# Difficulty on understanding conditional probability

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 $$P(Y>X+m | Y > X) = P(Y > m)$$

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I don't know how to show the long formula properly. Any idea? Thanks! – Ethan Mar 19 '11 at 14:12
Also is it possible to link to a comment? For example, the second comment after this reply math.stackexchange.com/questions/26386/… – Ethan Mar 19 '11 at 14:15

(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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Thanks! I should mention that m and k are nonnegative constants. In Part 3, (1) I was wondering if it is true when X is any random variable, not necessarily independent with Y? \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*} If $P(Y-k=m | Y > k)=P(Y=m)$, then the above becomes $\sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X) = P(Y=m) \sum_{k} P(X=k | Y > X) = P(Y=m)$? (2) Is the same conclusion true for exponential distribution? – Ethan Mar 19 '11 at 18:53
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 $Y-2=1$. – Henry Mar 20 '11 at 23:05