Conditional distribution of two independent exponential RVs

Question

I have two independent, exponential RVs $X$ and $Y$ that both have the same parameter. I am trying to find the distribution of $Y$ given that $X>Y$. So far, I have:

$$P(Y|X>Y) = P(Y=y|X>Y) = P(Y=y, X>Y)/P(X>Y)$$

and I don't know how to proceed at this point. I understand how to get the denominator, but the numerator is really confusing me. Could someone give me a hint?

Answer 1

Hint

$X>Y$ gives you the information that $Y$ is the smaller of the two. Thus the distribution of $Y$ given $X>Y$ is the distribution of the minimum of two independent exponentials, which can be computed by the usual methods for order statistics.

Answer 2

Let the probability density functions be $f_Y(y)=\lambda e^{-\lambda y}\mathbf 1_{0\leqslant y}$ and $f_X(x)=\mu e^{-\mu x}\mathbf 1_{0\leqslant x}$, and $X,Y$ be independent.

Then ths is where you start. The conditioned probability density function is: $$f_{Y\mid Y<X}(y) ~{= \dfrac{f_{Y}(y)~\mathsf P(X>y)}{\mathsf P(Y<X)}\\=\dfrac{\lambda e^{-\lambda y}\cdot e^{-\mu y}}{\int_0^\infty \lambda e^{-\lambda t}\cdot e^{-\mu t}\mathsf d t}\mathbf 1_{0\leqslant y}\\~~\ddots}$$

Answer 3

Well, the problem can be understood pretty easily the way I look at it by assuming an example. Suppose X and Y are random variables corresponding to the rolling of two dices. Then ask yourself a question that say the P(Y=2|X=1, 2, 3, 4, 5, 6); so if X really is greater than Y, then P(Y=y|X>y) = P(Y), if X > Y, else P(Y=y|X>y) = 0.