Density of Running Maximum of Drifted Brownian Motion Computation

Question

$\textbf{Proposition}$ The PDF of the Maximum of a Brownian Motion with Drift is given by $$ f_{M_t}(m)={\sqrt{\frac{2}{\pi t}}} \exp\left( - \frac{(m-at)^2}{2t} \right) -2a\exp\left({2am}\right) \mathcal{N}\left ( \frac{-m-at}{\sqrt{t}} \right) $$ where $\mathcal{N}$ is the CDF of the standard normal variable.

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?

Answer 1

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.

Answer 2

Sorry to bring this up atfer 7 yrs but I must point for the upcoming viewers that this formula in the question is not the pdf of the joint distribution of $M_t$ and $W_t$ but a derivative of $P(M_t<m, W_t<\omega)$ with respect to $\omega$.

The statement in the question is actually false. Here is a true derivation of the cdf of $M$, also with Girsanov's thm, where $\bar{M}_t$ is the driftless maximum of period $[0,t]$ and $\Phi$ is the cdf of a standard normal distribution.

$$\begin{align*} &P(M_t>\alpha)=E(1_{\{M_t>\alpha \}})=E(\exp{(\mu B_t-\frac{\mu^2}{2}t)}1_{\{\bar{M}_t>\alpha \}}) \\ =&\int_{-\infty}^\infty \exp{(\mu x-\frac{\mu^2}{2}t)}d P(X_t<x, \bar{M}_t>\alpha) \\ =&(\int_{-\infty}^\alpha+\int_{\alpha}^\infty) \exp{(\mu x-\frac{\mu^2}{2}t)}d P(X_t<x, \bar{M}_t>\alpha) \\ =&\int_{-\infty}^\alpha \exp{(\mu x-\frac{\mu^2}{2}t)}\frac{1}{\sqrt{t}}\Phi'(\frac{x-2\alpha}{\sqrt{t}})dx + \int_{\alpha}^\infty \exp{(\mu x-\frac{\mu^2}{2}t)}\frac{1}{\sqrt{t}}\Phi'(\frac{x}{\sqrt{t}})dx \\ =&\int_{-\infty}^\alpha \exp{(2\mu \alpha)}\frac{1}{\sqrt{t}}\Phi'(\frac{x-2\alpha-\mu t}{\sqrt{t}})dx + \int_{\alpha}^\infty \frac{1}{\sqrt{t}}\Phi'(\frac{x-\mu t}{\sqrt{t}})dx \\ =& \exp{(2\mu \alpha)}\Phi(\frac{-\alpha-\mu t}{\sqrt{t}})+1-\Phi(\frac{\alpha-\mu t}{\sqrt{t}}) \end{align*}$$