1
$\begingroup$

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!

$\endgroup$

1 Answer 1

2
$\begingroup$

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

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.