If $X_1,..., X_n$ are iid with distribution $Poi(\lambda)$, I want to find $\mathbb{E}[X_1\cdot X_2 \mid \sum_{i=1}^{n}X_i = u]$. I know I can turn this into
$$\mathbb{E}\left[X_1\cdot X_2 \mid \sum_{i=1}^{n}X_i = u\right] = \frac{\sum_{k=0}^{\infty}\mathbb{P}(X_1 \cdot X_2 = k, \sum_{i=1}^{n}X_i = u) \cdot k}{\mathbb{P}(\sum_{i=1}^{n}X_i = u)}$$ but I am not sure this really helps me. I know the sum is also Poisson-distributed with parameter $n \lambda$, so the denominator is just $\frac{(n\lambda)^ue^{-n\lambda}}{u!}$. It seems to me that $X_1 \cdot X_2$ shouldn't really have a nice distribution because at any given $k$ you have to check how many combinations of non-negative integers there are that multiply to that number and then what the respective probabilities would be, so I don't really know how to go on or if I went down the wrong path.
Hint: $$\mathbb E(X_1 X_2|\sum X_i)=\frac{\sum X_i (\sum X_i -1)}{n^2}$$
$\mathbb E(X_1 X_2)=\lambda^2$ and $\sum X_i$ is a complete and sufficient statistic. So $\mathbb E(X_1 X_2|\sum X_i)$ is UMVUE for $\lambda^2$. By UMVUE for $\lambda^k$ UMVUE for $\lambda^2$ is $\frac{\sum X_i (\sum X_i -1)}{n^2}.$