Is the distribution of one exponential will be smaller than a second one Uniform?

Question

I came by an expression which I am not sure I understand.

If:

$X_1 \sim exp(\lambda)$

$X_2 \sim exp(\lambda)$

Then:

$P(X_1<X_2|X_2) \sim Uniform(0,1)$

Where it is not clear to me what "$|X_2$" means, any idea how to read this? (thanks)

Answer 1

Recall that for every event $A$ and every random variable $X$, $P[A\mid X]$ is by definition a random variable $Y$ such that:

1. $Y$ is $X$-measurable
2. for every bounded measurable function $b$, $E[b(X)Y]=E[b(X)\mathbf 1_A]$

In the case at hand, $Y=P[X_1\lt X_2\mid X_2]$ is a random variable $a(X_2)$ (condition 1.) such that $E[a(X_2)b(X_2)]=E[b(X_2)\mathbf 1_{X_1\lt X_2}]$ for every bounded measurable function $b$ (condition 2.).

Using the independence of $X_1$ and $X_2$ and the identity $P[X_1\lt x]=1-\mathrm e^{-\lambda x}$ for every nonnegative $x$, one gets $Y=1-\mathrm e^{-\lambda X_2}$ hence the goal is to show that $Y=1-\mathrm e^{-\lambda X_2}$ is uniform on $(0,1)$.

But $F:x\mapsto1-\mathrm e^{-\lambda x}$ is the CDF of $X_2$ and one knows that, for every continuous random variable $X$ with CDF $F_X$, $F_X(X)$ is uniform on $(0,1)$. The result follows.

To rediscover this general fact in the case at hand, note that, for every $y$ in $(0,1)$, $[Y\leqslant y]=[X_2\leqslant x]$ with $\lambda x=-\log(1-y)$ hence $P[Y\leqslant y]=1-\mathrm e^{-\lambda x}$ yields $P[Y\leqslant y]=1-(1-y)=y$.

Exercise: Show that the result still holds if one replaces the hypothesis that $X_1$ and $X_2$ are i.i.d. with an exponential distribution of the same parameter by the hypothesis that they are i.i.d. with a continuous distribution.