Let $X$ and $Y$ be identically independent Poisson r.v. with parameter $\lambda$. Find $P(X=k|X+Y=n )$.
By defitnion,
$$ P(X=k|X+Y=n ) = P(X=k | Y=n-k) = \frac{ p_{XY}(k,n-k)}{p_Y(n-k)}$$
now since independence we have $p_X(k)p_Y(n-k)$ and so we have
$$ P(X=k|X+Y=n ) = p_X(X=k) $$ but this is not the answer I should get. What am I doing wrong in this problem?
It is not correct to state that $P(X=k\mid X+Y=n)=P(X=k\mid Y=n-k)$.
$$\begin{aligned}P\left(X=k\mid X+Y=n\right)P\left(X+Y=n\right) & =P\left(X=k\wedge X+Y=n\right)\\ & =P\left(X=k\wedge Y=n-k\right)\\ & =P\left(X=k\right)P\left(Y=n-k\right)\\ & =e^{-\lambda}\frac{\lambda^{k}}{k!}e^{-\lambda}\frac{\lambda^{n-k}}{\left(n-k\right)!}\\ & =e^{-2\lambda}\frac{\lambda^{n}}{n!}\binom{n}{k} \end{aligned} \tag1$$
Further if $X,Y\sim\mathsf{Poisson}\left(\lambda\right)$ are independent then $X+Y\sim\mathsf{Poisson}\left(2\lambda\right)$ leading to:$$P\left(X+Y=n\right)=e^{-2\lambda}\frac{\left(2\lambda\right)^{n}}{n!}\tag2$$
Based on $(1)$ and $(2)$ we find: $$P\left(X=k\mid X+Y=n\right)=\binom{n}{k}2^{-n}$$
