Rewriting probabilities as expectation

Question

Consider the stopping time $\tau_a:=\lbrace{t>0| W_t >a\rbrace}$, where $W_t$ is a Brownian Motion.

Define: $X_t:=W_{\tau_a+t}-W_{\tau_a}$. We have that $X_t$ is a Brownian Motion independent of $\mathcal{F}_{\tau_a} $.

Clearly we have: $\Pr(\tau_a\leq t, W_t\leq a)=\Pr(\tau_a\leq t,X_{t-\tau_a}+W_{\tau_a}\leq a)$ (1).

The question is, why can I write (1) in the following:

$$(1)=\mathbb{E}[1_{\lbrace{\tau_a≤t\rbrace}}\cdot\Pr(X_{t-\tau_a}\leq 0\mid\mathcal{F}_{\tau_a})]$$

I would like to understand every single passage.

Additionally, why is $X_{t-\tau_a}$ independent of $\mathcal{F}_{\tau_a} $?

Answer 1

You have $$ (1)=\mathbb E[1_{\{\tau_a\leq t\}}1_{\{X_{t-\tau_a}+W_{\tau_a}\leq a\}}] = \mathbb E[\mathbb E[1_{\{\tau_a\leq t\}}1_{\{X_{t-\tau_a}+W_{\tau_a}\leq a\}}|F_{\tau_a}]] = \mathbb E[1_{\{\tau_a\leq t\}}\Pr(X_{t-\tau_a}\leq 0]|F_{\tau_a})],$$ where you use that $W_{\tau_a}=a$ and measurability of $\tau_a$ wrt. $F_{\tau_a}$.

Moreover, $X_{t-\tau_a}=W_t-a$ is not independent of $F_{\tau_a}$.