Finding UMVUE for Poisson distribution using Rao Blackwell Let $X_1,X_2,\ldots,X_n$ be a random sample from a Poisson distribution with parameter $\lambda$. Let $\gamma(\lambda)=P(X\le 1)$.
Find UMVUE for  $\gamma(\lambda)$.
This is my attempt:
First I defined a indicator function as $T'=I_{(X_1\le 1)}=\begin{cases}
1,  & \text{if $X_1\le1$ } \\
0, & \text{otherwise}  \\
\end{cases}$  
which is unbiased estimator for  $\gamma(\lambda)$. Also since this belongs to one parameter exponential family $T=\sum X_i$ is a sufficient statistic for  $\gamma(\lambda)$.
Then by Rao Blackwell Theorem,
$E[T'|T=t]$ is a UMVUE for  $\gamma(\lambda).$
\begin{align}
& \operatorname E[T'\mid T=t]=1\cdot P[T'=1\mid T=t]+0\cdot P[T'=0\mid T=t] \\[10pt]
= {} & {P[X_1\le 1 \text{ and } X_1+ X_2+ X_3+\cdots+ X_n=t] }\over P[\sum X_i=t] 
\end{align}
Since at this moment$ X_1\le 1$ it becomes that $X_2+ X_3+\cdots X_n\ge t-1$.Then since $ X_1\le 1$ and $X_2+ X_3+\cdots+X_n\ge t-1$ are independent from one another
\begin{align}
& P[X_1\le 1 \text{ and } X_1+ X_2+ X_3+\cdots+X_n=t] \\[10pt]
= {} & P[X_1\le 1] \cdots P\left[\sum_{i=2}^n X_i\ge t-1\right]
\end{align}
Now I am stuck in computing $P[\sum_{i=2}^n X_i\ge t-1]$.   
$\sum_{i=2}^n X_i$ follows a poisson with parameter $(n-1) \lambda$.  In finding $P[\sum_{i=2}^n X_i\ge t-1]$ I came up to cumulative of Poisson
$e^{-(n-1)\lambda}\sum_{i=0}^{y-1}(({n-1)\lambda)}^i / i!$.Is there a way I can simplify this.Please help me to find a UMVUE for this problem.
 A: To simplify the notations, introduce $Y=X_2+\cdots+X_n$ and note that $(X_1,Y)$ is independent with Poisson distributions of parameters $\lambda$ and $\mu=(n-1)\lambda$ respectively and that $T=X_1+Y$ is Poisson with parameter $\lambda+\mu$. For every nonnegative integer $t$, $c(t)=E(T'\mid T=t)$ is
$$
c(t)=P(X_1\leqslant1\mid T=t)=\frac{P(X_1=0,Y=t)+P(X_1=1,Y=t-1)}{P(T=t)}.
$$
By independence,
$$
c(t)=\frac{P(X_1=0)P(Y=t)+P(X_1=1)P(Y=t-1)}{P(T=t)}.
$$
The Poisson distributions yield
$$
c(t)=\frac{\mathrm e^{-\lambda}\mathrm e^{-\mu}\mu^t/t!+\mathrm e^{-\lambda}\lambda\mathbf 1_{t\geqslant1}\mathrm e^{-\mu}\mu^{t-1}/(t-1)!}{\mathrm e^{-\lambda-\mu}(\lambda+\mu)^{t}/t!}=\frac{\mu^t+\lambda t\mu^{t-1}}{(\lambda+\mu)^{t}},
$$
and finally,
$$
c(t)=\left(\frac{n-1}n\right)^t\,\left(1+\frac{t}{n-1}\right).
$$
Thus, the UMVUE for $\gamma(\lambda)=P(X_1\leqslant1)=\mathrm e^{-\lambda}(1+\lambda)$ is
$$
\hat\gamma(\lambda)=c(T)=\left(\frac{n-1}n\right)^T\,\left(1+\frac{T}{n-1}\right).
$$
Edit: Since $T$ is Poisson with parameter $n\lambda$,
$$
E(c(T)^2)=\sum_{t=0}^\infty\mathrm e^{-n\lambda}\frac{(n\lambda)^t}{t!}\left(\frac{n-1}n\right)^{2t}\,\left(1+\frac{t}{n-1}\right)^2=\ldots
$$
