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?
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).
Proposition: Let $X: (\Omega,\mathcal{A}) \to (D,\mathcal{D})$ be a random variable. Assume that $\mathcal{X}$, $\mathcal{Y}$ are independent $\sigma$-algebras such that $X$ is $\mathcal{X}/\mathcal{D}$-measurable. If $\Psi: D \times \Omega \to \mathbb{R}$ is bounded and $\mathcal{D} \otimes \mathcal{Y}/\mathcal{B}(\mathbb{R})$-measurable, then $$\mathbb{E}(\Psi(X(\cdot),\cdot) \mid \mathcal{X}) = g(X)$$ for $$g(x) := \mathbb{E}(\Psi(x,\cdot)).$$
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 $\sigma \geq \tau$.
