Cramer-Rao lower bound for $g(\lambda)$ = $e^{-\lambda}$ when $x_i \sim Pois(\lambda)$ Let $x_1 ... x_n$ be $Pois(\lambda)$, $\lambda>0$ and $n \geq 2$. Use $W(X)=I(X_1=0)$ to estimate $g(\lambda)=e^{-\lambda}$. Additionally, $T(X)= \sum_{i=1}^n X_i$ is a sufficient statistic for $\lambda$.
I have already proved that $\phi(T(X))=(1-\frac{1}{n})^{\sum_{i=1}^n X_i}$ is an unbiased estimator of $e^{-\lambda}$.
I want to prove if $\phi(T(X))$ attains the Cramer-Rao lower bound, but i get lost trying to calculate it by applying the definition.
 A: First, let's calculate the variance of $\phi(T(X))$:
$$\begin{align}
\mathbb E[\phi(T(X))^2]&=\sum_{k=1}^\infty \frac{e^{-n\lambda}(n\lambda)^k}{k!}\cdot\left(1-\frac{1}{n}\right)^{2k}\\
&=e^{-n\lambda}\sum_{k=0}^\infty\frac{(n\lambda)^k}{k!}\left(1-\frac 2n+\frac 1{n^2}\right)^k\\
&=e^{-n\lambda}\sum_{k=0}^\infty\frac{1}{k!}\left(n\lambda-2\lambda+\frac{\lambda}{n}\right)^k\\
&=e^{-n\lambda}\cdot e^{n\lambda-2\lambda+\frac\lambda n}\\
&=e^{\lambda\left(\frac{1}{n}-2\right)}
\end{align}$$
And, using the fact that $\phi(T(X))$ is unbiased:
$$\begin{align}
\text{Var}(\phi(T(X)))&=\mathbb E[\phi(T(X))^2]-\mathbb E[\phi(T(X))]^2\\
&=e^{\lambda\left(\frac{1}{n}-2\right)}-e^{-2\lambda}\\
&=e^{-2\lambda}\cdot\left(e^{\lambda/n}-1\right)
\end{align}$$
Now, we have to compare this to the Cramér-Rao bound. The likelihood function is given by:
$$\begin{align}
\ell(\lambda)&=\sum_{i=1}^n\log\left(\frac{e^{-\lambda}\lambda^{X_i}}{X_i!}\right)=-n\lambda + \log(\lambda)T(X)-\sum_{i=1}^n\log(X_i!)
\end{align}$$
Then, we can get the Fisher information:
$$\begin{align}
\mathcal I_F&=-\mathbb E\left[\frac{\partial^2\ell(\lambda)}{\partial\lambda^2}\right]=\mathbb E\left[\frac{T(X)}{\lambda^2}\right]=\frac n\lambda
\end{align}$$
Finally, the Cramér-Rao bound for $g(\lambda)$ is:
$$\frac{g'(\lambda)^2}{\mathcal I_F}=\frac{e^{-2\lambda}}{n/\lambda}=\frac{\lambda}{n}e^{-2\lambda}$$
Which is different from the variance we previously found, so the estimator you found does not attain the Cramér-Rao bound. You can also prove, using Taylor expansion, that your estimator is asymptoticaly efficient, that is, the variance of the estimator is asymptotically equal to the Cramér-Rao bound.
Hope it was helpful!
