If $X_1,X_2,…,X_n\sim Pois(λ)$, find the UMVUE of $\exp(−2λ)$. I know based on lehmann scheffe theorem, the step is
(1)find $q(x)$ an unbiased estimater of $\exp(−2λ)$
(2) $T(X)$ is sufficient and complete
(3) $s(*)=\mathbb E(q(x)|t(x))$ is a UMVUE
(4) var$(s(*))$ <$\infty$, then it is the UMVUE
The answer IS: y=(-1)^x, when x is even y=1,when x is odd, y=-1.
But I still do not know step by step, how to make it work?
An easier approach is to take the approach in the two linked questions, using an initial unbiased estimator of $\exp(−2\lambda)$ to be $1$ if $X_1+X_2=0$ and to be $0$ if $X_1+X_2 \gt 0$.
Here is a sketch of an approach using an initial unbiased estimator of $\exp(−2\lambda)$ to be $+1$ if $X_1$ is even and to be $-1$ if $X_1$ is odd, which following your notation we might call $q(X_1)$:
1) $q(X_1)$ is an unbiased estimator as its conditional expectation, given $\exp(−2\lambda)$ and so $\lambda$, is $+1\left(e^{-\lambda}+e^{-\lambda}\frac{\lambda^2}{2!}+\cdots\right) -1\left(e^{-\lambda}\frac{\lambda^1}{1!}+e^{-\lambda}\frac{\lambda^3}{3!}+\cdots\right) = e^{-\lambda}\left(1-\frac{\lambda^1}{1!}+\frac{\lambda^2}{2!}-\frac{\lambda^3}{3!}+\cdots\right)=e^{-2\lambda} $
2) A complete sufficient statistic for a Poisson distribution is $T(\mathbf X)=\sum\limits_1^n X_i$
3) The conditional distribution of $X_1$, given $\sum\limits_1^n X_i=t$, is binomial with parameters $t$ and $\frac{1}{n}$, so you can show the conditional probability that $X_1$ is even is $\frac{1+\left(1-\frac2n\right)^t}{2}$ and that $X_1$ is odd is $\frac{1-\left(1-\frac2n\right)^t}{2}$, and thus $\mathbb E[q(X_1) \mid T(\mathbf X)=t] = \left(1-\frac2n\right)^t$ and this is then the uniformly minimum-variance unbiased estimator for $\exp(−2\lambda)$ using the Lehmann–Scheffé theorem
4) This UMVUE estimator is bounded: when $n=1$ it is $-1$ or $+1$, while for $n \ge 2$ it is bounded by $0$ and $1$. So it has a finite variance.
