# For a stopping time does $E[(\tau \wedge t)1_A]=E[(\tau \wedge s)1_A]$ for $s \le t$ and any $A \in \mathcal F_{\tau \wedge s }$?

I have the following question as stated in the title:

For a stopping time $$\tau$$ does $$E[(\tau \wedge t)1_A]=E[(\tau \wedge s)1_A]$$ for $$s \le t$$ and any $$A \in \mathcal F_{\tau \wedge s }$$?

Here $$\mathcal F_{\tau \wedge s }$$ is the $$\sigma$$-algebra of $$\tau \wedge s$$-past.

The context is that I'm trying to show that $$(B^2_{\tau \wedge t } - \tau \wedge t,\mathcal F_{\tau \wedge t })$$ is a martingale and the question is here is a second step in showing that $$E[\tau \wedge t|\mathcal F_{\tau \wedge s} ]=\tau \wedge s$$.

• Consider the case where $A=\Omega$ and $\tau$ is a constant stopping time with the value $\tau_0$. Then certainly we should not expect $\tau_0\wedge t=E[\tau\wedge t]=E[\tau\wedge s]=\tau_0\wedge s$ to hold in general. – Sangchul Lee Sep 25 '19 at 17:04

We have \begin{alignat*}{2} \mathbb{E}[(\tau \wedge t)\mathbf{1}_A] & = \mathbb{E}\Big[\mathbb{E}\left[(\tau \wedge t)\mathbf{1}_A\;|\;\mathcal F_{\tau \wedge s }\right]\Big]\quad\text{by the tower property}\\ & = \mathbb{E}\Big[\mathbb{E}\left[(\tau \wedge t)\;|\;\mathcal F_{\tau \wedge s }\right]\mathbf{1}_A\Big]\quad\text{by measurability}\\ & \neq \mathbb{E}\Big[\mathbb{E}\left[(\tau \wedge s)\;|\;\mathcal F_{\tau \wedge s }\right]\mathbf{1}_A\Big]\\ & = \mathbb{E}\left[(\tau \wedge s)\mathbf{1}_A\right], \end{alignat*} so answer seems no. Counterexample: $$\tau=\infty$$ almost surely.
If $$B$$ is a standard Brownian motion, then $$\mathbb E\left[ B^2_{t}\right] = {t}$$ and $$\mathbb E\left[ B^2_{t}\;\;|\;\; \mathcal{F}_s\right] = B_s^2 + {t-s}$$.
What you need to show seems $$\mathbb E\left[ B_{\tau\wedge t}^2 - {\tau\wedge t} \;\;|\;\; \mathcal{F}_{\tau\wedge s}\right] = B_{\tau\wedge s}^2 - {\tau\wedge s}$$ for which you need $$(B_t-B_s)\amalg\mathcal F_s$$ and $$(B_t-B_s)\sim B_{t-s}$$.
• What about if $\tau <\infty$ a.s.? – MrFranzén Sep 25 '19 at 16:34
• If $\tau>t$ a.s., it still doesn't work at least. But here it seems you have a BM? – thomasb Sep 25 '19 at 16:37
• Thank you for your answer. I realize now that I took the wrong approach. Of course I should apply the optional stopping theorem to the continous martingale $B_t-t$ and not $B_t$ and then trying to show the matringale property separately for $\tau \wedge t$. – MrFranzén Sep 25 '19 at 17:53