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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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In the summation, what is m,n? –  Gautam Shenoy Nov 19 '12 at 8:59
    
n should be infinity and k =1 –  user669083 Nov 19 '12 at 9:03
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2 Answers

up vote 3 down vote accepted

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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yes this is correct but can it be solved my way. –  user669083 Nov 19 '12 at 9:14
    
Yes. You want $P(X=m,X+Y=n)$ right? This is same as $P(X=m,Y=n-m)$ which is same as due to independence $P(X=m)P(Y=n-m)$. Can you do it now? –  Gautam Shenoy Nov 19 '12 at 9:17
    
it just gives me numerator equal to denominator and results in summation of 1+2+3+4...infinity. Where am I wrong ? –  user669083 Nov 19 '12 at 9:26
    
Your expression for P(X+Y = k) suggested above is incorrect. It should be $(k-1)(1-p)^{k-2}p^2$ –  Gautam Shenoy Nov 19 '12 at 9:31
    
Can you help me there ? As I supposed p(z=x+y) = Sigma for all x p(X=x,Y=z-x). Since these are independent, I supposed Sigma for all x p(X=x)p(Y=z-x) –  user669083 Nov 19 '12 at 9:35
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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 $$E[X|X+Y=k]=\frac{1}{k-1} \sum_{m=1}^{k-1}m = k/2$$

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