Show that the following 2 times are equal in distribution.

Question

Let $T_a=\inf \{t \in [0,\infty), B_t =a\}$, where $B_t$ is the Brownian motion. $T_a$ is a stopping time and has distribution:

$f_a (t) = \frac{a}{\sqrt{2 \pi t^3}} exp\big( \frac{-a^2}{2t} \big)$

Let $S_a= \sup\{t \in [0, \infty): B_t = at\}$. $S_a$ is not a stopping time.

Prove that $S_a = 1/T_a$ in distribution.

I am struggling a bit with the computation, can you help me ?

Answer 1

A starting point: The process $X$ defined by $X_t:= tB_{1/t}$ ($t>0$) and $X_0:=0$, is a Brownian motion.