# Prove that a stopped martingale is a martingale.

Let $$(M_t)$$ be a martingale w.r.t. $$(\mathcal F_t)_{t\geq 0}$$ and $$\tau$$ is a stopping time. I must prove that $$(M_{t\wedge \tau})$$ is a martingale. What should I prove ? That for $$s $$\mathbb E[M_{t\wedge \tau}\mid \mathcal F_s]=M_{s\wedge \tau}\quad \text{or}\quad \mathbb E[M_{t\wedge \tau}\mid \mathcal F_{s\wedge \tau}]=M_{s\wedge \tau} \ \ \ ?$$

I guess it's the second one, but in several post on MSE I saw that they tried to prove that $$\mathbb E[M_{t\wedge \tau}\mid \mathcal F_s]$$, so I'm a bit confused. What do you think ?

So, what I tried : Let $$s and $$F\in \mathcal F_{s\wedge \tau}$$. Then \begin{align*} \mathbb E[\mathbb E[M_{t\wedge \tau}\mid \mathcal F_{s\wedge \tau}]\boldsymbol 1_F]&=\mathbb E[M_{\tau}\boldsymbol 1_{F\cap \{\tau\leq t\}}]+\mathbb E[M_t\boldsymbol 1_{F\cap \{\tau>t\}}].\\ \end{align*} How can I continue ?

You want to prove $$\mathbb{E}[M_{t \wedge \tau} | \mathcal F_s] = M_{s \wedge \tau}$$ because you want to show $$(M_{t \wedge \tau})$$ is a martingale with respect to $$(\mathcal F_t)$$, not $$(\mathcal F_{t \wedge \tau})$$.
We will use that $$M$$ is a martingale iff $$\mathbb{E}[M_\sigma] = \mathbb{E}[M_0]$$ for all bounded stopping times $$\sigma$$. Letting $$M^\tau$$ denote the stopped process (so $$M^\tau_t = M_{t \wedge \tau}$$) and let $$\sigma$$ be a bounded stopping time, we have \begin{align*} \mathbb{E}[M^\tau_\sigma] &= \mathbb{E}[M_{\sigma \wedge \tau}] = \mathbb{E}[M_0] = \mathbb{E}[M_{0 \wedge \tau}] = \mathbb{E}[M^\tau_0] \end{align*} so $$M^\tau$$ is a martingale.
• What I don't understand is that $M_{t\wedge \tau}$ is not $\mathcal F_t-$measurable. So, how can $(M_{t\wedge \tau})$ can be a martingale w.r.t. $(\mathcal F_t)$ ? Apr 22, 2021 at 8:28
• $M_{t\wedge \tau}$ is $\mathcal F_t$ measurable, at least as long as we have some basic path regularity assumptions (e.g. right continuity of paths). If $\tau$ has only finitely many possible values it's clear that $M_{t \wedge \tau}$ is $\mathcal F_t$ measurable, and there exists a sequence of stopping times $(\tau_n)$ taking only finitely many possible values and decreasing to $\tau$. Each $M_{t \wedge \tau_n}$ is $\mathcal F_t$ measurable, and by right continuity $M_{t \wedge \tau_n} \rightarrow M_{t \wedge \tau}$ pointwise so $M_{t \wedge \tau}$ is also $\mathcal F_t$ measurable. Apr 22, 2021 at 14:09