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Let $X_t$ be a continuous non-negative martingale with $\lim_{t\to \infty}X_t=0$ a.s. and $X_0 = 1$.

Then how to show \begin{equation} P(\sup X_t\ge x)=\frac1x \end{equation} for any $x\ge1$ and $t\ge 0$?

I want to introduce stopping time $T_x$, but clearly I can not use optional stopping time here ($T_x$ is not a.s. finite). What should I do?

Filippo
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user98619
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    "clearly I cannot use optional stopping time here" Hmmm... why? – Did May 01 '14 at 06:51
  • @Did Maybe I am wrong?Because $T_x$ is even not a.s. finite...... – user98619 May 01 '14 at 11:29
  • Then do as one always do, first use $\min(t,T_x)$ then let $t\to\infty$. – Did May 01 '14 at 12:01
  • @Did It works,thanks! – user98619 May 01 '14 at 12:07
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    It looks like the OP was able to solve the problem, but I would like to know how $T_x$ is defined and to what martingale we are applying the OST... – Filippo Nov 07 '22 at 08:32
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    @Filippo The answer is found here , which means that this is a duplicate question. It is a generalization, given that $M_0$ is assumed to have a more arbitrary distribution than what is in this question. – Sarvesh Ravichandran Iyer Nov 07 '22 at 10:00
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    This is a duplicate question, a slightly more general one but close enough to this one. – Sarvesh Ravichandran Iyer Nov 07 '22 at 10:03
  • @SarveshRavichandranIyer Thank you very much! I am typing on my phone, I will study the answers as soon as I am back home. – Filippo Nov 07 '22 at 10:40
  • @SarveshRavichandranIyer I think that you sent the same link twice, didn't you? Did you do this on purpose or is there as second duplicate? – Filippo Nov 07 '22 at 10:54
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    @Filippo Sorry : the first comment was addressed to you, so that I could inform you that the question was answered very well elsewhere. The second comment is the default duplicate comment when one votes to close as a duplicate ("Does this answer your question?" ...) but I am pretty confident so I reworded that comment to make it an assertion. The link in both comments is the same, but one is meant for you to look at and the other for duplicate closers. – Sarvesh Ravichandran Iyer Nov 07 '22 at 11:04
  • @SarveshRavichandranIyer Thank you for the explanation! – Filippo Nov 07 '22 at 11:06
  • @Filippo You are welcome! – Sarvesh Ravichandran Iyer Nov 07 '22 at 11:06

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