Say $B$ a 2-dim. BM started from $0$ and let $\tau_D$ be its exit time from a bounded domain $D$. Let also $\sigma_i^{\epsilon}$ be the exit time from the circle of center $B_{\sigma_{i-1}}$ and radius $\epsilon$. I read a claim that given $n \lt \tau_D$:
$$ E[|B_{\sigma_n^{\epsilon}}|^2] = \sum_{j =1}^n E[|B_{\sigma^{\epsilon}_j} – B_{\sigma^{\epsilon}_{j-1}}|^2]$$
And I can't figure out why this would be true. Can someone shed light on this passage?
Best Answer
I will ignore the restriction $n < \tau_D$ for the reason written in the comments and just show that the result holds for a BM in the whole space with no consideration of a domain $D$.
You can write $$\mathbb{E}[|B_{\sigma_n^\varepsilon}|^2] = \mathbb{E}[(B_{\sigma_n^\varepsilon} - B_{\sigma_{n-1}^\varepsilon} + B_{\sigma_{n-1}^\varepsilon})^2] = \mathbb{E}[(B_{\sigma_n^\varepsilon} - B_{\sigma_{n-1}^\varepsilon})^2 + |B_{\sigma_{n-1}^\varepsilon}|^2 + 2(B_{\sigma_n^\varepsilon} - B_{\sigma_{n-1}^\varepsilon})B_{\sigma_{n-1}^\varepsilon}].$$
First, consider $$\mathbb{E}[(B_{\sigma_n^\varepsilon} - B_{\sigma_{n-1}^\varepsilon})B_{\sigma_{n-1}^\varepsilon}].$$ By the Strong Markov property, $B_t^{(n-1)} = B_{t + \sigma_{n-1}^\varepsilon} - B_{\sigma_{n-1}^\varepsilon}$ is a Brownian motion that is independent of $B_{\sigma_{n-1}^\varepsilon}$. Note that $\sigma_n^\varepsilon = \tau + \sigma_{n-1}^\varepsilon$ where $\tau$ is the exit time of the ball of radius $\varepsilon$ about $0$ of $B^{(n-1)}$. Therefore $$\mathbb{E}[(B_{\sigma_n^\varepsilon} - B_{\sigma_{n-1}^\varepsilon})B_{\sigma_{n-1}^\varepsilon}] = \mathbb{E}[B_\tau^{(n-1)} B_{\sigma_{n-1}^\varepsilon}] = \mathbb{E}[B_\tau^{(n-1)}]\mathbb{E}[B_{\sigma_{n-1}^\varepsilon}] = 0 \cdot \mathbb{E}[B_{\sigma_{n-1}^\varepsilon}] = 0 $$ where the third to last equality is by independence and the second to last is by symmetry of the BM $B^{(n-1)}$.
This means that $$\mathbb{E}[|B_{\sigma_n^\varepsilon}|^2] = \mathbb{E}[(B_{\sigma_n^\varepsilon} - B_{\sigma_{n-1}^\varepsilon})^2] + \mathbb{E}[|B_{\sigma_{n-1}^\varepsilon}|^2]$$ which forms the basis for a simple proof by induction yielding the desired result.