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 ?


1 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.

  • $\begingroup$ 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. $(\mathcal F_t)$ ? $\endgroup$
    – joshua
    Apr 22, 2021 at 8:28
  • 1
    $\begingroup$ $M_{t\wedge \tau}$ 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. $\endgroup$ Apr 22, 2021 at 14:09

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