# Probability that one Brownian motion hits 2 before the other hits 1

Let $$B_t$$ and $$W_t$$ be independent brownian motions. What is the probability that the first one hits 2 before the second one hits 1? Express the answer as an integral.

I am quite stuck on this. I have some ideas floating in my head like using the fact that we know the distribution of $$B_t^*$$, the maximum of $$B_s$$ over $$[0,t]$$, writing $$\mathbb{P}[B_t^* \geq 1, W_t^* < 2] = \mathbb{P}[B_t^* \geq 1] \mathbb{P}[W_t^* < 2]$$ and then we know both these probabilities, but then I don't see how to combine these probabilities for all times since these events are neither monotone nor disjoint for various t. Another idea is to let $$\tau_B$$ be the hitting time of $$1$$ for the first brownian motion and $$\tau_W$$ be the hitting time of $$2$$ for the second brownian motion. Let $$\tau = \tau_B \wedge \tau_W$$. We want $$\mathbb{P}[B_\tau = 1]$$, so the standard strategy is to construct a martingale and then use optional stopping to compute this probability, but here $$B_{\tau \wedge t}$$ need not be bounded so I am a little stuck.

• You have two independent getting times, you know marginal distributions for both of them and you need to find the probability that one is smaller. Same as asking what’s the probability that the outcome of a dice roll with 6 sides is higher than that of another independent 12 sides dice
– SBF
Commented May 8 at 15:11
• Oh right, this is just $\mathbb{Pr}[X < Y]$ where $X$ and $Y$ are independent and hence have joint measure the product measure. I totally missed that. Commented May 8 at 15:18

Since $$\tau_B$$ and $$\tau_W$$ are independent random variables and we can compute their density, $$\mathbb{Pr}[\tau_B < \tau_W]$$ is just $$\mathbb{Pr}[\tau_B < \tau_W] = \int\int_{x < y} f_X(x)f_Y(y) dxdy$$ where $$f_X(x)$$ and $$f_y(y)$$ are the marginals.