I have been working on a problem, which goes as follows: Given the statistical model $(\mathcal{X},\mathcal{B},\mathcal{P})$, where $\mathcal{P}=\{P_{\lambda}^{\otimes}:P_{\lambda}=Pos(\lambda), \lambda>0\}$, use the Rao-Blackwell theorem, to find an UMVU estimator (UMVUE) for $\lambda^2$ via $\hat{\gamma}: \mathbb{N}^n_0 \longrightarrow \mathbb{N}_0 $, $\hat{\gamma}(x)=x_1x_2$, with respect to the statistic $T:\mathbb{N}^n_0 \longrightarrow \mathbb{N}_0, T(x)=\sum_\limits{{i=1}}^n x_i$.
In other words we have to calculate the following expected value:
$$\hat{\gamma}^*(t):=\mathbb{E}_{\lambda}(\hat{\gamma}(X)|T(X)=t), \quad t \in \mathbb{N}_0$$
where $X=(X_1, \dots , X_n)$, ($X_1, \dots , X_n \stackrel{iid}{\sim}Pos(\lambda))$ is the random variable inducing the statistical model. We need $T$ to be a complete and sufficient statistic and $E_{\lambda}(\hat{\gamma})<\infty$, for $\hat{\gamma}^*$ being UMVUE.
I already solved this by applying Lehmann-Scheffe, to another unbiased estimator for $\lambda^2$ and the same statistic $T$, then concluded that $\hat{\gamma}^*$ has to be equal to my UMVUE, since UMVUE are unique.
I then tried to calculate the above expected value, but I am stuck evaluating it, here is what I have done so far:
For $t \in \mathbb{N_0}$:
\begin{align} \mathbb{E}_{\lambda}(\hat{\gamma}(X)|T(X)=t) & =\sum_\limits{x \in \mathbb{N}_0^n}x_1x_2\mathbb{P}_{\lambda}(X=x|T(X)=t)\\ & =\sum_\limits{x \in \mathbb{N}_0^n}x_1x_2\mathbb{P}_{\lambda}(X_1=x_1, \dots , X_n=x_n|\sum_\limits{{i=1}}^n X_i=t) \end{align}
I have concluded that: \begin{align}\mathbb{P}_{\lambda}(X_1=x_1, \dots , X_n=x_n|\sum_\limits{{i=1}}^n X_i=t) & = \frac{\mathbb{P}_{\lambda}(X_1=x_1, \dots , X_n=x_n, \sum_\limits{{i=1}}^n X_i=t)}{\mathbb{P}_{\lambda}(\sum_\limits{{i=1}}^n X_i=t)} \\ & = \frac{\mathbb{P}_{\lambda}(X_1=x_1, \dots , X_n=x_n)\chi_{\{\sum_\limits{{i=1}}^n X_i=t\}}}{e^{n\lambda}\frac{(n\lambda)^{t}}{t!}} \\ & =\frac{e^{n\lambda}\frac{\lambda^{t}}{x_1!\dots x_n!}\chi_{\{\sum_\limits{{i=1}}^n X_i=t\}}}{e^{n\lambda}\frac{(n\lambda)^{t}}{t!}} \\ &= \frac{t!}{x_1!\dots x_n!n^t}\chi_{\{\sum_\limits{{i=1}}^n X_i=t\}}\end{align}
I then tried to proceed with the evaluation :
\begin{align}\sum_\limits{x \in \mathbb{N}_0^n}x_1x_2\frac{t!}{x_1!\dots x_n!n^t}\chi_{\{\sum_\limits{{i=1}}^n X_i=t\}} & = \sum_\limits{x \in \{\sum_\limits{{i=1}}^n x_i=t\}}x_1x_2\frac{t!}{x_1!\dots x_n!n^t} \\ &= \sum_\limits{x_1,x_2 \in \{x_1,x_2>0, x_1+x_2 \leq t\}}\frac{x_1x_2}{x_1!x_2!}\sum_\limits{x \in \{\sum_\limits{{i=3}}^n x_i \\ =t-x_1-x_2\}}\frac{(t-x_1-x_2)!}{x_3!\dots x_n!n^{t-x_1-x_2}} \\ & =\sum_\limits{x_1,x_2 \in \{x_1,x_2>0, x_1+x_2 \leq t\}}\frac{x_1x_2}{x_1!x_2!}\\ &\stackrel{?}{=}\frac{t(t-1)}{n^2} \end{align}
My Question: Im not quite sure, about the last two equations, if the second last is correct, then the second sum would be equal to one, since we are summing over the support of a density function, but I doubt that the very last "equation" would follow, but this is what it should be equal to. Any help is appreciated!
