha -1}{\alpha }}.$

Applying Lemma 1 with

$x=\left(1+\epsilon \right){\theta }_{\alpha }{\left(\mathrm{log}\frac{{u}_{k}}{{a}_{{u}_{k}}}+\beta \mathrm{log}\mathrm{log}{u}_{k}+\left(1-\beta \right)\mathrm{log}\mathrm{log}{a}_{{u}_{k}}\right)}^{\frac{\alpha -1}{\alpha }},$

one can find a ${k}_{0}$ such that, for all $k\ge {k}_{0}$ ,

$\begin{array}{l}P\left(X\left(1\right)\le \frac{\left(1+\epsilon \right){\lambda }_{\beta }\left({u}_{k}\right)}{{a}_{{u}_{k}}^{\frac{1}{\alpha }}}\right)\\ \ge \frac{{c}_{0}}{2{\left(\mathrm{log}\left(\frac{{u}_{k}{\left(\mathrm{log}{u}_{k}\right)}^{\beta }{\left(\mathrm{log}{a}_{{u}_{k}}\right)}^{1-\beta }}{{a}_{{u}_{k}}}\right)\right)}^{1/2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×\mathrm{exp}\left\{-{\left(1+\epsilon \right)}^{\alpha /\left(\alpha -1\right)}\mathrm{log}\left(\frac{{u}_{k}{\left(\mathrm{log}{u}_{k}\right)}^{\beta }{\left(\mathrm{log}{a}_{{u}_{k}}\right)}^{1-\beta }}{{a}_{{u}_{k}}}\right)\right\},\end{array}$

where ${c}_{0}$ is some positive constant. Notice that

${\left(1+\epsilon \right)}^{\frac{\alpha }{\alpha -1}}=\left(1-{\epsilon }_{1}\right)<1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{for}\text{\hspace{0.17em}}\text{some}\text{ }\text{\hspace{0.17em}}{\epsilon }_{1}>0.$

Hence

$\begin{array}{l}P\left(X\left(1\right)\le \frac{\left(1+\epsilon \right){\lambda }_{\beta }\left({u}_{k}\right)}{{a}_{{u}_{k}}^{\frac{1}{\alpha }}}\right)\\ \ge \frac{{c}_{0}}{2{\left(\mathrm{log}\left(\frac{{u}_{k}{\left(\mathrm{log}{u}_{k}\right)}^{\beta }{\left(\mathrm{log}{a}_{{u}_{k}}\right)}^{1-\beta }}{{a}_{{u}_{k}}}\right)\right)}^{1/2}}\left(\frac{{a}_{{u}_{k}}}{{u}_{k}}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }×{\left(\frac{{u}_{k}}{{a}_{{u}_{k}}}\right)}^{{\epsilon }_{1}}\frac{1}{{\left({\left(\mathrm{log}{u}_{k}\right)}^{\beta }{\left(\mathrm{log}{a}_{{u}_{k}}\right)}^{1-\beta }\right)}^{\left(1-{\epsilon }_{1}\right)}}\\ =\frac{{c}_{0}}{2{\left(\mathrm{log}\left(\frac{{u}_{k}{\left(\mathrm{log}{u}_{k}\right)}^{\beta }{\left(\mathrm{log}{a}_{{u}_{k}}\right)}^{1-\beta }}{{a}_{{u}_{k}}}\right)\right)}^{1/2}}\left(\frac{{u}_{k+1}-{u}_{k}}{{u}_{k}}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }×{\left(\frac{{u}_{k}}{{a}_{{u}_{k}}}\right)}^{{\epsilon }_{1}}\frac{1}{{\left({\left(\mathrm{log}{u}_{k}\right)}^{\beta }{\left(\mathrm{log}{a}_{{u}_{k}}\right)}^{1-\beta }\right)}^{\left(1-{\epsilon }_{1}\right)}}.\end{array}$

Let ${1}_{k}={u}_{k}/{a}_{{u}_{k}}$ and ${m}_{k}={\left(\mathrm{log}{u}_{k}\right)}^{\beta }{\left(\mathrm{log}{a}_{{u}_{k}}\right)}^{1-\beta }$ . Note that 1k is non-decreasing and ${m}_{k}\to \infty$ as $k\to \infty$ . In turn one finds a ${k}_{1}\ge {k}_{0},$ such that

$\frac{{1}_{k}^{{\epsilon }_{1}}{m}_{k}^{{\epsilon }_{1}}}{{\left(log{1}_{k}{m}_{k}\right)}^{1/2}}\ge 1,\text{ }\text{ }\text{whenever}\text{ }\text{\hspace{0.17em}}k\ge {k}_{1}.$

