Let $B_t$ be a Brownian motion. Compute $P(\inf_{t \in [1,2]} B_t < 0 \mid B_1 > 0, B_2 > 0)$.
This is a practice interview question I found here. My attempts are below, and I would appreciate any hints.
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There is a prior question [that may or may not be related to the above question] that asks one to compute $P(B_1 > 0, B_2 > 0)$. I did this as follows: $P(B_1 > 0, B_2 > 0) = P(B_1 > 0, B_2 – B_1 > -B_1) = P(Z_1 > 0, Z_2 > -Z_1) = \frac{3}{8}$ by applying a symmetry argument to the $(Z_1, Z_2) \sim N(0, I_2)$ distribution. So I could solve the above question by computing $P(\inf_{t \in [0,1]} B_t < 0, B_1 > 0, B_2 > 0)$. But I don't think this is helpful.
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I know of the reflection principle, which implies $P(\inf \{t > 0 : B_t = a\} < 1) = 2 P(B_1 \le a) = 2 \Phi(a)$ for $a < 0$. This could be applied to the original problem via $P(\inf_{t \in [1, 2]} B_t < 0 \mid B_1 = b > 0) = 2 \Phi(-b)$, but I'm not sure this is the way to go.
Best Answer
By the reflection principle (by reflecting at the first zero after $t=1$), $$ \mathbb P[B_1>0,B_2>0, \inf_{t\in [1,2]}B_t<0] = \mathbb P[B_1>0,B_2<0]. $$ Note that I deleted $\inf_{t\in [1,2]}B_t<0$ from the right hand side since it is implied by the other conditions. Now your computations give the answer $\frac 1 8$.