Let $(B_t)_t$ be a standard Brownian motion, and $$ A = \sup\{t\leq 1\mid B_t =0 \},\qquad B = \inf\{ t\geq 1\mid B_t =0 \}. $$ I would like to show that $A$ and $B^{-1}$ are identically distributed and find their distribution. Could you please help me with this.
I would be grateful for any ideas or suggestions. Thanks.
The OP probably means $A=\sup\{t\leqslant1\mid B_t=0\}$ and $B=\inf\{t\geqslant1\mid B_t=0\}$, in any case this post answers the question modified accordingly.
Let $W_t=tB_{1/t}$, then $(W_t)_{t\geqslant0}$ is a Brownian motion and $W_t=0$ for $t\gt0$ iff $B_{1/t}=0$ hence $1/B=\sup\{t\leqslant1\mid W_t=0\}$. Thus, $1/B$ corresponds to the functional $A$ based on the paths of the Brownian motion $(W_t)_{t\geqslant0}$, in particular $A$ and $1/B$ are identically distributed.
To compute the distribution of $B$, note that, conditionally on $B_1=x$, $B$ is distributed as the first hitting time of $x$ by a standard Brownian motion, and that this first hitting time is distributed as $x^2T$, where $T=\inf\{t\geqslant0\mid B_t=1\}$. Thus, for every $t\gt1$, $$ f_B(t)=2\int_0^{+\infty}f_T\left(\frac{t}{x^2}\right)p_1(x)\frac{\mathrm dx}{x^2}, $$ where $f_B$ is the density of $B$, $f_T$ is the density of $T$ and $p_1$ is the density of $B_1$. Furthermore, $$ p_1(x)=\frac1{\sqrt{2\pi}}\mathrm e^{-x^2/2},\qquad f_T(t)=\frac1{t\sqrt{2\pi t}}\mathrm e^{-1/2t}. $$ This yields $$ f_B(t)=\frac{\mathbf 1_{t\gt1}}{\pi\sqrt{t}(1+t)}. $$ Finally, since $A$ is distributed like $1/B$, the usual change of variable yields $$ f_A(t)=\frac{\mathbf 1_{0\lt t\lt1}}{\pi\sqrt{t}(1+t)}. $$
