# Expected value of exit time of Brownian motion

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?

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

Hint:

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

• Simpler than I expected.Thank you very much! Nov 22, 2017 at 20:09
• @Blablablu You are welcome.
– saz
Nov 22, 2017 at 20:13
• 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? Nov 22, 2017 at 20:34
• @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.
– saz
Nov 22, 2017 at 20:37
• @Blablablu Ah, I see; sorry, I totally missed that. Doing it componentwise should indeed do the job.
– saz
Nov 22, 2017 at 20:45

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?

• 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. Feb 5, 2023 at 23:36