Harmonic functions and Brownian motion

Question

How can I prove that harmonic functions have the mean-value property using Brownian motion ${B_t}$?

I know that I need to use the fact that $B_{t\wedge\tau}$ is a martingale where $\tau$ is a stopping time which denotes the first time the Brownian motion leaves a ball with radius $r$ around $B_0$

Thanks

Answer 1

Fix a ball $B(x,r)$ and consider $\tau= \inf{t>0, |B_t-x| = r}$.

Now compute by Itô's formula $$f(B_\tau) = f(B_0) + \int_0^\tau f'(B_s)\, dB_s + \frac{1}{2}\int_0^\tau \Delta f(B_s)\, ds $$ Take expectations: $$\Bbb{E}[f(B_\tau)] = f(B_0) + \Bbb{E}\bigg[\int_0^\tau f'(B_s)\, dB_s\bigg] + \frac{1}{2}\int_0^\tau \Delta f(B_s)\, ds $$

Since $\Bbb{E}\big[\int_0^\tau f'(B_s)\, dB_s\big] = 0$ and $\Delta f\equiv 0$ we obtain that

$$\Bbb{E}[f(B_\tau)] = f(B_0)$$

Now it suffices to take a Brownian motion starting at $x$ and to note by rotation invariance that $$\Bbb{E}[f(B_\tau)] = \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} f(x)\, dS$$ and therefore you obtain the mean property

$$f(x) =\Bbb{E}[f(B_\tau)] = \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} f(x)\, dS$$

remark: It is important to note that $\tau$ is a finite almost surely stopping time and that $|B_\tau - x| = r$