Finding UMVUE, given complete and sufficient statistic Let $X_1,X_2,…,X_n(n≥3) $ be a random sample from Poisson(), where$ θ∈(0,∞)$ is unknown and let
$T=\sum _{i=1}^n X_i$
The UMVUE of $ e^{−2\theta}*\theta^3$ is
Since UMVMUE is a Complete Sufficient Statistics then
$E(ϕ(t))=e^{−2\theta}*\theta^3$
$\sum \phi(t)∗ \frac{e^{−nθ}*(nθ)^t}{t!}=e^{−2\theta}*\theta^3$
$\sum \phi(t)∗ \frac{(nθ)^t}{t!}=e^{−2\theta+n\theta}*\theta^3$
$\sum \phi(t)∗ \frac{(nθ)^t}{t!}=\sum \frac{(−2\theta+n\theta)^r}{r!}*\theta^3$
How do we proceed after this?
First, to use a standard notation, let's set
$$S=\sum_i X_i$$
and observe that $S$ is CSS, Complete and Sufficient Statistic.
Then let's construct an unbiased estimator for $g(\theta)=e^{-2\theta}\theta^3$
A very simple unbiased estimator is the following:
$$T=\mathbb{1}_{(X_1\cdot X_2=2)}$$
where it indicates that $T=1$ if $X_1\cdot X_2=2$ and it is zero elsewhere.
Immediately you can verify that $\mathbb{E}[T]=e^{-2\theta}\theta^3$
Now you can apply Rao-Blackwell theorem calculating
$$\mathbb{E}[T|S]$$
which is your UMVUE, given that $S$ is also complete.

$$\mathbb{E}[T|S]=\mathbb{P}[T=1|S=s]=\frac{e^{-2\theta}\theta^3\mathbb{P}[S=s|T=1]}{\mathbb{P}[S=s]}=$$
$$=\frac{e^{-2\theta}\theta^3\mathbb{P}\left[ \sum_{i=3}^n X_i=s-3\right]}{\frac{e^{-n\theta}(n\theta)^s}{s!}}=\frac{e^{-2\theta}\theta^3\cdot \frac{e^{-(n-2)\theta}[(n-2)\theta]^{s-3}}{(s-3)!}}{\frac{e^{-n\theta}(n\theta)^s}{s!}}=\frac{(n-2)^{s-3}}{n^s}\cdot\frac{s!}{(s-3)!}$$
First, to use a standard notation, let's set
$$S=\sum_i X_i$$
and observe that $S$ is CSS, Complete and Sufficient Statistic.
Then let's construct an unbiased estimator for $g(\theta)=e^{-2\theta}\theta^3$
A very simple unbiased estimator is the following:
$$T=\mathbb{1}_{(X_1\cdot X_2=2)}$$
where it indicates that $T=1$ if $X_1\cdot X_2=2$ and it is zero elsewhere.
Immediately you can verify that $\mathbb{E}[T]=e^{-2\theta}\theta^3$
Now you can apply Rao-Blackwell theorem calculating
$$\mathbb{E}[T|S]$$
which is your UMVUE, given that $S$ is also complete.

$$\mathbb{E}[T|S]=\mathbb{P}[T=1|S=s]=\frac{e^{-2\theta}\theta^3\mathbb{P}[S=s|T=1]}{\mathbb{P}[S=s]}=$$
$$=\frac{e^{-2\theta}\theta^3\mathbb{P}\left[ \sum_{i=3}^n X_i=s-3\right]}{\frac{e^{-n\theta}(n\theta)^s}{s!}}=\frac{e^{-2\theta}\theta^3\cdot \frac{e^{-(n-2)\theta}[(n-2)\theta]^{s-3}}{(s-3)!}}{\frac{e^{-n\theta}(n\theta)^s}{s!}}=\frac{(n-2)^{s-3}}{n^s}\cdot\frac{s!}{(s-3)!}$$