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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($\lambda_1$+$\lambda_2$) =$2\lambda - 2*\lambda=0$ 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=($\frac{2\lambda}{2\lambda+\lambda})=\frac{2}{3}$)

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

=> 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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    $\begingroup$ The information you need (explained in several different ways) is at stats.stackexchange.com/questions/429564/…. Although your answer is correct, the notation makes no sense at several points, which is what your professor may be objecting to. In particular, most people would interpret expressions like "$\operatorname{Bin}(10,2/3)$" as referring either to distributions or random variables, but not to expectations. $\endgroup$
    – whuber ♦
    Commented Feb 7, 2020 at 21:56
  • $\begingroup$ Cool - I've also had someone suggest E[Y-2X|X+Y=10]=E[Y-2X|Y=10-X]=E[10-X-2X]=10-3*E[X]=10-3*lambda (since E[X]=lambda). Would you know if those statements actually equivalent, or am I going down the wrong path? $\endgroup$
    – That One Dude Mike
    Commented Feb 7, 2020 at 23:23
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    $\begingroup$ The point where the conditional probability was replaced by a probability is invalid: it's tantamount to supposing the condition $X+Y=10$ doesn't affect anything. One way to see why not is to replace "10" everywhere by "0". You would wind up concluding $E[Y-2X\mid X+Y=0]=0-3\lambda$ but that's obviously incorrect because $X+Y=0$ implies $X=Y=0,$ whence $Y-2X=0$ whose expectation is $0,$ not $-3\lambda.$ $\endgroup$
    – whuber ♦
    Commented Feb 7, 2020 at 23:44

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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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