Suppose that $M:=\{M_t\}_{t\geq0}$ is a martingale adapted to some filtration $\mathcal{F}:=\{\mathcal{F}_t\}_{t\geq0}$ with $M_0\equiv0$ and that $\tau$ is an $\mathcal{F}_t$-stopping time. Suppose that $f:[0,\infty)\to[0,\infty)$ is a function of time. Is it ever the case that


My thinking is that if we define a process $N_t:=f(t)M_t$, then even though $N$ is no longer a martingale, since


it does have zero mean at every point in time.

I'm particularly concerned with the case $f(t)=1/t$. Any references, or key-words to search for are very welcomed (I'm really not sure where to start here). Many thanks in advanced.

  • 1
    $\begingroup$ Even if the process is a martingale, you cannot in general expect the stopped process to be a martingale. Optional Stopping Theorem gives you conditions for when a stopped martingale is a martingale. $\endgroup$
    – user126540
    Commented May 3, 2015 at 14:50
  • $\begingroup$ @Slungpue Yes, I know, but do you have any suggestions how to proceed from there? $\endgroup$
    – jkn
    Commented May 3, 2015 at 15:08
  • $\begingroup$ A correction btw: a stopped martingale is always a martingale, but $E[X_{\tau} \mid \mathcal F_t] = X_t$ doesn't hold in general for a martingale $X_t$. Anyway, I think in this case you result holds only if $f(t)M_t$ is a martingale (although this is only a proposition at this point) and then you have to apply OST. In that case the first step would be to find the set of functions for which $f(t)M_t$ is martingale. $\endgroup$
    – user126540
    Commented May 3, 2015 at 15:39
  • $\begingroup$ I'm not sure what you mean by "is it ever the case" since it seems obviously true in easy cases: Case 1: Suppose $f(t)=0$ for all $t$ (true in this case). Case 2: Define $f(t)$ arbitrarily, but define the stopping time $\tau$ as the first time $t$ when $M(t)=0$. $\endgroup$
    – Michael
    Commented May 12, 2015 at 2:38
  • $\begingroup$ Case 3: Define $f(t)$ arbitrarily, define $\mathcal{F}_t$ so that $\mathcal{F}_0$ includes a random exponential time $X$, but the process $M(t)$ is independent of $X$. However, make the stopping time $\tau=X$. So $E[f(\tau)M(\tau)] = E[f(X)M(X)]= E[f(X)E[M(X)|X]]=0$. $\endgroup$
    – Michael
    Commented May 12, 2015 at 2:40

1 Answer 1


In general, we cannot expect this. Just consider e.g. a Brownian motion $(M_t)_{t \geq 0}$ and the stopping time $$\tau := \inf\{t>0; M_t = 1\}.$$ Then $$\mathbb{E}(f(\tau) M_{\tau}) = \mathbb{E}f(\tau)=0 \iff f=0,$$ i.e. the claim holds only true for the (trivial) function $f:=0$.

  • $\begingroup$ I agree, this won't hold in general. But I'm wondering if it does under conditions similar to those needed for optional sampling. $\endgroup$
    – jkn
    Commented May 3, 2015 at 15:07
  • $\begingroup$ Hey, the time to award the bounty was running out, so I just awarded it to the only answer; yours. Thank you for replying in the first place. $\endgroup$
    – jkn
    Commented May 12, 2015 at 12:39
  • $\begingroup$ @jkn I'm sorry that I can't add any useful (sufficient) conditions/assumptions....in this generality, your question is really hard to answer. $\endgroup$
    – saz
    Commented May 12, 2015 at 12:45
  • $\begingroup$ It ok, you are right, it is hard to answer. I was hoping this may be a standard result that someone would recognise. Thanks for the effort anyway. $\endgroup$
    – jkn
    Commented May 13, 2015 at 6:43

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