Find the distribution of the maximum of a Wiener Process with negative drift So.. what I have now is
Let $M=max\{W_t; 0\leq t <\infty\}$ since $W_0=0$, $M\geq 0$ with probability 1. So,
$P(M>x)=P(T_x<\infty)$ where $T_x$ is the stopping time, so we now use the martingale $e^{2\mu W_t/\sigma^2}$ and since it is a martingale and uniformly bounded, we have
$E[e^{2\mu W_{T_x}/\sigma^2}]=1$, so...
$E[e^{2\mu W_{T_x}/\sigma^2}]=e^{2\mu x / \sigma^2}P(T_x<\infty)+0 P(T_x=\infty) \Rightarrow P(M>x)=e^{2x\mu/\sigma^2}$
But I'm not sure if everything I did makes sense..
 A: Let 
$$W_t := \sigma B_t - \mu \cdot t$$
be a Brownian motion with drift $\mu>0$ and
$$T_x := \inf\{t \geq 0; W_t \geq x\}.$$
Using the stationarity and independence of the increments of the Brownian motion $(B_t)_{t \geq 0}$, it is not difficult to see that $\left(\exp\left[\frac{2\mu W_t}{\sigma^2}\right]\right)_{t \geq 0}$ is a martingale. In particular, we find by the optional stopping theorem
$$\mathbb{E} \bigg( \exp \bigg[ \frac{2\mu W_{t \wedge T_x}}{\sigma^2} \bigg] \bigg) = 1$$
for any $t \geq 0$. Obviously,
$$\begin{align*}\mathbb{E} \bigg( \exp \bigg[ \frac{2\mu W_{t \wedge T_x}}{\sigma^2} \bigg] \bigg) &= \mathbb{E} \bigg( \exp \bigg[ \frac{2\mu W_{t}}{\sigma^2} \bigg] \cdot 1_{\{T_x=\infty\}} \bigg) + \mathbb{E} \bigg( \exp \bigg[ \frac{2\mu W_{t \wedge T_x}}{\sigma^2} \bigg] \cdot 1_{\{T_x<\infty\}} \bigg) \\ &=:I_1+I_2. \end{align*}$$
For $I_2$ we see that the integrand is bounded by $\exp\left(\frac{2\mu x}{\sigma^2}\right)$ and that $T_x \wedge t \to T_x$ as $t \to \infty$. Therefore, the dominated convergence theorem yields
$$I_2 \stackrel{t \to \infty}{\to} \mathbb{E} \bigg( \exp \bigg[ \frac{2\mu W_{T_x}}{\sigma^2} \bigg] \cdot 1_{\{T_x<\infty\}} \bigg) = \exp \left(\frac{2\mu x}{\sigma^2} \right) \cdot \mathbb{P}(T_x<\infty).$$
It remains to prove that $I_1 \to 0$ as $t \to \infty$. First of all, we recall that
$$\frac{B_t}{t} \to 0 \qquad \text{as} \, t \to \infty.$$
This means that
$$W_t = t \cdot \left( \sigma \frac{B_t}{t}- \mu \right) \to - \infty, \qquad t \to \infty.$$
Invoking again the dominated convergence theorem, we see that $I_1 \to 0$ as $t \to \infty$. Hence
$$1 = \lim_{t \to \infty} (I_1+I_2) = \exp \left(\frac{2\mu x}{\sigma^2} \right) \cdot \mathbb{P}(T_x<\infty),$$
i.e.
$$\mathbb{P}(T_x<\infty) = \exp \left(-\frac{2\mu x}{\sigma^2} \right).$$
