Prove that a stopped martingale is a martingale.

Question

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 $s<t$ $$\mathbb E[M_{t\wedge \tau}\mid \mathcal F_s]=M_{s\wedge \tau}\quad \text{or}\quad \mathbb E[M_{t\wedge \tau}\mid \mathcal F_{s\wedge \tau}]=M_{s\wedge \tau} \ \ \ ?$$

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 $s<t$ 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 ?

Answer 1

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.