Let $\tau$ be a stopping time w.r.t a filtration $(F_n)_n$ and $(X_n)_n$ a discrete time martingale w.r.t the same filtration.

Usually in the exercises, in order to apply the optional stopping thm, we have to show that $$P(\tau < \infty)=1$$

Then, we are allowed to write $X_{n \wedge \tau} \rightarrow X_{\tau}$ a.s.

Indeed, since $\tau$ is finite, as $n$ gets larger and larger we have that $\exists N \in \mathbb{N}: \forall n > N: \tau< n$ and hence $$n \wedge \tau \rightarrow \tau$$

My questions then is: if $\tau = \infty$ a.s., why doesn't that hold? I mean, I can't figure out what could be the value of $lim_{n} (n \wedge \tau)$

  • $\begingroup$ Please define $X_\tau$ $\endgroup$
    – J. De Ro
    Jan 11, 2020 at 19:48
  • $\begingroup$ $(X_n)_n$ is a generic martingale. In the exercises often we consider the stopped martinale $(X_{n \wedge \tau})_n$ and the we want to take the limit as $n$ goes to $\infty$. $\endgroup$
    – andereBen
    Jan 11, 2020 at 19:50
  • $\begingroup$ That does not answer my question. How is $X_\tau$ defined? One possibility is that it is $(X_\tau)(\omega) := X_{\tau(\omega)}(\omega)$ and then you need to say what $X_\infty$ should be? $\endgroup$
    – J. De Ro
    Jan 11, 2020 at 19:53
  • $\begingroup$ Yes for me $X_{\tau}$ is exactly what you wrote. But I can't understand what is $X_{\infty}$ $\endgroup$
    – andereBen
    Jan 11, 2020 at 19:54
  • 1
    $\begingroup$ I think that's why you need $\tau < \infty$ a.s., so you don't have to worry about $X_\infty$. Another possible definition is $X_\tau = \sum_{n=0}^\infty X_n I_{\{\tau = n\}}$ so basically this says that we define $X_\tau$ to be $0$ if $\tau = \infty$. $\endgroup$
    – J. De Ro
    Jan 11, 2020 at 19:57

1 Answer 1


If $\tau = \infty$ a.s., we have $X_{\tau\land n} = X_n$ a.s. and this will not (necessarily) converge to $X_\tau$.

  • $\begingroup$ In your equality $n$ is fixed, so yes: $n \wedge \tau = n$ a.s. But what happens if we take $n \rightarrow \infty$? $\endgroup$
    – andereBen
    Jan 11, 2020 at 19:55
  • $\begingroup$ Anything can happen. $(X_n)_n$ doesn't even have to converge. $\endgroup$
    – J. De Ro
    Jan 11, 2020 at 19:58
  • $\begingroup$ Okay, thanks. If I assume that, for instance, the stopped process is uniformly bounded (and $\tau$ is not a.s. finite), i.e. $X_{n \wedge \tau} < C$, then I know that the martingale converges a.s. and in $L^1$, by a convergence theorem. So the problem would be now that I'm not allowed to say that $X_{n \wedge \tau} \rightarrow X_{\tau}$ a.s., right? $\endgroup$
    – andereBen
    Jan 11, 2020 at 20:04
  • $\begingroup$ In general you need that $\tau < \infty$ a.s. to conclude $X_{n \land \tau} \to X_\tau$ a.s.. If you don't have that you will run into problems. $\endgroup$
    – J. De Ro
    Jan 11, 2020 at 20:16
  • $\begingroup$ Clear, thank you. $\endgroup$
    – andereBen
    Jan 11, 2020 at 20:18

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