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

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?

Contents

$$begin{eqnarray*} P(X|X+Y=n) &=& frac{P(X=x, Y=n-x)}{P(Z=n)}\ &=& frac{P(X=x)P(Y=n-x)}{P(Z=n)} \ &=& {n choose x} left( frac{lambda_1}{lambda_1+lambda_2} right)^x left( frac{lambda_2}{lambda_1+lambda_2} right)^{n-x} end{eqnarray*}$$
Which is a binomial pmf with $$p = left( frac{lambda_1}{lambda_1+lambda_2} right)$$ and expected value $$E(X|X+Y=n) = np = n left( frac{lambda_1}{lambda_1+lambda_2} right)$$