Suppose X and Y are independent Poisson random variables with respective parameters $lambda$ and 2$lambda$. Find $E[Y-2*X|X+Y=10]$

So, I've tried to tackle the problem several different ways, including just assuming the sum of to Poisson distributions = P($lambda_1$+$lambda_2$) =$2lambda – 2*lambda=0$ and also that

$E[Y-2*X|X+Y=10]$ = $E[Y|X+Y=10]$+$E[-2*X|X+Y=10]$

= $E[Y|X+Y=10]$-2$E[X|X+Y=10]$

and

$E[Y|X+Y=10]$= Bin(n=X+Y=10,p=($frac{2lambda}{2lambda+lambda})=frac{2}{3}$)

$E[X|X+Y=10]$= Bin(n=X+Y=10,p=($frac{lambda}{2lambda+lambda})=frac{1}{3}$)

=> Bin(10,$frac{2}{3}$) – Bin(10,$frac{1}{3}$) = $frac{20}{3}$-2*$frac{10}{3}$ = 0

However, my professor seems to be suggesting that this does not equal 0. Is there a step I'm missing from this process or a part of this probability I'm failing to understand?

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#### Best Answer

We can look at a more general case that I believe will help solve this problem.

begin{eqnarray*} P(X|X+Y=n) &=& frac{P(X=x, Y=n-x)}{P(Z=n)}\ &=& frac{P(X=x)P(Y=n-x)}{P(Z=n)} \ &=& {n choose x} left( frac{lambda_1}{lambda_1+lambda_2} right)^x left( frac{lambda_2}{lambda_1+lambda_2} right)^{n-x} end{eqnarray*}

Which is a binomial pmf with $p = left( frac{lambda_1}{lambda_1+lambda_2} right)$ and expected value $E(X|X+Y=n) = np = n left( frac{lambda_1}{lambda_1+lambda_2} right)$

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