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I have some confusion when understanding the concept of conditional probability.

  1. Given any two random variables $X$ and $Y$ and any two constants $m$ and $k$, Is it true that $P(Y-X=m | Y > X) = P(Y = m+k | Y > k)? $

    My guess is it is not always true, because
    $\begin{align*} P(Y-X=m | Y > X) & = \sum_{k} P(Y-X=m, X=k | Y > X) \\ & = \sum_{k} P(Y-X=m | X=k, Y > X) P(X=k | Y > X) \\ & = \sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X). \end{align*}$

  2. What if $X$ and $Y$ are i.i.d.? Even further, how about when $X$ and $Y$ are i.i.d. with some memoryless distribution, i.e. exponential or geometric distribution?

    In my previous post under the setting of $X$ and $Y$ being i.i.d. with geometric distribution, Henry wrote $P(Y-X=m | Y > X) = P(Y = m+k | Y > k),$ about which he said in his comment it is true because "$k$ is merely shorthand for" $X$. But I still don't quite understand that yet.

  3. If Y is a random variable subject to a memoryless distribution, i.e. exponential or geometric distribution and X is any other random variable with any distribution, is it true that $P(Y>X+m | Y > X) = P(Y > m)$

Thanks for your help!

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    Also is it possible to li$n$k to a comme$n$t? For example, the seco$n$d comment after this reply http://math.stackexchange.com/questions/26386/questions-about-geometric-distribution/26412#264122011-03-19

1 Answers 1

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(Other readers should follow the links to see the context. $X$ and $Y$ are discrete random variables taking values on the positive integers)

  1. No they are not true in general.

  2. What is true, if $X$ and $Y$ are independent, is $Pr(Y-X=m | Y > X, X=k) = Pr(Y = m+k | Y > k) $

  3. If $Y$ is memoryless (i.e. geometric) then $Pr(Y=m+k | Y > k) = Pr(Y = m) $, so if $X$ and $Y$ are independent $Pr(Y=m+X | Y > X, X = k) = Pr(Y = m) $ and since that is true for all $k$ $Pr(Y-X=m | Y > X) = Pr(Y = m) $

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    Are you sure you can say P(Y-X=m | X=k, Y > X) = P(Y-k=m | Y > k) if $X$ and $Y$ are not independent? Suppose $X$ is $1,2,3,4$ with equal probability and $Y= X \pm 1$ with equal probability. Then let $m=1$ and $k=2$. We have P(Y-X=1 | X=2, Y > X) = 1 since the only possibility given $X=2$ and Y > X is $(2,3)$. But P(Y-2=1 | Y > 2) = 1/2 since the possibilities given Y > 2 are $(2,3)$, $(3,4)$, $(4,5)$ and $(4,3)$ and only when $Y=3$ do we have $Y-2=1$.2011-03-20