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: $$f_{(M(t)|W(t)=x)}(y) = \frac{2(2a-x)}{t}e^{-2(a^2-ax)}, \ a \geq \max\{0,x\}$$


Here is an answer to the joint distribution, which confirms your answer:

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.


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