Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question has been solved.

I'm working on a problem where I need to show $$\limsup_{t \rightarrow \infty} X_t \leq \sqrt{c}\quad \text{a.s.}$$ where $X_t$, $t \geq 0$ is a stochastic process and $c$ is a given constant.

My plan is to show that for every $m \in \mathbb{N}$, $$\limsup_{t \rightarrow \infty} X_t \leq \sqrt{c} + \frac{1}{m}\quad \text{a.s.}$$

I've got another related stochastic process $Y_t$. I've managed to prove that $$ X_t > \sqrt{c} + \frac{1}{m} \quad \text{ implies } \quad Y_t \geq (1+t)^{\frac{1}{m}} $$ and that for any sequence $t_n$ with $t_n \rightarrow \infty$ $$ \lim_{n \rightarrow \infty} Y_{t_n} \quad \text{is finite a.s.} $$

It's clear that if $\limsup_{t \rightarrow \infty} X_t \leq \sqrt{c} + \frac{1}{m}$ for some $\omega$ in the sample space, then we can find a sequence $t_n \rightarrow \infty$, depending on $\omega$, with $$ X_{t_n} > \sqrt{c} + \frac{1}{m} \quad \text{at $\omega$} $$ so $$ Y_{t_n} > (1+t_n)^{\frac{1}{m}} \rightarrow \infty \quad \text{as $n \rightarrow \infty$} \quad \text{at $\omega$} $$ hence $\omega$ is not in the probability one set where $\lim_{n \rightarrow \infty} Y_{t_n}$ converges.

Since $t_n$ depends on $\omega$, this is not much use in finding a probability 1 set where $$\limsup_{t \rightarrow \infty} X_t \leq \sqrt{c} + \frac{1}{m}$$

I'm stuck here. I'd appreciated any help. Thanks.

In case more information is needed, this is problem 7.5.6 in Durrett, 2nd ed. $$X_t = \dfrac{B(t)}{2(1+t)\ln(1+t)},$$ $$Y_t = (1+t)^{-1/2}\exp\left(\frac{B(t)^2}{2(1+t)}\right),$$ $B_t$ is Brownian motion, and $c = \frac{1}{\sqrt{2}}$. The first part of the problem was to show $Y_t$ is a martingale, which I've done.

share|cite|improve this question
Please do not deface your question, even if getting an answer does not interest you anymore. – Did Apr 14 '13 at 7:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.