Find UMVUE of $P(X_1+X_2+X_3 =2)$ Where $X_1, ... X_n \sim$ Poisson$(\lambda)$ (independent) Let $X_1, ... , X_n \sim $ Poisson$(\lambda)$ (independence is assumed) Find the UMVUE $T(X_1,...,X_n) $ of $g(\lambda) := P(X_1+X_2+X_3=2)$
So we have $g(\lambda) := P(X_1+X_2+X_3=2) = e^{-3\lambda}\frac{(3\lambda)^2}{2!}$
$S(\underline{X}) = \sum_{i=1}^n X_i$ is complete and sufficient. and $S(\underline{X}) \sim $ Poisson$(n\lambda)$
So I want to find $h$ such that  $T(\underline{X}) = h(S(\underline{X}))$.
Setting up we have $\displaystyle E[h(S)] = e^{-3\lambda}\frac{(3\lambda)^2}{2!} \implies \sum_{k=0}^\infty \frac{e^{-n\lambda(n\lambda)^k}}{k!}h(S) = e^{-3\lambda}\frac{(3\lambda)^2}{2!}$
The next step I made was to get a sum on the RHS so I have and moved the $e^{-n\lambda} $ over:
$\displaystyle \sum_{k=0}^\infty \frac{(n\lambda)^k}{k!}h(k) = \sum_{k=0}^n\frac{((n-3)\lambda)^k}{k!}\frac{(3\lambda)^2}{2!} = \sum_{k=0}^\infty \frac{(n\lambda)^k}{k!} \bigg( \sum_{j=0}^k {k \choose j}(n\lambda)^{-j}(-3\lambda)^{j}\frac{(3\lambda)^2}{2!} \bigg)$
However, I don't think I have computed this correctly, from what I have above I should be able match to get this $T(\underline{X})$
Looking for input if I did this correctly.
EDIT: I have $h(S)$ but I should have $h(k)$ so I will change this
You want to solve for $h(\cdot)$ where for every $\lambda&gt;0$,
$$E[h(S)]=\sum_{k=0}^\infty h(k)\frac{e^{-n\lambda}(n\lambda)^k}{k!}=c\,e^{-3\lambda}\lambda^2$$
for some positive constant $c$.
That is,
$$
\sum_{k=0}^\infty \frac{h(k)n^k}{k!}\lambda^k=c\,e^{(n-3)\lambda}\lambda^2
=c\sum_{j=0}^\infty \frac{(n-3)^j}{j!}\lambda^{j+2}
=\sum_{k=2}^\infty \frac{c(n-3)^{k-2}}{(k-2)!}\lambda^{k}
$$
Now compare coefficients of $\lambda^k$.
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By definition, the answer is given by
\begin{align}
&amp;\bbox[10px,#ffd]{\sum_{k_{1} = 0}^{\infty}{\lambda^{k_{1}}\expo{-\lambda} \over k_{1}!}
\sum_{k_{2} = 0}^{\infty}{\lambda^{k_{2}}\expo{-\lambda} \over k_{2}!}
\sum_{k_{3} = 0}^{\infty}{\lambda^{k_{3}}\expo{-\lambda} \over k_{3}!}\
\overbrace{\bracks{z^{2}}z^{k_{1} + k_{2} + k_{3}}}^{\ds{\large \delta_{k_{1} + k_{2} + k_{3}, 2}}}}
\\[5mm] = &amp;\
\expo{-3\lambda}\bracks{z^{2}}\bracks{\sum_{k = 0}^{\infty}{\pars{\lambda z}^{k} \over k!}}^{3}
\\[5mm] = &amp;\
\expo{-3\lambda}\
\underbrace{\bracks{z^{2}}\expo{3\lambda z}}
_{\ds{{\pars{3\lambda}^{2} \over 2!}}} =
\bbx{\large{9 \over 2}\,\lambda^{2}\expo{-3\lambda}} \\ &amp;
\end{align}