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$X$ and $Y$ are independent and geometrical distributed with same parameter $p$. How to find $E(X|X+Y=k)$ for all $k =$ $2,3,4$....

I thought $E(X|X+Y=k) = \sum_{x=1}^{k-1} xp(x,x+y=k)/p(x+y=k)$ $n$ is infinity

now, I found $p(x+y= k) = p^2 (1-p)^{k-2}$ Am I on the correct path ? How do I find $p(x,x+y=k)$ ?

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    Ok, can we find $Var(Y|X+Y)$ from here?2017-05-11

2 Answers 2

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Here is a simple method:

Because the variables are iid, $E[X|X+Y = k] = E[Y|X+Y = k]$.

Now $ E[X|X+Y = k] + E[Y|X+Y = k] = E[X+Y|X+Y = k] = k$

Thus $E[X|X+Y = k] = k/2$.

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    Thanks Gautam, I got the mistake.2012-11-19
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Your method: $E[X|X+Y=k]=\sum_{m=1}^{k-1} m \frac{P(X=m,X+Y=k)}{P(X+Y=k)}$

$P(X=m,X+Y=k) = p^2(1-p)^{k-2}$

$P(X+Y=k) = \sum_{m=1}^{k-1}P(X=m,X+Y=k) = (k-1)p^2(1-p)^{k-2} $

Simplify to get $E[X|X+Y=k]=\frac{1}{k-1} \sum_{m=1}^{k-1}m = k/2$