brownian motionconditional-expectationprobability theorystochastic-processesstopping-times
Let $W$ be a Brownian motion with filtration $(F_t)$. Let $\tau$ be a stopping time.
It is well-known by the strong Markov property that the law of $W_{\tau+t}-W_\tau$ given $F_\tau$ is normal with mean zero, variance $t$.
I am interested in a very minor extension of this result and I would like to prove that the law of $W_{\tau\vee t}-W_\tau$ given $F_\tau$ is normal with mean zero and variance $(\tau\vee t)-\tau$. How can I do it?
$$\{\tau_x\leq t\}=\{\max_{k\leq t}B_k\geq x\}.$$
$$\mathcal{F}_t=\sigma(B_s: 0\leq s\leq t).$$
By definition of filtration, $\{B_k\geq x\}\in \mathcal{F}_t$ for all $k\in [0,t]$. So it's a stopping time.
More details:
The function $B(t,\omega):=B_t(\omega)$ is measurable with respect to the product sigma algebra $\mathcal{B}\times \mathcal{F}_t$, where $\mathcal{B}$ is the borel sigma algebra (generated by open intervals). Hopefully this is something you've proven in your class (or perhaps taken as a given). Otherwise, see Lemma 1.1 of this for details. Then just like in classical real analysis, if $f_n$ is measurable, then so is $\sup_n f_n, \max_n f_n$, etc.
No, it's not at all obvious.
If we interpret $\mathcal{F}_t$ as "information up to time $t$", it's not surprising that the hitting time of an open interval $(a,b)$ is not an $\mathcal{F}_t$-stopping time. For instance if $B_t(\omega)=a$ for some $t>0$, then the information about the past, i.e. $(B_s(\omega))_{s \leq t}$, is not enough to decide whether $\tau(\omega)=t$; we need a small glimpse into the future. 
Making this intuition rigorous is, however, not easy. One possibility is to apply Galmarino's test which states that a mapping $\tau: \Omega \to [0,\infty]$ is a stopping time (with respect to $(\mathcal{F}_t)_{t \geq 0}$) if and only if $$\tau(\omega)=t, B_s(\omega) = B_s(\omega') \, \, \text{for all $s \leq t$} \implies \tau(\omega')=t, \tag{1}$$ see this question. If $\tau$ is the hitting time of an open interval $(1)$ is not satisfied; the easiest way to see this is to consider the canonical Brownian motion, i.e. consider $(B_t)_{t \geq 0}$ as a process on the space of continuous mappings.
Let me finally remark that there are other ways to prove that the filtration generated by a Brownian motion is not right-continuous, see, for instance, this answer.
Best Answer
If $\tau$ is a (finite) stopping time, then the process
$$B_s := B_{s+\tau}-B_{\tau}, \qquad s \geq 0$$
is a Brownian motion which is independent of $\mathcal{F}_{\tau}$. As
$$W_{\max\{t,\tau\}}-W_{\tau} = B_{\max\{0,t-\tau\}},$$
we have
$$\mathbb{E}\big(f(W_{\max\{t,\tau\}}-W_{\tau}) \mid \mathcal{F}_{\tau} \big) = \mathbb{E} \big( f(B_{\max\{0,t-\tau\}}) \mid \mathcal{F}_{\tau} \big)$$
for any bounded measurable function $f$. Since $\max\{0,t-\tau\}$ is $\mathcal{F}_{\tau}$-measurable and $(B_s)_{s \geq 0}$ is independent from $\mathcal{F}_{\tau}$, it follows (see below for details) that
$$\mathbb{E}\big(f(W_{\max\{t,\tau\}}-W_{\tau}) \mid \mathcal{F}_{\tau} \big) = g(\max\{0,t-\tau\}) \tag{1}$$ where $$g(s) := \mathbb{E}f(B_s).$$
Using this identity for $f(x) := \exp(ix \xi)$ with $\xi \in \mathbb{R}$ fixed, we get
\begin{align*} \mathbb{E} \left( \exp \left[i \xi (W_{\max\{t,\tau\}}-W_{\tau}) \right] \mid \mathcal{F}_{\tau} \right) &= \exp \left(- \frac{\max\{0,t-\tau\}}{2} \xi^2 \right) \\ &= \exp \left(- \frac{\max\{t,\tau\}-\tau}{2} \xi^2 \right) \end{align*}
which proves the assertion.
To prove $(1)$ rigorously, we can use the following property of conditional expectation (which you can find, in this particular formulation, in the book on Brownian motion by Schilling & Partzsch, Lemma A.3).
To prove $(1)$ we choose the objects as follows:
Let's check the assumptions of the proposition: As already mentioned earlier on, the Brownian motion $(B_s)_{s \geq 0}$ is independent from $\mathcal{F}_{\tau}$, i.e. $\mathcal{X}$ and $\mathcal{Y}$ are independent. Moreover, $\tau$ is $\mathcal{F}_{\tau}$-measurable (i.e. $\mathcal{X}$-measurable) and therefore $X$ is $\mathcal{X}$-measurable. Moreover, the progressive measurability of $(B_s)_{s \geq 0}$ implies that $\Psi$ is $\mathcal{D} \otimes \mathcal{Y}/\mathcal{B}(\mathbb{R})$-measurable.
Since we have verified all assumptions, we may apply the above proposition and this gives exactly $(1)$.
Remark: The above reasoning shows, more generally, that $W_{\sigma}-W_{\tau}$ given $\mathcal{F}_{\tau}$ is Gaussian with mean $0$ and variance $\sigma-\tau$ for any stopping time $\sigma$ which is $\mathcal{F}_{\tau}$-measurable and satisfies $\sigma \geq \tau$.