# 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.

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!