Let $X$, $Y$ be two discrete Poisson Random Variables such that $X \sim \mathcal P(\lambda)$ and $X \sim \mathcal P(\mu)$.
We are asked to evaluate $\mathbb E (X^2+Y^2|X+Y=n)$
We know the sum of $X$ and $Y$ : $$X+Y \sim \mathcal P(\lambda+\mu)$$ I tried using this methode :
$X^2 + Y^2 = (X+Y)^2 - 2XY$
We get by replacing in the expectation and by the linearity of the expectation :
$$\begin{align*} \mathbb E (X^2+Y^2|X+Y=n) &= \mathbb E ((X+Y)^2-2XY|X+Y=n)\\ &= \mathbb E ((X+Y)^2|X+Y=n)-2\mathbb E(XY|X+Y=n)\\ &= \mathbb E (n^2|X+Y=n)-2\mathbb E(XY|X+Y=n) \hspace{1cm} \text{since }X+Y=n \\ &= n^2 -2\mathbb E(XY|X+Y=n) \end{align*} $$ And here I got stuck calculating the term $\mathbb E(XY|X+Y=n)$
Also this idea $$\mathbb E (X^2+Y^2|X+Y=n)=\mathbb E (X^2|X+Y=n) +\mathbb E (Y^2|X+Y=n)$$ didn't work quite well. Maybe I still have some problems understanding the concept.
Thank you !
It depends on what you know about Poisson distributions.
If $X$ and $Y$ are independent then the conditional distribution for $X$ given $X+Y=n$ is $\mathrm{Bin}(n, \frac{\lambda}{\lambda+\mu})$, and similarly for $Y$ is $\mathrm{Bin}(n, \frac{\mu}{\lambda+\mu})$
So $\mathbb E[X \mid X+Y=n]=n\frac{\lambda}{\lambda+\mu}$, $\mathrm{Var}(X \mid X+Y=n)= n\frac{\lambda}{\lambda+mu}(1-\frac{\lambda}{\lambda+\mu})$ and $\mathbb E[X^2 \mid X+Y=n]=n\frac{\lambda}{\lambda+\mu} +(n^2-n)\left(\frac{\lambda}{\lambda+\mu}\right)^2 $ and similarly $E[Y^2 \mid X+Y=n]=n\frac{\mu}{\lambda+\mu} +(n^2-n)\left(\frac{\mu}{\lambda+\mu}\right)^2$
Thus $E[X^2 +Y^2\mid X+Y=n] = n +(n^2-n)\frac{\lambda^2+\mu^2}{(\lambda+\mu)^2}= n^2 -\frac{2\lambda\mu}{(\lambda+\mu)^2} n(n-1) $