Let $W_t$ be a standard Wiener process, $a$ some real number, and $\chi (x)$ the indicator function. I am trying to calculate the following expectation: $$ \mathbb E \left[e^{-\mu W_T } \chi \left( {\min_{0 \leq t \leq T} W_t \leq a} \right) \right] $$ I believe I have to use the reflection principle in some way. I have tried reasoning that conditioned on that $W_t$ hits $a$ at some point prior to $T$, the distribution of $W_T$ has to be centred around $a$. However, its variance is going to be $T-\tau$ where $\tau = \inf \lbrace s:W_s \leq a \rbrace$, so we get something like $$ = \mathbb E \left[ \mathbb E \left[ e^{-\mu W_T } \chi \left( {\min_{0 \leq t \leq T} W_t \leq a} \right) \bigg| \min_{0 \leq t \leq T} W_t \leq a\right] \right]\\ = \mathbb E \left[ \mathbb E \left[ e^{-\mu W_T } \bigg| \min_{0 \leq t \leq T} W_t \leq a\right] \right] $$ and following the said reasoning the inner expectation is $$ \int_{\mathbb R} e^{-\mu x} \frac{1}{\sqrt{2 \pi}(T-\tau)}\exp{\left(\frac{-(x-a)^2}{2(T-\tau)} \right)}dx $$ and the one would have to find the distribution of $\tau$ to calculate the next expectation etc. So even if this were to be correct, it looks quite messy.
I suspect there is supposed to be a simpler way, still involving the reflection principle? Any help highly appreciated.
Conditioning on $\tau = \inf \left\lbrace s : W_s \leq a \right\rbrace$ yields two different situations. Either $W_0 \leq a$, and then it's a simple matter of taking the expectation at $\tau = 0$, or $W_0 \gt a$.
In the second case, since $\tau$ is the first hitting time, we know that $W_\tau = a$, so we get:
$E \left[ e^{-\mu a} E\left [e^{W_T - W_\tau} \mid W_\tau = a\right] P(\tau) \chi(\tau <= T) \right] = \\ E \left[ e^{-\mu a + \frac{\mu^2 (T - \tau)}{2}} P(\tau) \chi(\tau <= T) \right]$
We're left with integrating over the first hitting times, with the indicator truncating at T. Since the process is driftless, $P(\tau)$ is not an Inverse Gaussian, but it is a Lévy distribution with parameters $(0,\frac{a^2}{\mu^2})$. Which can be easily transformed into an Inverse Gamma or a Scaled Inverse Chi-squared.
I don't believe it is possible to simplify the expectation further, though at least it is one-dimensional.