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 since $X$ and $-X$ have the same distribution. Thank you for your help
Distribution of a stopping time
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probability-theory
brownian-motion
1 Answers
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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). $
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0I refute that there is an *ambiguity*. The two occurrences of $\mathbb P$ are concerned with events $Z$-measurable and $Y$-measurable respectively. As mentioned in the post, $Z$ is distributed like $X$ under $\mathbb P_x$ and $Y$ like $X$ under $\mathbb P_0$, hence $Z_0=x$ $\mathbb P$-almost surely and $Y_0=0$ $\mathbb P$-almost surely. The subscripts $x$ and $0$ to $\mathbb P_x$ and $\mathbb P_0$ refer to the process $X$. – 2012-07-12