= Reflection identity for Brownian exit from a half-disc
{title2=$\mathbb P(T<S)=\mathbb P(\operatorname{Re}B_T>0)-\mathbb P(\operatorname{Re}B_T<0)$}
For <planar Brownian motion> starting on the positive real axis inside a disk, let $T$ be circular exit and $S$ first contact with the imaginary axis. Reflecting the path after $S$ fixes the disk and swaps positive and negative circular exits. The <Strong Markov property> makes their probabilities equal on $S<T$. Therefore $\mathbb P(T<S)=\mathbb P(\operatorname{Re}B_T>0)-\mathbb P(\operatorname{Re}B_T<0)$.
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