$E(T(X))$ where $A:={n(\bar X-\mu)^2\over \sigma^2}\sim\chi^2_{(1)}$ ; $B:={\sum\limits_i(X_i-\bar X)^2\over \sigma^2}\sim \chi^2_{(n-1)}$

Question

$X_1,X_2,\cdots X_n$ is a random sample from $N(\mu,\sigma^2)$.

I have to find out

an unbiased estimator of ${\mu^2\over \sigma^2}$

My work:

Initially I take $T(X)={\bar X^2\over {1\over n}\sum(X_i-\bar X)^2}$ I need to find $E(T(X))$.

I know

$A:={n(\bar X-\mu)^2\over \sigma^2}\sim\chi^2_{(1)}$ ; $B:={\sum\limits_i(X_i-\bar X)^2\over \sigma^2}\sim \chi^2_{(n-1)}$

But how to find out $E(T(X))$ if I know $E(A)$ and $E(B)$ seperately?

My thoughts:

The fact that $\bar X$ is independent of $\sum\limits_i(X_i-\bar X)^2$might be of help.

Some modification of ${A\over B}$ might lead to the $F$ distribution.

Please help me proceed.

Answer 1

Note that $C=n\bar{X}^2/\sigma^2$ follows a noncentral $\chi^2$ distribution with parameters $\nu=1$ and $\lambda=n(\mu/\sigma)^2$. Therefore,$(n-1)C / B = n(n-1)\bar{X}^2/(\sigma^2 B)$ follows a noncentral $F$ distribution: $$\mathbb{E}\left(\frac{n(n-1)\bar{X}^2}{\sum_i(X_i-\bar{X})^2}\right) = \frac{(n-1)(1+n(\mu/\sigma)^2)}{n-3}.$$ Rewriting yields: $$\mathbb{E}\left(\frac{n\bar{X}^2}{\sum_i(X_i-\bar{X})^2}\right) = \frac{1}{n-3}+\frac{n}{n-3}\frac{\mu^2}{\sigma^2}.$$ Therefore your unbiased estimator is: $$\frac{(n-3)\bar{X}^2}{\sum_i(X_i-\bar{X})^2} - \frac{1}{n}.$$

Answer 2

An answer has already been selected, but here is something I'd like to add: Suppose you have $X\sim\chi^2_p$, $Y\sim \chi^2_q$ with $X,Y$ independent and you want $E(X/Y)$. Then $E(X/Y)=E(X)E(1/Y)$.

$E(X)$ is straightforward to obtain. But how to obtain $E(1/Y)$? A straight way is to observe that if $Y\sim Gamma$ then $1/Y$ is also "certain" $Gamma$, (inverted Gamma), and if you write its pdf, you can see that its expectation can be easily evaluated.