Therefore, for all $k\ge {k}_{1}$ , we have

$\begin{array}{l}P\left(X\left(1\right)\le \frac{\left(1+\epsilon \right){\lambda }_{\beta }\left({u}_{k}\right)}{{a}_{{u}_{k}}^{\frac{1}{\alpha }}}\right)\\ \ge {c}_{0}\frac{\left({u}_{k+1}-{u}_{k}\right)}{2{u}_{k}{\left(\mathrm{log}{u}_{k}\right)}^{\beta }{\left(\mathrm{log}{a}_{{u}_{k}}\right)}^{1-\beta }}={c}_{0}\frac{\left({u}_{k+1}-{u}_{k}\right)}{2{u}_{k}}{\left(\frac{\mathrm{log}{a}_{{u}_{k}}}{\mathrm{log}{u}_{k}}\right)}^{\beta }\frac{1}{\mathrm{log}{a}_{{u}_{k}}}\\ \ge {c}_{0}\frac{\left({u}_{k+1}-{u}_{k}\right)}{2{u}_{k}}\left(\frac{\mathrm{log}{a}_{{u}_{k}}}{\mathrm{log}{u}_{k}}\right)\frac{1}{\mathrm{log}{a}_{{u}_{k}}}={c}_{0}\frac{\left({u}_{k+1}-{u}_{k}\right)}{2{u}_{k}\mathrm{log}{u}_{k}}.\end{array}$ (5)

Observe that

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

From the fact that ${\int }_{{k}_{1}}^{\infty }\frac{\text{d}t}{t\mathrm{log}t}=\infty$ and from (4), (5), and (6) one gets

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

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

$P\left(Y\left({u}_{k}\right)\le \left(1+\epsilon \right){\lambda }_{\beta }\left({u}_{k}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}i.o\right)=1$

which establishes (3).

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

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

Define a subsequence $\left({t}_{k}\right)$ , such that

${a}_{{t}_{k}}=\left({t}_{k+1}-{t}_{k}\right)/{\left(\mathrm{log}{t}_{k}\right)}^{\left(1-\beta \right)\left(1+\epsilon \right)},\text{\hspace{0.17em}}k=1,2,\cdots$ (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

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

Note that

$\left({A}_{t}\text{ }i.o.,t\to \infty \right)\subset \left({B}_{k}\text{ }i.o.,k\to \infty \right).$

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

$\left({B}_{k}\text{ }i.o.,k\to \infty \right)\subset \left({C}_{k}\text{ }i.o.,k\to \infty \right).$

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

$P\left({C}_{k}\text{ }i.o.\right)=0.$ (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

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

The fact that ${a}_{t}/t$ is non-increasing as $t\to \infty$ implies that

$\frac{{a}_{{t}_{k+1}}}{{t}_{k+1}}\le \frac{{a}_{{t}_{k}}}{{t}_{k}}\text{ }\text{ }\text{or}\text{ }\text{ }\frac{{a}_{{t}_{k+1}}}{{a}_{{t}_{k}}}\le \frac{{t}_{k+1}}{{t}_{k}}.$

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

${a}_{{t}_{k+1}}/{a}_{{t}_{k}}\le \left(1+{\epsilon }_{1}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{for}\text{\hspace{0.17em}}\text{all}\text{ }\text{\hspace{0.17em}}k\ge {k}_{2}.$

Let $\left(1-\epsilon \right)\right){\left(1+{\epsilon }_{1}\right)}^{1/\alpha }=\left(1-{\epsilon }_{2}\right)$ . Then, for $k\ge {k}_{2}$ ,

$P\left({C}_{k}\right)\le P\left(X\left(1\right)\le \left(1-{\epsilon }_{2}\right){\theta }_{\alpha }{\left(\mathrm{log}\frac{{t}_{k+1}}{{a}_{{t}_{k+1}}}{\left(\mathrm{log}{t}_{k+1}\right)}^{\beta }{\left(\mathrm{log}{a}_{{t}_{k+1}}\right)}^{1-\beta }\right)}^{\left(\alpha -1\right)/\alpha }\right).$

From lemma 1, we can find a ${k}_{3}\left(\ge {k}_{2}\right)$ such that for all $k\ge {k}_{3}$ ,

$\begin{array}{c}P\left({C}_{k}\right)\le {c}_{1}{\left(\mathrm{log}\frac{{t}_{k+1}}{{a}_{{t}_{k+1}}}{\left(\mathrm{log}{t}_{k+1}\right)}^{\beta }{\left(\mathrm{log}{a}_{{t}_{k+1}}\right)}^{1-\beta }\right)}^{-\frac{1}{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×\mathrm{exp}\left\{{\left(1-{\epsilon }_{2}\right)}^{\alpha /\left(\alpha -1\right)}\left(\mathrm{log}\frac{{t}_{k+1}}{{a}_{{t}_{k}}}{\left(\mathrm{log}{t}_{k+1}\right)}^{\beta }{\left(\mathrm{log}{a}_{{t}_{k+1}}\right)}^{1-\beta }\right)\right\},\end{array}$

