How can we show $\operatorname E\left[Y_\tau;t\ge\tau\mid\tau=s\right]=1_{\{\:t\:\ge\:s\:\}}\operatorname E\left[Y_s\right]$?

Question

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a probability space;
• $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$;
• $(Y_t)_{t\ge0}$ be a real-valued $(\mathcal F_t)_{t\ge0}$-progressive integrable process on $(\Omega,\mathcal A,\operatorname P)$;
• $\tau$ be an $(\mathcal F_t)_{t\ge0}$-stopping time on $(\Omega,\mathcal A)$.

Let $s,t\ge0$. Can we show that $$\operatorname E\left[Y_\tau;t\ge\tau\mid\tau=s\right]=1_{\{\:t\:\ge\:s\:\}}\operatorname E\left[Y_s\right]\tag1?$$

I've seen this identity, but have no idea why it holds. Shouldn't we have $$\operatorname E\left[Y_\tau;t\ge\tau\mid\tau=s\right]=1_{\{\:t\:\ge\:s\:\}}\operatorname E\left[Y_s;\tau=s\right]\tag2$$ instead?

Answer 1

You are right. (1) is wrong. Here is a counter-example: let $(Y_t)$ be Brownian motion and $\tau=\inf \{t \geq 0: Y_t=1\}$. Then $Y_{\tau}=1$. If we take $s \leq t$ the LHS of (1) is $1$ and RHS is $0$.