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I am trying to solve Robert Hogg's mathematical anaysis 6th exercise 6.2.11. The problem says. Let $\bar{X}$ be the mean of a random sample of size n from a $N(\theta,\sigma^2)$ distribution, $-\infty<\theta<\infty$, $\sigma^2>0$. Assume that $\sigma^2$ is known. Show that $\bar{X}-\frac{\sigma^2}{n}$ is an unbiased estimator of $\theta^2$ and find its efficiency.

The first part pretty easy, the efficiency part sames very conter-intuitive.

My solution:

Since $\bar{X}$ be the mean of a random sample of size n from a $N(\theta,\sigma^2)$ by central limit theorem we know that $Var(\bar{X})=E(\bar{X}^2)-[E(\bar{X})]^2=\frac{\sigma^2}{n}$ and $E(\bar{X})=\theta$,.

Therefore, $E(\bar{X}^2)=\frac{\sigma^2}{n}+\theta^2$.

$E(\bar{X}^2-\frac{\sigma^2}{n})=E(\bar{X}^2)-\frac{\sigma^2}{n}=\frac{\sigma^2}{n}+\theta^2-\frac{\sigma^2}{n}=\theta^2$

Therefore, $\bar{X}^2-\frac{\sigma^2}{n}$ is an unbiased estimator of $\theta^2$

For the second part, first I need to calculate the variance of $\bar{X}^2-\frac{\sigma^2}{n}$

$Var(\bar{X}^2-\frac{\sigma^2}{n})=Var(\bar{X}^2)$

By CTL we know that $\frac{\bar{X}-\theta}{\frac{\sigma}{\sqrt{n}}}$ has a N(0,1) distribution. Further the square of standard normal has a $\chi^2(1)$ distribution.

Therefore,$ Var((\frac{\bar{X}-\theta}{\frac{\sigma}{\sqrt{n}}})^2)=1/2*4=2$, i.e variance of $\chi^2(1)=2$

$\therefore Var(\frac{\bar{X}^2-2\theta\bar{X}+\theta^2}{\sigma^2/n})=2$

Solve this equation:

$\frac{n^2}{\sigma^4}Var(\bar{X}^2)-\frac{4\theta^2n^2}{\sigma^4}Var(\bar{X})+0=2$, we already know what is $Var(\bar{X})$

$\therefore Var(\bar{X}^2)=\frac{2\sigma^4}{n^2}+\frac{4\theta^2\sigma^4}{n}$

By this step, I get the variance of the $\bar{X}^2-\frac{\sigma^2}{n}$ which is $\frac{2\sigma^4}{n^2}+\frac{4\theta^2\sigma^4}{n}$

Since $\bar{X}^2-\frac{\sigma^2}{n}$ is an unbiased estimator of $\theta^2$, we need to find the fisher information of $\theta^2$.

We first calculate the $I(\theta)$ for $\bar{X}$

$f(\bar{X};\theta)=\frac{1}{\sqrt{2\pi}\frac{\sigma}{n}}exp(-\frac{1}{2}\frac{(\bar{X}-\theta)^2}{\sigma^2/n})$

$logf(\bar{X};\theta)=log\frac{1}{\sqrt{2\pi}\frac{\sigma}{n}}-\frac{1}{2}\frac{(\bar{X}-\theta)^2}{\sigma^2/n}$

$\frac{\partial logf(\bar{X};\theta)}{\partial \theta}=\frac{n}{\sigma^2}(\bar{X}-\theta)$

Therefore, $I(\theta)=Var(\frac{\partial logf(\bar{X};\theta)}{\partial \theta})=Var(\frac{n}{\sigma^2}(\bar{X}-\theta))=Var(\frac{n}{\sigma^2}\bar{X})=\frac{n^2}{\sigma^4}Var(\bar{X})=\frac{n^2}{\sigma^4}.\frac{\sigma^2}{n}=\frac{n}{\sigma^2}$

Next we will use formular $I(g(\theta))=\frac{I(\theta)}{(g'(\theta))^2}$ to calculate $I(\theta^2)$

$I(\theta^2)=\frac{I(\theta)}{((\theta^2)')^2}=\frac{\frac{n}{\sigma^2}}{4\theta^2}=\frac{n}{4\theta^2\sigma^2}$

Efficiency=$\frac{\frac{1}{nI(\theta^2)}}{Var(\bar{X}^2-\frac{\sigma^2}{n})}=\frac{4\theta^2\sigma^2}{n^2}.\frac{n^2}{2\sigma^4+4\theta^2\sigma^2n}=\frac{1}{\frac{2\sigma^2}{4\theta^2}+n}$

We can see when sample becomes bigger efficiency becomes smaller, this is very counter-intuitive to me.

Please help me to find which part is wrong for the calculation.

Thank you very much.

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  • $\begingroup$ Maybe you've N(theta,sigma-squared) distribution, so please edit it $\endgroup$ May 3, 2015 at 9:01

1 Answer 1

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Thanks you guy's responses. I found where the problem was.

I should use $f(X;\theta)$ to calculate the $I(\theta)$, but not $f(\bar{X};\theta)$

Now $I(\theta)=\frac{1}{\sigma^2}$ and $I(\theta^2)=\frac{1}{4\theta^2\sigma^2}$

So Efficiency=$\frac{\frac{1}{nI(\theta)}}{Var(\bar{X}^2+\frac{\sigma^2}{n})}=\frac{4\theta^2\sigma^2}{n}.\frac{n^2}{2\sigma^4+4\theta^2\sigma^2n}=\frac{n}{\frac{\sigma^2}{2\theta^2}+n}=\frac{1}{1+\frac{\sigma^2}{2n\theta^2}}$

When n becomes bigger efficiency close to 1.

Now I think the solution is correct.

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