# Continuous-time Martingale and Brownian Motion Supremums

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.

• If your question concers only the first part of the exercise, then why post the other two unrelated parts of the exercise? The first part is solved here – saz Apr 15 '19 at 17:28

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.