# Unachievable Cramer-Rao lower bound

I'm working on a problem in Statistics, as follows:

Let $$X_1, ..., X_n$$ be a random sample from a Poisson distribution with parameter $$\theta$$. Denote $$T_n = \sum_{i=1}^n X_i$$.

a) Show that the sample mean $$\overline{X} = T_n/n$$ is an efficient estimator.

b) Suppose that $$g(\theta) = P(X=0) = e^{-\theta}$$. For the minimal variance unbiased estimator $$\hat{g}(\theta) = (1-\frac{1}{n})^{T_n}$$, prove that the Cramer-Rao lower bound is not achievable.

I've got part (a) just fine by showing that $$Var(\overline{X}) = \frac{1}{nI(\theta)}$$, where $$I(\theta)$$ denotes the Fisher information -- that is, the variance of $$\overline{X}$$ attains the Cramer-Rao lower bound.

I'm struggling with (b). I tried to approach it the same way as (a) by showing that $$Var(\hat{g}(\theta)) \neq \frac{1}{nI(\hat{g}(\theta))}$$. However, when I try to compute the Fisher information for $$I(\hat{g}(\theta)) = -E(\frac{d^2}{d \theta^2} log(\hat{g}(\theta))$$, I run into a problem -- the first derivative of $$log(\hat{g}(\theta))$$ with respect to $$\theta$$ ends up being zero, since there are no $$\theta$$'s involved in the formula for $$\hat{g}(\theta)$$.

How can I refine my logic for part (b) ?

Thanks!

For part (a), I would rather just show that $$\overline X$$ satisfies the condition of equality in the Cramer-Rao inequality.

For $$x=(x_1,\ldots,x_n)$$ with $$x_i\in\{0,1,\ldots\}$$ for all $$i$$, pmf of $$(X_1,\ldots,X_n)$$ is

$$p_{\theta}(x)=\frac{e^{-n\theta}\theta^{n\bar x}}{\prod_{i=1}^n (x_i!)}$$

Therefore,

$$\frac{\partial}{\partial\theta}\ln p_{\theta}(x)=\frac{n}{\theta}(\bar x-\theta)\tag{*}$$

This is precisely the equality condition, i.e. $$\frac{\partial}{\partial\theta}\ln p_{\theta}(x)$$ is proportional to $$T(x)-\theta$$ with $$T(x)=\bar x$$.

Since $$\overline X$$ is unbiased for $$\theta$$, equation $$(*)$$ implies that $$\overline X$$ is the minimum variance unbiased estimator of $$\theta$$ with variance of $$\overline X$$ attaining the Cramer-Rao lower bound for $$\theta$$.

For (b), first find the CR lower bound for $$g(\theta)=e^{-\theta}$$. It is given by $$\text{CRLB}(g(\theta))=\frac{(g'(\theta))^2}{I(\theta)}\quad,$$

where $$I(\theta)=\frac{n}{\theta}$$ is the information within the whole sample.

That is, $$\text{CRLB}(g(\theta))=\frac{\theta e^{-2\theta}}{n}$$

Now with $$a=1-\frac1n$$ for $$n>1$$,

\begin{align} \operatorname{Var}_{\theta}(a^{T_n})&=\operatorname{E}_{\theta}\left[(a^{T_n})^2\right]-\left(\operatorname{E}_{\theta}[a^{T_n}]\right)^2 \\&=\operatorname{E}_{\theta}\left[(a^2)^{T_n}\right]-(g(\theta))^2\tag{**} \end{align}

Since $$T_n=\sum\limits_{i=1}^n X_i\sim \mathsf{Poisson}(n\theta)$$, it is true that $$\operatorname{E}(c^{T_n})=e^{n\theta(c-1)}$$ for any constant $$c$$.

So $$(**)$$ I think reduces to $$\operatorname{Var}_{\theta}(a^{T_n})=\exp\left[n\theta(a^2-1)\right]-e^{-2\theta}=\cdots=e^{-2\theta}(e^{\theta/n}-1)$$

Finally take the ratio $$\operatorname{Var}_{\theta}(a^{T_n})/\text{CRLB}(g(\theta))$$:

$$\frac{\operatorname{Var}_{\theta}(a^{T_n})}{\text{CRLB}(g(\theta))}=\frac{n(e^{\theta/n}-1)}{\theta}=\frac{n}{\theta}\left(\frac{\theta}{n}+\frac{\theta^2}{2n^2}+\cdots\right)=1+\frac{\theta}{2n}+\cdots>1$$

• While this calculation is not valid for $n=1$, the same conclusion holds while working separately with a single observation. Dec 2, 2019 at 15:03