Suppose $T_x=\inf \{ t \mid W_t = x\}$. Where $(W_t)$ is a standard Wiener process. What is the distribution of $$T_a-T_{b}$$ for $a, b\in\mathbb R$? Does it have any relation to $T_{a-b}$?
2026-04-03 20:35:30.1775248530
Distribution of difference of first hitting times of Brownian motion
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Yes, assuming that $0<b<a$, the random variables $T_a-T_b$ and $T_{a-b}$ have the same distribution. This follows from the strong Markov property of Brownian motion at time $T_b$. Note that if $b<0<a$ then $T_a-T_b$ is negative with positive probability.