Joint distribution of $W(t)$ Brownian motion and $B(s)=\int_0^t \operatorname{sgn} ( W(s) ) dW(s)$

Question

Let $(W(t))$ be a brownian motion and $B(s)=\int_0^t \operatorname{sgn} ( W(s) ) dW(s)$. Does one know the joint distribution $(W(s),B(s))$ for a given $s$? I know some related theory like Tanaka's formula and the joint distribution of the absolute value of W and the local time.

Answer 1

Tanaka's formula shows that $B=|W|-L$ where $L$ is the local time at $0$ of $W$, hence $(|W|,B)=(|W|,|W|-L)$. Now, $(|W|,L)$ coincide in distribution with $(M-W,M)$ where $$ M(s)=\sup\{W(u)\mid0\leqslant u\leqslant s\}, $$ hence $(|W|,B)$ coincide in distribution with $(M-W,-W)$.

The joint distribution of $(M,W)$ follows from the observation, due to Désiré André and called the reflection principle, that, for every $m\geqslant0$ and every $w\leqslant m$, $$ P[M(s)\geqslant m,W(s)\leqslant w]=P[W(s)\geqslant2m-w]. $$ Assuming without loss of generality that $s=1$, differentiating this twice yields that $(M(1),W(1))$ has the density $f$ defined by $$ f(m,w)=2(2m-w)\varphi(2m-w)\mathbf 1_{w\leqslant m}\mathbf 1_{m\geqslant 0}, $$ where $\varphi$ is the standard normal density. By a change of variable, $(M(1)-W(1),-W(1))$ has the density $g$ defined by $$ g(x,y)=f(x-y,-y)=2(2x-y)\varphi(2x-y)\mathbf 1_{x\geqslant0}\mathbf 1_{x\geqslant y}. $$ Finally, the sign of $W$ is symmetric and independent of $(|W|,B)$ hence the density $h$ of $(W(1),B(1))$ is $$ h(x,y)=(2|x|-y)\varphi(2|x|-y)\mathbf 1_{|x|\geqslant y}. $$ Likewise, by scaling, the density $h_s$ of $(W(s),B(s))$ is $$ h_s(x,y)=\frac1s h\left(\frac{x}{\sqrt{s}},\frac{y}{\sqrt{s}}\right). $$ In particular, $W(s)$ and $B(s)$ are not independent (except if $s=0$).