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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 $\sup_{t \geq 0} \left(W_t - \frac12 a t\right)$?

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    $$P\left( \sup_{t \geq 0} \left(W_t - \frac12 a t\right)>x\right)=P\left(\sup_{t\geq 0}N_t>e^{ax}\right)= 1 \wedge\frac{N_0}{e^{ax}}=e^{-ax}\qquad (x>0)$$ – Did Apr 18 '15 at 18:04

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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 $$ F(x) = P \left( \sup_{t \geq 0} \left( W_t - \frac12 a t \right) \leq x \right) = \left( 1 - e^{-ax} \right) 1_{\{x \geq 0\}}$$