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Let $(B_t)$ a Brownian motion. I want to prove that for all $L\geq 0$, $$\mathbb P(\sup_{0\leq s\leq t}B_s\geq L)=2\mathbb P(B_t\geq L).$$

The proof start by : let $\tau=\inf\{t\geq 0\mid B_t= L\}$. Then, $$\mathbb P(B_t\geq L)=\mathbb P(B_t\geq L, \tau\leq t),$$

but I don't understand this equality... For me, we have $$\mathbb P(B_t\geq L)\geq \mathbb P(B_t\geq L,\tau\leq t),$$ but I don't get the equality.

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1 Answer 1

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I guess that $\mathbb P\{B_0=0\}=1$ (otherwise your equality is not correct). By continuity of $t\mapsto B_t$, you have that $\{B_t\geq L\}\subset \{\tau\leq t\}$. Therefore $$\{B_t\geq L\}\cap \{\tau\leq t\}=\{B_t\geq L\},$$ and thus, your result.

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