Conditional distribution of $X$ given $X<Y$ where $X,Y$ are exponential variables

Question

Let $X$ and $Y$ be independent, exponential random variables whose rates are denoted by $\lambda$ (for $X$) and $\mu$ (for $Y$). Find the conditional distribution for $X$ given $X < Y$.

I'm a bit stumped at how to start this problem. I know that $P(X < Y) = \frac{\lambda}{\lambda + \mu}$, and I suspect that I need to use this information for the derivation I need to find, but I am not sure what to do.

Answer 1

We can derive the conditional cdf

$$P\left(X\leq x\mid X<Y\right) = \frac{P\left(X\leq x, X< Y\right)}{P\left(X<Y\right)}.$$

The denominator is, indeed, $\frac{\lambda}{\lambda+\mu}$. The numerator is the same as $P\left(X \leq \min\left(x,Y\right)\right)$, and can be calculated through conditioning:

\begin{eqnarray*} P\left(X \leq \min\left(x,Y\right)\right) &=& \mathbb{E}\left(P\left(X \leq \min\left(x,Y\right)\right)\mid Y\right)\\ &=& \int^x_0 \left(1-e^{-\lambda y}\right)\mu e^{-\mu y}dy + \left(1-e^{-\lambda x}\right)\int^{\infty}_x \mu e^{-\mu y}dy, \end{eqnarray*}

which can be computed somewhat tediously but straightforwardly. Once you finish it, you will find that the conditional distribution of $X$ given $X < Y$ is $Exp\left(\lambda+\mu\right)$, that is, the conditional distribution of $X$ given $X <Y$ behaves like the unconditional distribution of $\min\left(X,Y\right)$.