Continuous-time Martingale and Brownian Motion Supremums

Question

I am reading through Le Gall's book on Brownian Motion, Martingales, and Stochastic Calculus. I just read through the chapter on optional stopping of martingales, but I cannot solve the first exercise:

1. Let $M$ be a martingale on a probability space with a complete filtration such that $M$ has continuous stopping times, and $M_0 = x \in \mathbb{R}_+$. We assume $M_t \geq 0$ for every $t \geq 0$, and that $M_t \to 0$ when $t \to \infty \; $ a.s. Show that, for every $y > x,$ $$P(\sup_{t \geq 0} M_t \geq y) = \frac{x}{y}$$

2. Give the law of $\sup_{t \leq T_0} B_t$, when $B$ is a Brownian motion started from $x > 0$ and $T_0 = \inf \{t \geq 0 : B_t = 0 \}$

3. Assume now that $B$ is a Brownian motion started from $0$, and let $\mu > 0$. Using an appropriate exponential martingale, show that $\sup_{t\geq 0}(B_t - \mu t)$ is exponentially distributed with parameter $2\mu$.

Now, I am able to upper bound the probability $P(\sup_{t \geq 0} M_t \geq y) \leq \frac{x}{y}$ by using one of the Doob martingale inequalities. However, I am not sure how to finish off the proof of 1. I am assuming that I would have to use optional stopping, but I cannot make a useful stopping time.

Answer 1

1. Let $T_y=\inf\left\{t\ge0\;;\; M_t\ge y\right\}$, we have : $$ \mathbb{E}\left[M_0\right] = \mathbb{E}\left[M_{T_y}.\mathbf{1}_{\left\{T_y<\infty\right\}} + M_\infty.\mathbf{1}_{\left\{T_y=\infty\right\}}\right] $$ but $M_\infty=0$, so $$ x = y . \mathbb{E}\left[\mathbf{1}_{\left\{T_y<\infty\right\}}\right] = y . \mathbb{P}\left(T_y<\infty\right) = y . \mathbb{P}\left(\sup_{t\ge 0} M_t\ge y\right) $$ i.e. $$ \mathbb{P}\left(\sup_{t\ge 0} M_t\ge y\right) = \frac{x}{y} $$

2. For a standard Brownian Motion, the reflexion principle gives $$ \mathbb {P} \left(\sup _{0\leq s\leq t}W(s)\geq a\right) = 2\mathbb {P} \left(W(t)\geq a\right) $$ from which you can get the density probability function by derivating : $$ f_{\sup _{0\leq s\leq t}W(s)}(a) = \sqrt{\frac{2}{\pi t}} \mathbf{e}^{-\frac{a^2}{2t}}. $$ Now, $B = x + W$ so for $a\ge x$ $$ f_{\sup _{0\leq s\leq T_0}B(s)}(a) = \sqrt{\frac{2}{\pi T_0}} \mathbf{e}^{-\frac{(a-x)^2}{2T_0}}. $$

3. We need to prove that $\mathbb{P}\left(\sup_{t\ge 0} B(t)-\mu t \geq y\right) = \mathbf{e}^{-2\mu y}$. But \begin{alignat*}{2} \mathbb{P}\left(\sup_{t\ge 0} B(t)-\mu t \geq y\right) & = \mathbb{P}\left(\sup_{t\ge 0} \left(2\mu B(t)-\frac{1}{2}4\mu^2 t \right) \geq 2\mu y\right) \\ & = \mathbb{P}\left(\sup_{t\ge 0} \mathbf{e}^{2\mu B(t)-\frac{1}{2}4\mu^2 t }\geq \mathbf{e}^{2\mu y}\right), \end{alignat*}

and $\left(\mathbf{e}^{2\mu B(t)-\frac{1}{2}4\mu^2 t}\right)_{t\ge 0}$ is a martingale that verifies the conditions in question 1. Thus $$ \mathbb{P}\left(\sup_{t\ge 0} \mathbf{e}^{2\mu B(t)-\frac{1}{2}4\mu^2 t }\geq \mathbf{e}^{2\mu y}\right) = \mathbf{e}^{-2\mu y} $$ which is the result.