# Derivation of Joint and Conditional density of a Brownian Motion and its Maximum

Given $$W(t)$$ and $$M(t)=\max\limits_{0\leq s\leq t}W(s)$$ where $$\{W(t),\ t\geq 0\}$$ is the standard brownian motion, compute their joint distribution and the conditional distribution of $$M(t)$$ given $$W(t).$$

Any chance I can get a second check on the following results, and that I haven't lost a negative sign anywhere?

$$f_{(W(t), \ M(t))}(x,a) = \frac{2(2a-x)}{t^{3/2}}.\phi\bigg(\frac{2a-x}{\sqrt{t}}\bigg), \ \ x \leq a, \ \ a\geq 0$$ and what the conditional evaluates to be: $$f_{(M(t)|W(t)=x)}(y) = \frac{2(2a-x)}{t}e^{-2(a^2-ax)}, \ a \geq \max\{0,x\}$$

It's obvious that $$M_t$$ is non-decreasing in $$t$$. We define the hitting time of level $$a$$ by

$$T_a = inf \{ t \geq 0 : W_t = a\}$$

We can assign $$T_a = \infty$$ if $$a$$ is never reached. Therefore, $$\{M_t \geq a\} = \{ T_a \leq t\}$$ . Taking $$T = T_a$$ in the reflection principle (we will get to this in a second), for $$a \geq 0, a \geq x$$ and $$t \geq 0$$,

\begin{align} P\left[M_t \geq a, W_t \leq x\right] &= P\left[T_a \leq t, W_t \leq x \right] \\ &= P[T_a \leq t, 2a-x \leq \tilde{W_t}] \\ &= P[2a-x \leq \tilde{W_t}] \\ &= 1 - \Phi\left( \frac{2a-x}{\sqrt{t}}\right) \\ \end{align}

Lemma: Let $$a \gt 0$$. If the Brownian motion start at zero, then $$P[T_a \lt t] = 2P[W_t \gt a]$$ proof: If $$W_t \gt a$$ then then by continuity of the Brownian path, $$T_a < t$$. Moreover, by symmetry, $$P[W_t-W_{T_a} \gt 0 \mid \ T_a \lt t] =1/2$$. Thus

\begin{align} P[W_t \gt a] &= P[T_a \lt t, W_t-W_{T_a} \gt 0] \\ &= P[T_a \lt t] P[W_t-W_{T_a} \gt 0 \mid T_a \lt t] \\ &= \frac{1}{2}P[T_a \lt t] \end{align}

Reflection Principle Let $$\{W_t, t \geq 0\}$$ be a standard Brownian motion and let $$T$$ be a stopping time, and define $$\tilde{W_t} = \begin{cases} W_t & t \leq T\\ 2W_T-W_t & t \gt T\end{cases}$$ Then $$\{\tilde{W_t}, t \geq 0 \}$$ is a standard Brownian mottion.