Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The question is to prove $$P\{\sup_{t\geq 0}M_{t}>x\mid \mathcal{F}_{0}\}=\min\left\{1,\frac{M_{0}}{x}\right\},$$ where $M$ is a positive continuous martingale which converges to 0 almost surely as t tends to infinity. I'd like to use the optional sampling theorem in order to prove this. I therefore introduce the stopping time $$\tau=\inf\{t\geq 0\mid M_{t}\geq x\}$$

As $M$ is a martingale we also have that the stopped process is a martingale $$M_{0}=\mathbb{E}[M_{0}\mid\mathcal{F}_{0}]=\mathbb{E}[M_{\tau}\mid\mathcal{F}_{0}]$$ $$=\mathbb{E}[M_{\tau}\mathbb{1}_{\{\tau<\infty\}}\mid\mathcal{F}_{0}]+\mathbb{E}[M_{\tau}\mathbb{1}_{\{\tau=\infty\}}\mid\mathcal{F}_{0}]$$ $$=\mathbb{E}[M_{\tau}\mathbb{1}_{\{\tau<\infty\}}\mid\mathcal{F}_{0}]$$ $$=x\mathbb{P}(\sup_{t}M_{t}\geq x \mid\mathcal{F}_{0})=x\mathbb{P}(\sup_{t}M_{t}> x \mid\mathcal{F}_{0})+x\mathbb{P}(\sup_{t}M_{t}= x \mid\mathcal{F}_{0})$$

Here is where I get stuck. I'd like to prove that $\mathbb{P}(\sup_{t}M_{t}= x \mid\mathcal{F}_{0})$ is equal to $0$. Is there anyone that could help me prove this or correct me if i'm approaching this in the wrong manner?

share|cite|improve this question

If $P(S_t\geqslant x\mid\mathcal F_0)=M_0/x$ for every $x\gt M_0$, then $P(S_t\gt x\mid\mathcal F_0)=M_0/x$ for every $x\gt M_0$.

To wit, let $x\gt M_0$. Then, $[S_t\geqslant y]\subseteq[S_t\gt x]\subseteq[S_t\geqslant x]$ for every $y\gt x$ hence, applying the hypothesis, one sees that $M_0/y\leqslant P(S_t\gt x\mid\mathcal F_0)\leqslant M_0/x$. Considering the limit $y\to x$, $y\gt x$, one gets the result.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.