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 $k\ge {k}_{3}$ ,

$\begin{array}{l}P\left({C}_{k}\right)\le {c}_{1}{\left(\mathrm{log}\frac{{t}_{k+1}}{{a}_{{t}_{k}}}{\left(\mathrm{log}{t}_{k+1}\right)}^{\beta }{\left(\mathrm{log}{a}_{{t}_{k+1}}\right)}^{1-\beta }\right)}^{-1/2}{\left(\frac{{a}_{{t}_{k+1}}}{{t}_{k}}\right)}^{\left(1+{\epsilon }_{3}\right)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left({\left(\mathrm{log}{t}_{k+1}\right)}^{\beta }{\left(\mathrm{log}{a}_{{t}_{k+1}}\right)}^{1-\beta }\right)}^{-\left(1+{\epsilon }_{3}\right)}.\end{array}$

Since

${\left({a}_{{t}_{k+1}}/{t}_{k+1}\right)}^{\left(1+{\epsilon }_{3}\right)}\le {\left({a}_{{t}_{k}}/{t}_{k}\right)}^{\left(1+{\epsilon }_{3}\right)}\le {a}_{{t}_{k}}/{t}_{k},$

then from (8) and for all $k\ge {k}_{3}$ , we have

$P\left({C}_{k}\right)\le {c}_{1}{\left(log\frac{{t}_{k}}{{a}_{{t}_{k}}}{\left(log{t}_{k}\right)}^{\beta }{\left(log{a}_{{t}_{k}}\right)}^{1-\beta }\right)}^{-1/2}\left(\frac{{a}_{{t}_{k}}}{{t}_{k}}\right){\left({\left(log{t}_{k}\right)}^{\beta }{\left(log{a}_{{t}_{k}}\right)}^{1-\beta }\right)}^{-\left(1+{\epsilon }_{3}\right)}.$

$\begin{array}{c}P\left({C}_{k}\right)\le {c}_{1}{\left(\mathrm{log}\frac{{t}_{k}}{{a}_{{t}_{k}}}{\left(\mathrm{log}{t}_{k}\right)}^{\beta }{\left(\mathrm{log}{a}_{{t}_{k}}\right)}^{1-\beta }\right)}^{-1/2}\left(\frac{{t}_{k+1}-{t}_{k}}{{t}_{k}}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×\frac{1}{{\left(\mathrm{log}{t}_{k}\right)}^{1+{\epsilon }_{3}}}\frac{1}{{\left(\mathrm{log}{a}_{{t}_{k+1}}\right)}^{\left(1-\beta \right)\left(1+{\epsilon }_{3}\right)}}\\ \le {c}_{1}\left(\frac{{t}_{k+1}-{t}_{k}}{{t}_{k}}\right)\frac{1}{{\left(\mathrm{log}{t}_{k}\right)}^{\left(1+{\epsilon }_{3}\right)}}.\end{array}$

Observe that

${\int }_{{k}_{4}}^{\infty }\frac{\text{d}t}{t{\left(\mathrm{log}t\right)}^{\left(1+{\epsilon }_{3}\right)}}\ge \underset{k={k}_{4}}{\overset{\infty }{\sum }}\frac{\left({t}_{k+1}-{t}_{k}\right)}{{t}_{k+1}{\left(\mathrm{log}{t}_{k+1}\right)}^{\left(1+{\epsilon }_{3}\right)}},$

and

$\frac{\left({t}_{k+1}-{t}_{k}\right)}{{t}_{k+1}{\left(\mathrm{log}{t}_{k+1}\right)}^{\left(1+{\epsilon }_{3}\right)}}\simeq \frac{\left({t}_{k+1}-{t}_{k}\right)}{{t}_{k}{\left(\mathrm{log}{t}_{k}\right)}^{\left(1+{\epsilon }_{3}\right)}}.$

Hence

$\underset{k={k}_{4}}{\overset{\infty }{\sum }}\frac{\left({t}_{k+1}-{t}_{k}\right)}{{t}_{k}{\left(\mathrm{log}{t}_{k}\right)}^{\left(1+{\epsilon }_{3}\right)}}<\infty .$

Now we get ${\sum }_{k={k}_{4}}^{\infty }P\left({C}_{k}\right)<\infty$ , 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  , 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.

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