$\textbf{Proposition}$ The $pdf$ of the Maximum of a Brownian Motion with Drift is given by
$$
f_{M_t}(m)={\sqrt{\frac{2}{\pi t}}} \mathrm{e}^\frac{(m-at)^2}{2t}-2a\mathrm{e}^{2am}\mathcal{N}\left ( \frac{-m-at}{\sqrt{t}} \right)
$$
Starting from the joint density of a drifted Brownian Motion $W_t$ and its running maximum $M_t=\sup_{s\le{t}}W_s$ I would like to compute the marginal density of $M_t$.
The joint density, which can be recovered through an application of Girsanov's theorem, reads
$$
f_{M_t,W_t}(m,\omega)=\mathrm{e}^{-\frac{a^2t}{2}+a\omega}\frac{2(2m-\omega)}{t \sqrt{2\pi t}}{\mathrm{e}^{-\frac{(2m-\omega)^2}{2t}}}.
$$
Applying the Theorem of Tonelli-Fubini, together with the observation that $M_t\le m \implies W_t\le m$, leads to the evaluation of
$$
f_{M_t}(m)=\int_{-\infty}^{m} \mathrm{e}^{-\frac{a^2t}{2}+a\omega}\frac{2(2m-\omega)}{t \sqrt{2\pi t}}{\mathrm{e}^{-\frac{(2m-\omega)^2}{2t}}}d\omega.
$$
Integrating by parts yields
$$
f_{M_t}(m)={\sqrt{\frac{2}{\pi t}}} \mathrm{e}^\frac{(m-at)^2}{2t}-\frac{2a}{\sqrt{2 \pi t}}\mathrm{e}^{-\frac{a^2t}{2}}\int_{-\infty}^{m} \mathrm{e}^{a\omega}{\mathrm{e}^{-\frac{(2m-\omega)^2}{2t}}}d\omega
$$
How do I solve this last integral?
Best Answer
So continuing the computation gives:
\begin{align} \frac{2a}{\sqrt{2 \pi t}}\mathrm{e}^{-\frac{a^2t}{2}}\int_{-\infty}^{m} \mathrm{e}^{a\omega}{\mathrm{e}^{-\frac{(2m-\omega)^2}{2t}}}d\omega &=\\ \frac{2a}{\sqrt{2 \pi t}}\int_{-\infty}^{m} \mathrm{e}^{-\frac{(\omega-at)^2}{2t}}\mathrm{e}^{\frac{-2m^2+\omega m}{t}}d\omega. \end{align}
Now the change of variable $y=\frac{\omega-at}{\sqrt{t}}$ yields:
$$ 2a\int_{-\infty}^\frac{m-at}{\sqrt{t}} \frac{1}{\sqrt{2 \pi}}exp\left(\frac{y^2}{2}-\frac{4m^2-2mat+2m\sqrt{t}y}{2t} \right)dy $$
Then
$$ 2a\mathrm{e}^{2am}\int_{-\infty}^\frac{m-at}{\sqrt{t}} \frac{1}{\sqrt{2 \pi}}\mathrm{e}^{-\frac{(y\sqrt{t}-2m)^2}{2t}} dy. $$
A second change of variable $z=\frac{y\sqrt{t}-2m}{\sqrt{t}}$ finally gives
$$ 2a\mathrm{e}^{2am}\int_{-\infty}^\frac{-m-at}{\sqrt{t}} \frac{1}{\sqrt{2 \pi}}\exp{\left( -\frac{z^2}{2}\right)} dz, $$
and the term in the integral is the density $f_N (z)$ of a standard normal variable.