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In some thesis, while they verified the proposed method, they just mentioned that

$$ E[X|X+Y] = (X+Y)\lambda_X/(\lambda_X+\lambda_Y) $$

where X, Y are independent Poisson variables with means $ \lambda_X, \lambda_Y $. But I can't make out the equation. Is there any way to prove the solution? Kindly let me know.

Thank you for considering my request.

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I finally solved the problem by myself. The solve is shown below:

Previously, we should know that

  1. $P[X+Y=k]=e^{-\lambda_x-\lambda_y}\frac{(\lambda_x+\lambda_y)^k}{k!} $

verification : Poisson Distribution of sum of two random independent variables $X$, $Y$

  1. $P[X=k|X+Y=i]= \frac{P[X=k,Y=i-k]}{P[X+Y=i]} $

Now, let get started from

$$ E[X=k|X+Y=i]= \sum_{k=0}^{i}k\frac{P[X=k,Y=i-k]}{P[X+Y=i]} = \sum_{k=0}^{i}k\frac{P[X=k]P[Y=i-k]}{P[X+Y=i]}=\sum_{k=0}^{i}k\frac{\frac{e^{-\lambda_x}\lambda_x^k}{k!}\frac{e^{-\lambda_y}\lambda_y^{(i-k)}}{(i-k)!}}{\frac{e^{-\lambda_x-\lambda_y}(\lambda_x+\lambda_y)^i}{i!}}=\sum_{k=1}^{i}\frac{i!}{(k-1)!(i-k)!}\frac{\lambda_x^k\lambda_y^{(i-k)}}{(\lambda_x+\lambda_y)^i}$$ $ k=0$ is vanished. $$ = \frac{i\lambda_x}{(\lambda_x+\lambda_y)^i}\sum_{k=1}^{i}\frac{(i-1)!}{(k-1)!(i-k)!}\lambda_x^{k-1}\lambda_y^{(i-k)}=\frac{(X+Y)\lambda_x}{\lambda_x+\lambda_y}$$

$\because \sum_{k=1}^{i-1}\frac{(i-1)!}{(k-1)!(i-k)!}\lambda_x^{k-1}\lambda_y^{(i-k)}=(\lambda_x+\lambda_y)^{i-1}$ $, i=X+Y$

$\therefore E[X|X+Y]=\frac{(X+Y)\lambda_x}{\lambda_x+\lambda_y}$

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