Let $M$ be a positive, continuous martingale that converges a.s. to zero as $t$ tends to infinity. I now want to prove that for every $x>0$ $$ P\left( \sup_{t \geq 0 } M_t > x \mid \mathcal{F}_0 \right) = 1 \wedge \frac{M_0}{x}. $$ My approach: I thought that rewriting the conditional probability to an expectation would help so we obtain that we must prove: $$ \mathbb{E} \left[ 1_{\sup_{t \geq 0 } M_t > x} \mid \mathcal{F}_0 \right] = 1 \wedge \frac{M_0}{x}.$$ A hint to me was given that I should consider stopping the process when it gets above $x$. Thus a stopping time that would do this is $\tau = \inf\{t\geq 0 : M_t>x\}$. But now I'm stuck as I want to apply optional sampling results but we have an indicator which complicates things. How could I proceed from this? Any help is appreciated!
1 Answers
For fixed $k \in \mathbb{N}$ define a bounded stopping time $\tau_k$ by
$$\tau_k := \inf\{t \geq 0; M_t > x \} \wedge k = \tau \wedge k.$$
Then, by the optional stopping theorem, we have
$$\mathbb{E}(M_{\tau_k} \mid \mathcal{F}_0) = M_0. \tag{1}$$
On the other hand, since $(M_t)_{t \geq 0}$ is a continuous martingale, we know that
$$M_{\tau_k} = \begin{cases} x 1_{\{\tau<k\}} + M_{k} 1_{\{\tau \geq k\}}, & M_0 \leq x, \\ M_0, & M_0>x \end{cases}. \tag{2}$$
Combining $(1)$ and $(2)$ yields
$$M_0 1_{\{M_0>x\}} + 1_{\{M_0 \leq x\}} x \mathbb{P}(\tau < k \mid \mathcal{F}_0) + \mathbb{E}(M_k 1_{\{\tau \geq k\}} \mid \mathcal{F}_0) = M_0.$$
Finally, the claim follows by letting $k \to \infty$. (For the second term use monotone convergence, for the third one dominated convergence and the fact that $M_k \to 0$ as $k \to \infty$.)
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I'm sorry but how could I find a uniform bound on $M_k 1_{{ \tau \geq k }}$ to apply the DCT? – Apr 20 '15 at 15:14
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3@Rodel By definition $M_k 1_{{\tau \geq k}} \leq \max{x,M_0}$. – saz Apr 20 '15 at 15:54
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@saz Thanks for the high-quality answer (as always). I actually found a solution here (question 5 part (a)): https://www0.maths.ox.ac.uk/system/files/coursematerial/2015/3091/65/pbsheet3_B82_sol.pdf but it did not explain the expansion of $M_0$ so I was really confused. – Math1000 Jul 24 '16 at 07:36
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@saz Thanks for the answer. Can I ask why $M_{\tau_k}=x$ on the set ${M_0\le x}\cap{\tau<k}?$ I understand that on this set $M_{\tau_k}=M_\tau,$ but wouldn't $M_\tau>x$ by the definition of $\tau$, that is, $\inf{t \geq 0; M_t > x }$? If we had "$\ge$" instead of "$>$", this would be clear by continuity, but I'm not sure how to get $M_\tau=x$ from this definition. – Mar 25 '21 at 19:38
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1@sodaowl $M_{\tau}=x$ follows from the continuity of the sample paths. Just suppose that $M_{\tau}(\omega)>x$ for some $\omega$, then, by the continuity of the sample paths, there is some $\delta>0$ such that $M_s(\omega)>x$ for all $s \in (\tau(\omega)-\delta,\tau(\omega)+\delta)$. In particular, $M_s(\omega)>x$ for, say, $s=\tau(\omega)-\delta$. This, however, is a contradiction to the definition of $\tau$ since $\tau(\omega)$ is the infimum of all times $s$ with $M_s(\omega)>x$. – saz Mar 26 '21 at 07:21
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I don't understand why the final equation implies the desired result. I thought that we intuitively we have $$\lim_{k\to\infty} P(\tau<k\mid F_0)=1$$ and $$\lim_{k\to\infty}E(M_k 1_{{\tau \geq k}} \mid F_0)=0$$ but I assume that my intuition is wrong, because the desired result does contain a conditional probability. – Filippo Nov 09 '22 at 07:52