# Density of running supremum of Brownian motion until a stopping time

I am stuck on an exercise in my book:

The question relies on the following fact:

Let $$M$$ be a continuous, non-negative local martingale such that $$M_0=1$$ and $$M_t \rightarrow 0$$ almost surely as $$t \rightarrow \infty$$. Then, for each $$a>1$$, $$\mathbb{P} \bigg( \sup_{t \geq 0} M_t > a \bigg) = \frac{1}{a}.$$

It asks us to find the density of

(i) $$\, \sup_{0 \leq t \leq \tau(-b)} W_t$$, $$\quad$$where $$\tau(-b) = \inf \{t \geq 0 \, | \, W_t < -b \}$$ and $$b>0$$.

(ii) $$\sup_{t \geq 0} (W_t - \lambda t),$$ $$\quad$$ for $$\lambda >0$$.

My main problem is how to associate $$W_t$$ to a continuous local martingale $$M$$ satisfying the aforementioned properties. The Doolean exponential $$M_t := \exp \{ W_t - \frac{t}{2} \}$$ does not work.

• Why does it not work? (Two related questions: math.stackexchange.com/q/1240559 and math.stackexchange.com/q/1241149)
– saz
Apr 26, 2015 at 6:42
• @saz Thanks. But none of the two questions touch upon the first part of the question, i.e. finding $\sup_{0 \leq t \leq \tau(-b)} W_t$. Apr 26, 2015 at 9:12
• Note that you don't mention where you are stuck. You just write "[..] doesn't work". So how is the reader supposed to know whether you are talking about the first or the second part? Adding some more information on your progress/problems/thoughts would be helpful.
– saz
Apr 26, 2015 at 9:17
• @saz But the Doolean can only tackle the situation of Brownian motion with drift. Without drift, what $M_t$ would you suggest? Apr 26, 2015 at 9:20
• What about $M_t := \frac{W_{t \wedge \tau(-b)}+b}{b}$?
– saz
Apr 26, 2015 at 9:24

Part (i): Use $$M_t := \frac{W_{t \wedge \tau(-b)} + b}{b}.$$
How to come up with this choice? Well, we are interested in $$\sup_{0 \leq t \leq \tau(-b)} W_t = \sup_{t \geq 0} W_{t \wedge \tau(-b)}.$$ We know that $(W_{t \wedge \tau(-b)})_{t \geq 0}$ is a martingale and that $W_{t \wedge \tau(-b)} \to -b$ as $t \to \infty$. This means that, in order to get a martingale $(M_t)_{t \geq 0}$ satisfying the assumption $M_t \to 0$ as $t \to \infty$, we have to shift the process. Adding $b$ yields a new martingale
$$\tilde{M}_t := W_{t \wedge \tau(-b)} +b \to 0 \qquad \text{as t \to \infty}.$$
By the definition of $\tau(-b)$, we have $\tilde{M}_t \geq 0$, i.e. $(\tilde{M}_t)_{t \geq 0}$ satisfies all the assumptions except $\tilde{M}_0=1$. Scaling the martingale with the factor $\frac{1}{b}$, we are done.
Part (ii): Consider the martingale $$M_t := \exp \left( \lambda W_t - \frac{\lambda^2}{t} \right) = \exp \left( \lambda \left[ W_t - \frac{\lambda}{2} t \right] \right).$$ (See also this question and this question.)