Brownian motion minimium

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 diff erent 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

• One place to start would be to consider the distribution of $X_1(t)-X_2(t)$. – Math1000 Jun 12 '18 at 6:12
• @Math1000 $X_1(t) - X_2(t) \sim N((\mu_1 - \mu_2)t, 4t)$. Show that $P(X_1(t) - X_2(t) \neq 0)$. Isn't this probability equal to 1? – stutsy Jun 12 '18 at 8:35
• More precisely, you need to show that $$\tau := \inf\{t>0: X_1(t)-X_2(t)=0\}$$ is finite almost surely. Is $\tau$ a stopping time? – Math1000 Jun 12 '18 at 10:09
• @Math1000 How would you prove that? Does this use the hint where the minimum is required? – stutsy Jun 12 '18 at 11:32