# Brownian motion and exit time

brownian motionprobability theory

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?

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.