Let {$X_1(t), t \geq 0$} and {$X_2, t \geq 0$} be two independent Brownian Motion processes with drift $\mu_1$ and $\mu_2$, respectively, and same variance parameter $\sigma^2 = 1$. Assume that $X_1(0) = x_1$ and $X_2(0) = x_2$, with $x_1 > x_2$. Find the probability that the two processes never cross each other. (Note: the answer may be different for the cases $\mu_1 - \mu_2 < 0$, $\mu_1 - \mu_2 = 0$ and $\mu_1 - \mu_2 > 0$. Give details.) (Hint: For a Brownian Motion with positive drift $\mu > 0$ and variance parameter $\sigma^2$, P(M < -y) =$ e^{-2\mu y/ \sigma^2}$ for y > 0, where M = min{$X(t); t \geq 0$}.)
Not sure how to start this question and finish this. Thanks