For a Brownian motion $(B_t)_{t \geq 0}$, do we have $E[(B_\tau - B_\sigma)^2]=E[B_\tau^2 - B_\sigma^2]$ for stopping times $\sigma \leq \tau$?

Question

Let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t \geq 0}, P)$ be a filtered probability space and let $(B_t)_{t \geq 0}$ be a Brownian motion on that space. The question is if the following is true:

For two bounded stopping times $\sigma \leq \tau$, we have $$E[(B_\tau-B_\sigma)^2] = E[B_\tau^2 - B_\sigma^2]$$

If this is true, how can I see that it holds?

Thanks a lot in advance!

Answer 1

For a Brownian motion $(B_t)_{t \geq 0}$ it is well-known that $B_{\tau} \in L^2(\mathbb{P})$ for any bounded stopping time $\tau$, this follows e.g. from Wald's equation and ensures that all the expectations which appear in the remaining part of the answer are finite.

Recall the optional stopping theorem for martingales with continuous sample paths:

Let $(M_t,\mathcal{F}_t)_{t \geq 0}$ be a martingale with continuous sample paths. If $\sigma,\tau$ are bounded stopping times such that $\sigma \leq \tau$ then $$\mathbb{E}(M_{\tau} \mid \mathcal{F}_{\sigma}) = M_{\sigma}$$ (in particular, $M_{\tau} \in L^1(\mathbb{P})$).

The Brownian motion is a martingale with continuous sample paths, and therefore we may apply the above result to obtain $$\mathbb{E}(B_{\tau} \mid \mathcal{F}_{\sigma}) = B_{\sigma}$$ for any two bounded stopping times $\sigma \leq \tau$. Combining this with the fact that $B_{\sigma}$ is $\mathcal{F}_{\sigma}$-measurable, we find that

$$\begin{align*} \mathbb{E}(B_{\sigma} B_{\tau}) &= \mathbb{E} \big[ \mathbb{E}(B_{\sigma} B_{\tau} \mid \mathcal{F}_{\sigma}) \big] \\ &= \mathbb{E} \big[B_{\sigma} \mathbb{E} (B_{\tau} \mid \mathcal{F}_{\sigma}) \big] \\ &= \mathbb{E}(B_{\sigma}^2). \end{align*}$$

Hence,

$$\mathbb{E}((B_{\tau}-B_{\sigma})^2) = \mathbb{E}(B_{\tau}^2)-2 \mathbb{E}(B_{\tau} B_{\sigma}) + \mathbb{E}(B_{\sigma}^2) = \mathbb{E}(B_{\tau}^2)-\mathbb{E}(B_{\sigma}^2).$$