1
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ $X_1$ 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. $\endgroup$ – Michael Hardy Aug 18 '13 at 22:19
  • $\begingroup$ 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… $\endgroup$ – Anaphory Aug 18 '13 at 23:17
0
$\begingroup$

I don't completely understand your question but the standard way of doing this might be through conditional probability:

\begin{align} P(X_1>X_2)&= \sum _xP(X_1 >X_2\;|\;X_2 =x)\times P(X_2 =x)\\ &= \sum _xP(X_1 >x)\times P(X_2 =x)\\ \end{align}

$\endgroup$
  • 1
    $\begingroup$ Your last line assumes independence, which was not mentioned in the question. $\endgroup$ – Michael Hardy Aug 18 '13 at 22:21
  • $\begingroup$ Since he is asking for approximations, I assume he wants to discretize his domain as it is. In any case, it is trivial to extend the statement to the continuous case. The sum is replaced by integral and discrete weight by the PDF. $\endgroup$ – Inquest Aug 18 '13 at 22:40
  • $\begingroup$ 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 $P(X_i>X_2)+P(X_2>X_1) = 1$ :) $\endgroup$ – Thomas Andrews Aug 18 '13 at 22:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.