Distribution of a transformed Brownian motion Let $W$ be a standard Brownian motion. From an earlier proven result I know that $N_t = \exp\left\{a W_t - \frac12 a^2 t \right\}$ defines a martingale on the natural filtration of $W$ for all $a \in \mathbb{R}$. I now want to prove that for every $a>0$ the random variable 
$$ \sup_{t \geq 0} \left(W_t - \frac12 a t\right),$$
has an exponential distribution with parameter $a$.
To prove this I want to use another earlier proven result namely that 
$$ P\left( \sup_{t \geq 0 } M_t > x \mid \mathcal{F}_0 \right) = 1 \wedge \frac{M_0}{x} \quad\text{a.s.,} $$
for every $x>0$ and positive, continuous martingale $M_t$ that converges a.s. to zero as $t \rightarrow \infty$. 
Yet I don't really see how I could combine these results as we are dealing with an exponential martingale $N_t$ which is indeed continuous and positive (does it converge to zero if $t$ tends to infinity?) hence we could use the statement:
$$ P\left( \sup_{t \geq 0 } \exp\left\{a W_t - \frac12 a^2 t \right\} > x \mid \mathcal{F}_0 \right) = 1 \wedge \frac{1}{x} \quad\text{a.s.,}$$
but what does this say about the distribution of  $\sup_{t \geq 0} \left(W_t - \frac12 a t\right)$?
 A: First notice that $N_t$ defines positive, continuous martingale on the natural filtration of $W$ as it is the exponential of a scaled standard Brownian motion. Also notice that we have
$$N_t = \frac{e^{a W_t}}{e^{\frac12 a^2 t}} \rightarrow 0 \text{ as }t \rightarrow \infty \text{ a.s.,}$$
as $N_t$ defines a positive martingale it must have a finite limit which must equal zero. (correct me if I'm wrong).
Therefore we can apply the result same as @Did did thus for all $x>0$,
\begin{align*}
 P \left( \sup_{t \geq 0} \left( W_t - \frac12 a t \right) > x \right) &= P\left( \sup_{t\geq 0} e^{a\left(W_t - \frac12 a t\right)} > e^{a x} \right) \\
&= P\left( \sup_{t\geq 0} N_t > e^{a x} \right) = 1 \wedge \frac{N_0}{e^{ax}}\\
&= 1 \wedge \frac{e^{a W_0 - \frac12 a^2\cdot 0}}{e^{ax}} = e^{-ax},
\end{align*}
as $W_0 \equiv 0$. We can rewrite
$$ P \left( \sup_{t \geq 0} \left( W_t - \frac12 a t \right) \leq x \right) = 1 - P \left( \sup_{t \geq 0} \left( W_t - \frac12 a t \right) > x \right) = 1 - e^{-ax},$$
which is the distribution function of a Exp$(a)$ probability distribution if $x>0$. 
If $x \leq 0$ then simply note that we can still use the statement as $e^{ax}>0$ for all $x \in \mathbb{R}$ to obtain that the probability we first derived must be 1 hence $P \left( \sup_{t \geq 0} \left( W_t - \frac12 a t \right) \leq x \right) = 0$ for all $x \leq 0$.
Hence $\sup_{t \geq 0} \left( W_t - \frac12 a t \right) \sim \text{Exp}(a)$ as it's distribution function is given by 
$$ F(x) = P \left( \sup_{t \geq 0} \left( W_t - \frac12 a t \right) \leq x \right) = \left( 1 - e^{-ax} \right) 1_{\{x \geq 0\}}$$
