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Let $B_t$ be a Brownian motion. Let $\tau$ be a stopping time with $\mathbb E(\tau)< \infty $

Show that $\mathbb E(B_\tau)=0$

I know that since $B_t$ is a martingale we have $\mathbb E(B_{\tau \wedge n})=B_0=0$ but I can't see how I could use any covergence theorem to take the limit into the expactation. Is this the right approach at all? Any ideas?

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Since $(B_t^2-t)_{t \geq 0}$ is a martingale, it follows from the optional stopping theorem that

$$\mathbb{E}(B_{\tau \wedge n}^2) = \mathbb{E}(\tau \wedge n).$$

This implies

$$\mathbb{E}((B_{\tau \wedge n}-B_{\tau \wedge m})^2) = \mathbb{E}(\tau \wedge n)-\mathbb{E}(\tau \wedge m) \xrightarrow[]{m,n \to \infty} 0.$$

This shows that $(B_{\tau \wedge n})_{n \geq 1}$ is an $L^2$-Cauchy sequence and so $B_{\tau \wedge n} \to B_{\tau}$ in $L^2$. Hence, in particular, $B_{\tau \wedge n} \to B_{\tau}$ in $L^1$ and so $$\mathbb{E}(B_{\tau \wedge n}) \xrightarrow[]{n \to \infty} \mathbb{E}(B_{\tau}).$$

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    $\begingroup$ Thank you very much! Just a question: why exactly does it imply $\mathbb{E}((B_{\tau \wedge n}-B_{\tau \wedge m})^2) = \mathbb{E}(\tau \wedge n)-\mathbb{E}(\tau \wedge m)$? Why do we have the square of the difference and not the difference of the squares? $\endgroup$ – PeterGarder Nov 21 '17 at 18:06
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    $\begingroup$ @PeterGarder For any (square-integrable) martingale $(M_n)_{n \in \mathbb{N}}$ we have $$\mathbb{E}(M_n M_m \mid \mathcal{F}_m) = M_m^2 \tag{1}$$ for any $m \leq n$. On the other hand, $$\mathbb{E}((M_n-M_m)^2) = \mathbb{E}( \mathbb{E}((M_n-M_m)^2 \mid \mathcal{F}_m)).$$ Now expand the square and use $(1)$ to show that the right-hand side equals $\mathbb{E}(M_n^2)-\mathbb{E}(M_m^2)$. $\endgroup$ – saz Nov 21 '17 at 18:21
  • $\begingroup$ makes perfect sense now. Thanks again! $\endgroup$ – PeterGarder Nov 21 '17 at 18:29
  • $\begingroup$ @PeterGarder You are welcome. $\endgroup$ – saz Nov 21 '17 at 18:35

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