Joint distribution of Brownian motion and its running maximum when time is different

Question

Let $B_t$ be a standard Brownian motion and $B_t^*=\max_{s\leq t}B_s$. The joint distribution of $(B_t,B_t^*)$ is well known and its density function is given by $$ f(x,y)=\dfrac{2(2y-x)}{t}\cdot\dfrac{1}{\sqrt{2\pi t}}\exp\left(-\dfrac{(2y-x)^2}{2t} \right) $$ if $x\leq y$ and $y\geq 0$. My question is that is there a formula for the joint distribution of $(B_s,B_t^*)$ when $s\leq t$?

Answer 1

If you consider the filtration $\mathcal{F}_s$

• the increments of the Brownian motion for $t>s$ are independent from $\mathcal{F}_s$

Moreover, since $\mathbb{E}(B_t|\,\mathcal{F}_s)=B_s$:

• $B^*_s\geq B_s\Rightarrow \mathbb{E}(B^*_t|\,\mathcal{F}_s)=B^*_s$
• $B^*_s=B_s\Rightarrow \mathbb{E}(B^*_t|\,\mathcal{F}_s)=B_s.$

This means that $$ \mathbb{P}(\mathbb{E}(B^*_t|\,\mathcal{F}_s)\geq y, B_s\leq x )=\mathbb{P}(B^*_s\geq y, B_s\leq x) $$

and

$$ f_{B^*_t,\,B_s|\,\mathcal{F}_s}(x,y|\mathcal{F}_s)=f_{B^*_s,\,B_s}(x,y). $$

If you consider instead the filtration $\mathcal{F}_t$, then $B_s$ is not random anymore, but a realized value, hence you would not have a joint distribution, but merely a conditional one. I will try and expand on such distribution later on when I will have more time.

Hope this helps.