Let $M_t$ be a $\mathcal F_t$ martingale. That is, for $t>s$ we have $$\mathbb{E}[M_t\mid\mathcal F_s]=M_s,\quad \text{a.s.}$$ Assume $\tau$ is a $\mathcal F_t$ stopping time. Is $(M_{\tau\wedge t})_{t\ge 0}$ a $\mathcal F_t$ martingale? In other words is it true that $$\mathbb{E}[M_{\tau\wedge t}\mid\mathcal F_s]=M_{\tau\wedge s},\quad \text{a.s.}$$ hold?

My thoughts: If the deterministic time $s$ can be considered as $\mathcal F_s$ stopping time, I guess this should follow from optional sampling theorem, but i'm not too sure about this, any help appreciated.

  • $\begingroup$ You are mentioning the name of a result which is much stronger than the question you ask. If your problem is to be sure that every constant time is a stopping time, then yes they are. $\endgroup$
    – Did
    Sep 4, 2017 at 22:09
  • $\begingroup$ @Did Any correct answer to the main question is acceptable. Feel free to answer my main question in what ever fasion you like and I will accept the answer if it seems correct. Thank you. :-) $\endgroup$
    – noidea
    Sep 4, 2017 at 22:27
  • $\begingroup$ Urgent homework? $\endgroup$
    – Did
    Sep 4, 2017 at 22:53
  • $\begingroup$ @Did No, not at all. This seems btw to be a way too trivial question for it to be a homework problem. If you have a look at my profile you will notice that I have been active during the summer to learn some probability theory. Thank you for your input. $\endgroup$
    – noidea
    Sep 4, 2017 at 23:19
  • $\begingroup$ @Did I currently read some more about this, could this simply follow from the fact that each martingale is also a local martingale? $\endgroup$
    – noidea
    Sep 5, 2017 at 10:14

1 Answer 1


Assume $M$ is a supermartingale. First, for a fixed $n\ge 1$, let $$ D_n = \left\{\frac{k}{2^n}, k= 0,1,2,\ldots\right\}\subset D_{n+1}\subset \cdots $$ be the set of non-negative dyadic rationals of order no greater than $n$. It follows that $$ M = (M_t,\mathcal{F}_t; t\in D_n) $$ is a super martingale (discrete time).

Second, we construct a stopping time $T_n$ such that $T_n\ge T$ and $T_n$ only take values in $D_n$. Indeed, let $$ T_n(\omega) = \inf\left\{t\in D_n; t\ge T(\omega)\right\}. $$ Then $T_n\ge T_{n+1} \ge \cdots $ and $T_n$ is a stopping time. Fix $0\le s\le t$, we wish to show that $$ \mathbb{E}\left[M_{t\wedge T}|\mathcal{F_s}\right]\le M_s $$ almost surely. Similarly define $$ t_n = \inf\left\{u\in D_n; u\ge t\right\} \ge t_{n+1}\ge\cdots $$ and $$ s_n = \inf\left\{u\in D_n; u\ge s\right\} \ge s_{n+1}\ge\cdots $$ It follows from the discrete time optional sampling theorem that $$ \mathbb{E}\left[M_{t_n\wedge T_n}|\mathcal{F}_{s_m}\right] \le M_{s_m \wedge T_n} $$ for any integers $m\ge n$. Letting $m\to\infty$, we have $s_m \to s$ decreasing and $\mathcal{F}_{s_m}\to\mathcal{F}_s$. By Lévy's Downward theorem and the continuity of process $M$, we have $$ \mathbb{E}\left[M_{t_n\wedge T_n}|\mathcal{F}_{s}\right] \le M_{s_m \wedge T_n} $$ Observe that $(M_{t_n\wedge T_n},\mathcal{F}_{t_n\wedge T_n})$ is a backward martingale, whence it is uniformly integrable. Letting $n\to\infty$, we arrive at $$ \mathbb{E}\left[M_{t\wedge T}|\mathcal{F}_{s}\right] \le M_{s\wedge T}. $$ This completes the proof.

(from Prof. Hui Wang at Dam Brown University: Continuous time process and Brownian motion)


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .