Distribution of a stopping time

Question

For $c\in\mathbb{R}$ we define the stopping time $\tau_c:=\inf\{t>0;X_t>c\}$ if $c\ge 0$ otherwise $\tau_c:=\inf\{t>0;X_t<c\}$. Let $X$ be a Markov Process on a metric space $E$ with initial distribution $\mu$. We denoty with $\mathbb{P}_\mu$ the distribution on $E^{[0,\infty)}$ of $X$ under $P$.Now suppose $X$ is the canonical Brownian motion on the Space $C[0,\infty)$ with the raw filtration generated by $X$. By symmetry Brownian motion is equally likely to hit 0 starting from x as it hits x starting from 0. Now I want to prove formally that the distribution of $\tau_0$ under $\mathbb{P}_x$, (the $x$ denoted the dirac measure at $x$.) and $\tau_x$ under $\mathbb{P}_0$ has the same distribution. What I know is that $\mathbb{P}_0[\tau_{-x}< s]=\mathbb{P}_0[\tau_x <s]$ since $X$ and $-X$ have the same distribution. Thank you for your help

Answer 1

Assume that $Z$ is distributed like $X$ under $\mathbb P_x$. Then $Y=x-Z$ is distributed like $X$ under $\mathbb P_0$, hence $\tau_0$ for $Z$ is $\tau_x$ for $Y$ and you are done.

Formally, for every $s\gt0$, $[\tau^Z_0\leqslant s]=[\tau^Y_x\leqslant s]$ hence $$ \mathbb P_x(\tau_0\leqslant s)=\mathbb P(\tau^Z_0\leqslant s)=\mathbb P(\tau^Y_x\leqslant s)=\mathbb P_0(\tau_x\leqslant s). $$