• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $(\mathcal F_t)_{t\ge0}$ be a filtration of $\mathcal A$
  • $M$ be a continuous local $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$ with $M_0=0$

Let $\varepsilon>0$. Note that $$\tau:=\inf\left\{t\ge0:\left|M_t\right|\ge\varepsilon\right\}$$ is an $\mathcal F$-stopping time with $$\left|M^\tau\right|\le\varepsilon\;.\tag1$$

How can we conclude that $M^\tau$ is an $\mathcal F$-martingale?

If $\tau$ would be bounded, then $M^\tau$ would be a local $\mathcal F$-martingale by the optional stopping theorem and hence be an $\mathcal F$-martingale by $(1)$. However, since $\tau$ isn't bounded, I don't know we can conclude.

  • 1
    $\begingroup$ The boundedness can't be relevant here, as the definition of martingale contains only fixed times. $\endgroup$
    – zhoraster
    Jun 10, 2017 at 15:19
  • $\begingroup$ For me this is basically the definition of the term "local martingale". What is your definition? $\endgroup$ Jun 10, 2017 at 15:20
  • 2
  • $\begingroup$ @zhoraster What I've meant is the following: If $\sigma,\tau$ are stopping times, $\tau$ is bounded and $M$ is a martingale, then $M_{\sigma\wedge\tau}=\text E[M_\tau\mid\mathcal F_\sigma]$. This can be used to show that $M^\tau$ is a martingale. Now I would use this to proof that if $\tau$ is a bounded stopping time and $M$ is a local martingale, then $M^\tau$ is a local martingale. Indeed, if $(\sigma_n)_n$ is a localizing sequence, then $(M^\tau)^{\sigma_n}=(M^{\sigma_n})^\tau$ is a martingale by the former claim. $\endgroup$
    – 0xbadf00d
    Jun 10, 2017 at 15:28
  • $\begingroup$ @NateEldredge $M$ is a local martingale iff there is a sequence $(\sigma_n)_n$ of increasing stopping times with $\sigma_n\to\infty$ such that $M^{\sigma_n}$ is a martingale for all $n$. $\endgroup$
    – 0xbadf00d
    Jun 10, 2017 at 15:29

1 Answer 1


Let $s < t$. We have to show that $E[M_t^\tau \mid \mathcal{F}_s] = M_s^\tau$ almost surely.

For each $n$ we know, as you mentioned, that $(M^\tau)^{\sigma_n} = M^{\tau \wedge \sigma_n}$ is a martingale, where $\sigma_n$ is the localizing sequence of stopping times. Thus $E[M^{\tau \wedge \sigma_n}_t \mid \mathcal{F}_s] = M^{\tau \wedge \sigma_n}_s$ a.s. (*)

(If you are worried about the boundedness of stopping times here, you can assume without loss of generality that each $\sigma_n$ is bounded, by replacing it with $\sigma_n \wedge n$.)

Now $M_s^{\tau \wedge \sigma_n} = M_{s \wedge \tau \wedge \sigma_n}$. By assumption $\sigma_n \uparrow \infty$ a.s., so $M_s^{\tau \wedge \sigma_n} \to M_s^\tau$ almost surely. The same holds with $t$ in place of $s$. Moreover, $|M_s^{\tau \wedge \sigma_n}| \le \varepsilon$ a.s. for all $n$.

By the conditional dominated convergence theorem, with dominating function $\varepsilon$, we can conclude $E[M^{\tau \wedge \sigma_n}_t \mid \mathcal{F}_s] \to E[M^{\tau}_t \mid \mathcal{F}_s]$ a.s. So we pass to the a.s. limit on both sides of (*) and get the result.

Note that this same argument shows the following version of the optional stopping theorem, which is really good to know:

Let $M_t$ be a continuous martingale and $\tau$ a (not necessarily bounded!) stopping time. Suppose that the process $M_{t \wedge \tau}$ is bounded, i.e. there is a constant $C$ such that $|M_{t \wedge \tau}| \le C$ a.s. for all $t$. (You might say: "$M$ is bounded up to time $\tau$.") Then $E[M_\tau] = E[M_0]$.

This is the version that Evan Aad was referring to in the question I linked in my comment. More generally, it also holds if $\{M_{t \wedge \tau} : t \ge 0\}$ is uniformly integrable. As a general principle for martingales, anything that's true for bounded stopping times $\tau$ will also be true if the process is bounded up to $\tau$.

  • $\begingroup$ 1. How do you conclude that $(M^\tau)^{\sigma_n}$ is a martingale? I guess the argument should be that $$(M^\tau)^{\sigma_n\:\wedge\:n}=(M^{\sigma_n})^{\tau\:\wedge\:n}\;.$$ 2. You're using the continuity of $M$ in your proof, but you don't assume continuity of the $M$ in your version of the OST. Did you forget to add this assumption? $\endgroup$
    – 0xbadf00d
    Jun 11, 2017 at 14:18
  • $\begingroup$ @0xbadf00d: (1) $\sigma_n \wedge \tau \wedge n$ is a bounded stopping time, so as you know, $M^{\sigma_n \wedge \tau \wedge n}$ is a martingale. (2) I do not think I am using the continuity of $M$ in my proof. Note that the sequence $\tau \wedge \sigma_n$ not only converges a.s. to $\tau$, it is eventually constant. $\endgroup$ Jun 12, 2017 at 1:59
  • $\begingroup$ (1) I'm sorry, but I still don't get it. First of all, if I got you right, $C$ is a constant (not an (integrable) random variable) and we even assume "$|M^\tau_t|\le C$ for all $t\ge0$ a.s." (which is stronger than "$|M^\tau_t|\le C$ a.s. for all $t\ge0$", unless $C$), right? (2) I guess you want to use that $(M^\tau)^n=M^{\tau\:\wedge\:n}$ is a martingale (noting that $\tau\wedge n$ is bounded) by the OST. However, the continuous-time version of the OST doesn't hold, unless $M$ (and the filtration $\mathcal F$!) is right-continuous (a.s.). But you don't want to assume (right-)continuity ... $\endgroup$
    – 0xbadf00d
    Jun 18, 2017 at 19:12
  • $\begingroup$ (3) In order to shorten this, I would appreciate, if you could provide a proof of your statement, since I really want to understand why it holds. Thank you very much in advance! $\endgroup$
    – 0xbadf00d
    Jun 18, 2017 at 19:12
  • $\begingroup$ @0xbadf00d: Oh, good point, I didn't think about hypotheses of the OST. This isn't a good time for me to study it carefully so I will just add the word "continuous". $\endgroup$ Jun 18, 2017 at 19:22

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