I am trying to calculate either the probability density function or the cumulative distribution function of the following conditional distribution:

$X_1$ given $X_1 > X_2$, where $X_1$ and $X_2$ are i.i.d exponential(1).

I have been trying many different ways to calculate the below probability, which is the CDF: $$ P(X_1 \le t | X_1 > X_2) $$ I feel like I'm missing something when trying to do this. I probably am not aware of the specific techniques that should be used when we condition on an event that is entirely described by 2 Random variables, rather than the usual conditioning like $X_1 | X_2 = a$ (where $a$ is a constant).

I have been trying to see whether any textbooks may explain how to do this calculation, but to no avail. I'll be very grateful for any guidance on how to proceed. Thank you so much.


1 Answer 1


The joint density of $(X_1, X_2)$ is easy:

$$f_{X_1, X_2}(x,y) = e^{-x} e^{-y} \mathbb 1 (x > 0) \mathbb 1 (y > 0). \tag{1}$$

Now, given that $X_1 > X_2$, what does the support look like? It is simply $$0 < X_2 < X_1 < \infty.$$

The joint density of $(X_1, X_2)$ given this condition that $X_1 > X_2$ is proportional to $(1)$, but now it must be divided by the probability that $X_1 > X_2$. By an obvious symmetry argument, $\Pr[X_1 > X_2] = 1/2$. So the joint density now looks like this:

$$f_{(X_1, X_2) \mid X_1 > X_2}(x,y) = 2e^{-x} e^{-y} \mathbb 1 (x > y > 0). \tag{2}$$

All that is left now is to compute the marginal density of $X_1$ from this joint density, which is accomplished by integrating over the permissible values of $X_2$:

$$f_{X_1 \mid X_1 > X_2}(x) = \int_{y=0}^\infty 2e^{-x} e^{-y} \mathbb 1 (x > y > 0) \, dy = 2e^{-x} \int_{y=0}^x e^{-y} \, dy. \tag{3}$$ The upper limit of integration changes from $\infty$ to $x$ as a result of the indicator function condition $x > y > 0$. The rest is straightforward. Your answer will be a function of $x$ only.

As an exercise, what is the same conditional density when $X_1$ and $X_2$ remain independent, but no longer identically distributed exponential random variables? More generally, if $$X_1 \sim \operatorname{Exponential}(\lambda_1), \quad X_2 \sim \operatorname{Exponential}(\lambda_2),$$ where the parametrization is by rate, then what is $f_{X_1 \mid X_1 > X_2}$? (Your question is the special case $\lambda_1 = \lambda_2 = 1$.)

As a further exercise, what is the conditional density in the case $\lambda_1 = \lambda_2 = 1$ of $X_1$ given $|X_1 - X_2| \le m$ for some fixed $m > 0$?

  • $\begingroup$ This explanation is so intuitive and I can understand what's happening now. I had been going nowhere by trying to find P(X1 <= t | X1 > X2) by using P(X1 <= t, X1 > X2 | X2 = a), essentially introducing a constant a in order to condition on some fixed value taken by X2, and then comparing a with t. That method led me nowhere. Your explanation helps me approach this starting with the joint density to get the conditional distribution. I'll work out the exercises now. Thank you so much for your time. $\endgroup$ Sep 14, 2023 at 9:18

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