) 1 β a u k ) } ,

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.

Cite this paper

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

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