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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?

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  • $\begingroup$ Note that $P[Y|X>Y]$ does not make sense since $Y$ is not an event. I think you want to compute $P[Y\leq y|X>Y]$, for $y>0$. And this cannot be computed without knowing the joint distribution for $(X,Y)$. $\endgroup$ – Michael Feb 27 '18 at 0:08
  • $\begingroup$ I am trying to find the distribution of Y given X>Y, hence P(Y=y|X>Y). $\endgroup$ – mistersunnyd Feb 27 '18 at 0:12
  • $\begingroup$ $P[Y=y|X>Y]$ does not constitute a distribution function. For example, if $X$ and $Y$ happen to be independent, then clearly $P[Y=y|X>Y]=0$ for all $y \in \mathbb{R}$. Of course, no independence assumptions have been given so it is impossible to know the joint distribution of $(X,Y)$, as I said before. $\endgroup$ – Michael Feb 27 '18 at 0:13
  • $\begingroup$ Ah I'm very sorry, but X and Y are independent. Let me edit my question. $\endgroup$ – mistersunnyd Feb 27 '18 at 0:14
  • $\begingroup$ As Michael said, $P(Y=y\mid X>Y) = 0$ for any $y$. What you might actually want / have use for is $P(Y\le y\mid X>Y)$ or a conditional density. $\endgroup$ – spaceisdarkgreen Feb 27 '18 at 0:16
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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.

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  • $\begingroup$ Hmm I still fail to see what I should be computing in the numerator. I am trying to find what distribution (Gamma, Poisson, Exponential, etc) Y given X>Y is, and I am not sure where to start for the numerator. I know how to find the expected value of this conditional thing, but I just don't see how I'm supposed to find another distribution. $\endgroup$ – mistersunnyd Feb 27 '18 at 1:29
  • $\begingroup$ @mistersunnyd Well, I'm telling you how to compute the whole thing (forget the numerator, it's just the whole thing times $1/2$ to cancel off the denominator). Conditional on $X>Y,$ $Y$ has the distribution of the minimum of two independent exponentials. $\endgroup$ – spaceisdarkgreen Feb 27 '18 at 2:09
  • $\begingroup$ Ok I see, so Y=min(X,Y)~exp(2). The pdf would be 2e^-2y, and that multiplied by 1/2 would be e^-2y. This doesn't look like an exponential, but could it be a Gamma? $\endgroup$ – mistersunnyd Feb 27 '18 at 2:19
  • $\begingroup$ @mistersunnyd I think you're confused with the $1/2$ thing. The conditional pdf of $Y$ given $Y<X$ is $2e^{-2y}.$ This is an exponential with mean $1/2.$ $\endgroup$ – spaceisdarkgreen Feb 27 '18 at 2:28
  • $\begingroup$ Ah I see. Thank you for your help! $\endgroup$ – mistersunnyd Feb 27 '18 at 2:30
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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: $$f_{Y\mid Y<X}(y) ~{= \dfrac{f_{Y}(y)~\mathsf P(X>y)}{\mathsf P(Y<X)}\\=\dfrac{\lambda e^{-\lambda y}\cdot e^{-\mu y}}{\int_0^\infty \lambda e^{-\lambda t}\cdot e^{-\mu t}\mathsf d t}\mathbf 1_{0\leqslant y}\\~~\ddots}$$

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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.

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