Suppose $X$ follows exponential distribution with a positive parameter $\lambda$ and $Y$ is a positive continuous random variable, independent of $X$. Then what is the conditional distribution of $X-Y$ given $X > Y$?
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1$\begingroup$ The distribution of $X-Y$ truncated to the positive half-line. $\endgroup$– Xi'anCommented Oct 30, 2020 at 7:15
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$\begingroup$ It's quite straightforward (trivial, even) -- just use the lack of memory property. $\endgroup$– Glen_bCommented Oct 30, 2020 at 8:47
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$\begingroup$ @Glen_b I know the memorylessness property of exponential random variable. But here $Y$ is also a random variable. Does it matter? $\endgroup$– Van TomCommented Oct 30, 2020 at 11:02
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$\begingroup$ It makes no difference $\endgroup$– Glen_bCommented Oct 30, 2020 at 12:59
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1 Answer
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Consider the distribution of $X\vert X>Y$ or the memorylessness. Then see if it is easy to change it to $X-Y\vert X>Y$
But here Y is also a random variable. Does it matter?
If you know $\mathbb{P}(A \vert B)$ and $B$ is itself a random variable, then you can find the probability of $\mathbb{P}(A)$ as a compound distribution or by using the law of total probability
$$\mathbb{P}(A) = \sum_{\forall B} \mathbb{P}(A \vert B)\mathbb{P}(B) $$
if $ \mathbb{P}(A \vert B) = f(A)$ is a function independent of $B$ then it can be taken out of the sum and you get
$$\mathbb{P}(A) = \sum_{\forall B} \mathbb{P}(A \vert B)\mathbb{P}(B) = f(A)\sum_{\forall B} \mathbb{P}(B) = f(A) $$
Similarly when $\mathbb{P}(X-Y\vert X>Y, Y)$ is independent from $Y$ then you know $\mathbb{P}(X-Y\vert X>Y)$
answered Oct 30, 2020 at 11:20
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$\begingroup$ Just complementing, here he is using the Law of total probability. $\endgroup$– igorkfCommented Oct 30, 2020 at 12:07
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$\begingroup$ @igorkf I knew that I was using some sort of 'well known law with a name', but I couldn't get the name. $\endgroup$ Commented Oct 30, 2020 at 12:11