# $\tau$ be a stopping time, $(M_t)_{t\ge 0}$ a martingale. Is $(M_{\tau\wedge t})_{t\ge 0}$ a martingale?

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.

• 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.
– Did
Sep 4, 2017 at 22:09
• @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. :-) Sep 4, 2017 at 22:27
• Urgent homework?
– Did
Sep 4, 2017 at 22:53
• @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. Sep 4, 2017 at 23:19
• @Did I currently read some more about this, could this simply follow from the fact that each martingale is also a local martingale? Sep 5, 2017 at 10:14

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.