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Let $B_t$ be $n$-dimensional Brownian motion. Let $\tau$ be the stopping time $\tau=\inf(t\in \mathbb R_+: |B_t-x| \ge r)$ with $x \in \mathbb R^n $ and $r>0$ i.e. the first exit time from a ball with radius $r$ around some point $x$. Assume we already know $\mathbb E(\tau)<\infty$

Show $\mathbb E(\tau)=\begin{cases} \frac{r^2-|x|^2}{n}, & \text{$|x|<r$} \\ 0, & \text{else} \end{cases}$

I was already able to show that $\mathbb E(\tau)=\frac 1 n \mathbb E(|B_\tau|^2)$ but I don't know how to compute $\mathbb E(|B_\tau|^2)$. I am used to hitting times where I can simply plug in the value for the stopped process but I am not sure what to do here. I can show the case where it equals $0$ but what about the other? It makes sense that it will be $r^2-|x|^2$ but what would be the right way to show this?

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  • $\begingroup$ Are we to assume that $B_0=x$? $\endgroup$ Nov 23, 2017 at 16:00
  • $\begingroup$ @JohnDawkins No, I think we have $B_0=0$. That is the reason why $\mathbb E(\tau)=0$ for $|x| \ge r$ $\endgroup$
    – Blablablu
    Nov 23, 2017 at 18:13

2 Answers 2

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Hint:

$$\mathbb{E}(B_{\tau}^2) = \mathbb{E}[((B_{\tau}-x)+x)^2]=\mathbb{E}(|B_{\tau}-x|^2) + 2x (\underbrace{\mathbb{E}(B_{\tau})}_{0}-x) + x^2$$

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  • $\begingroup$ Simpler than I expected.Thank you very much! $\endgroup$
    – Blablablu
    Nov 22, 2017 at 20:09
  • $\begingroup$ @Blablablu You are welcome. $\endgroup$
    – saz
    Nov 22, 2017 at 20:13
  • $\begingroup$ just one more thing, with the absolute value we end up with: $\mathbb{E}(|B_{\tau}|^2) = \mathbb{E}[((|B_{\tau}|-|x|)+|x|)^2]=\mathbb{E}(||B_{\tau}|-|x||^2) + 2|x| \mathbb{E}(|B_{\tau}|)-|x|) + |x|^2=\mathbb{E}(||B_{\tau}|-|x||^2)-|x|^2$ But in the very last term the triangle inequality is not in the desired direction. Which part am I missing? $\endgroup$
    – Blablablu
    Nov 22, 2017 at 20:34
  • $\begingroup$ @Blablablu Why would you want to do it this way ...? Just note that $$\mathbb{E}(|B_{\tau}|^2) = \mathbb{E}(B_{\tau}^2)$$ and then proceed as my hint suggests. $\endgroup$
    – saz
    Nov 22, 2017 at 20:37
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    $\begingroup$ @Blablablu Ah, I see; sorry, I totally missed that. Doing it componentwise should indeed do the job. $\endgroup$
    – saz
    Nov 22, 2017 at 20:45
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I want to try another solution to see if it is correct:

I will use the optimal stoping time to the martingale $B_t^2-n\cdot t$ which is a martingale because is the sum of $(B_t^i)^2-t$ martingales (This is one of the tipical examples in al the text books).

Because $B_t^2-n\cdot t$ is a martingale we have that: $|x|^2-n\cdot0=E[B_{\tau_r}^2-n\cdot\tau_r]=|a|^2-n\cdot E[\tau_r]$, so $E[\tau_r]=\frac{|a|^2-|r|^2}{n}$.

Is it correct?

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  • $\begingroup$ Maybe some argument is missing because the optional stopping Theorem only gives that $(B_{\min(t,\tau_r)}^2-n\min(t,\tau_r))_{t\in[0,\infty)}$ is a martingale. $\endgroup$ Feb 5, 2023 at 23:36

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