Given probability distributions, how to calculate $P(X_1>X_2)$?

I have two independent random variables $X_1$ and $X_2$, that give me the time step when two events occur, so they give values in $\mathbb{N}$.

Event 1 occurs every time step with probability $ε$, so $P(X_1\leq n) = 1-(1-ε)^n$, whereas $X_2$ has a cumulative distribution function that starts steeper than the one of $X_1$, but then gets somewhat “saturated” and grows only very slowly from then on.

How can I from these cumulative distribution functions obtain (or estimate; I don't have a closed form for $P(X_2)$ which I could then use to calculate an integral/series, and some “good” numerical estimate would be sufficient anyway) the probability $P(X_1<X_2)$ that event 1 occurs before event 2?

• $X_1$ and $X_2$ would be random variables, not probability distributions. Each of them would have a probability distribution. But they would also have a joint probability distribution. And the answer would depend on what that is. One of the simplest ways of specifying that would be to say that they're independent, if in fact they are. – Michael Hardy Aug 18 '13 at 22:19
• I was thinking that my formulation was off, but somehow I was stuck with it. $X_1$ and $X_2$ are indeed independent random variables. I have not talked in these terms for too long… – Anaphory Aug 18 '13 at 23:17

• Assuming they are independent, this formula can be rewritten in a fashion that is interesting. If $\phi_i(x)=P(X_i<x)$ is the cumulative distribution function, then a variation of the above argument gives you $$\int_{-\infty}^{\infty} \phi_2(x)d\phi_1$$. Assuming continuous differentiability (or perhaps some lighter condition) on the $\phi_i$, you can use integration by parts to see that $P(X_i>X_2)+P(X_2>X_1) = 1$ :) – Thomas Andrews Aug 18 '13 at 22:42