Distribution of Brownian motion at Exponential time I'm trying to determine a probablity of the form
$$
\mathbb{P}(\sup_{0 \leq t \leq Y} W_t > a),
$$
where W is a brownian motion with drift d, and parameter $\sigma^2$ and independent of $Y \sim Exp(\lambda)$. By using the reflection principle for Wiener processes, I believe I may rewrite the above as
$$
2\mathbb{P}(W_{Y_n}>a).
$$
However, at this point I don't know how to move further, since I'm unsure of the distribution of $W_{Y_n}$. Any help is greatly appreciated!
Thanks in advance!
 A: We have
$$
\begin{align}
\mathbb{P}(\sup_{0 \leq t \leq Y} W_t > a) &= \mathbb{E}(\mathbb{I}_{\{\sup_{0 \leq t \leq Y} W_t > a\}})  \\
&= \mathbb{E}(\mathbb{E}(\mathbb{I}_{\{\sup_{0 \leq t \leq Y} W_t > a\}}|Y))  \\
&= \mathbb{E}(\mathbb{P}(\sup_{0 \leq t \leq Y} W_t > a|Y))  \\
&= \int_{y\in \Bbb R^+}\mathbb{P}(\sup_{0 \leq t \leq Y} W_t > a|Y=y)d\mathbb{P}(Y\le y) \\
&= \int_{y\in \Bbb R^+}\mathbb{P}(\sup_{0 \leq t \leq Y} W_t > a|Y=y)f_Y(y)dy \tag{1} \\
\end{align}
$$
As $Y$ and $W_t$ are independent, we can write
$$\sup_{0 \leq t \leq Y} W_t|\{ Y=y\} =\sup_{0 \leq t \leq y} W_t  \tag{2}$$
Remark: the independency here plays a very important role because if not, we need to rewrite $(2)$ as
$$\sup_{0 \leq t \leq Y} W_t|\{ Y=y\} =\sup_{0 \leq t \leq y} (W_t|\{Y = y \})  $$
and then determine the marginal distribution  $W_t|\{Y = y \}$ which is usually not easy.
Return back to $(2)$, we have $\sup_{0 \leq t \leq y} W_t $ is equal in distribution to $|W_y|$ and so
$$\mathbb{P}(\sup_{0 \leq t \leq Y} W_t > a|Y=y) = 2\mathbb{P}(W_y>a) =2 \Phi\left(-\frac{a}{\sqrt{y}}\right) \qquad ,a>0$$
with $\Phi(.)$ the cumulative probability function of $\mathcal{N}(0,1)$.
From $(1)$, we have
$$
\begin{align}
\mathbb{P}(\sup_{0 \leq t \leq Y} W_t > a) &=2\lambda \int_0^{+\infty}\Phi\left(-\frac{a}{\sqrt{y}}\right)e^{-\lambda y}dy \tag{3}
\end{align}
$$
Q.E.D
