Given $X_1,\ldots, X_7$ are i.i.d r.vs which follow $N(\mu, \sigma^2)$.
(a) Find the UMVUE of $\frac{\mu}{\sigma^2}$.
(b) Compute variance of UMVUE in part (a)
(c) Find the lower bound of the variance of unbiased estimator of $\frac{\mu}{\sigma^2}$ for $n$ i.i.d variables $X_1, X_2,…,X_n$ follows $N(\mu, \sigma^2)$.
My thought: First, we realize that $E(S^2) = \sigma^2$ where $S^2 = \frac{\sum_{i=1}^{7} (X_i – \overline{X})^2}{6}$ is a sample variance. Thus, $E(\frac{1}{S^2}) = \frac{A}{\sigma^2}$ (I could not see how to compute this constant $A$). Since the sample mean and sample variance of $X_1,\ldots, X_7$ are independent, we have: $E(\frac{\overline{X}}{S^2}) = E(\overline{X})E(\frac{1}{S^2}) = \frac{\mu}{A}$, so $\frac{\overline{X}}{AS^2}$ is the unbiased estimator of $\frac{\mu}{\sigma^2}$.
Now, this unbiased estimator could be expressed as a complete sufficient statistics $(\sum_{i=1}^{7} X_i, \sum_{i=1}^{7} X_i^2)$ because $S^2 = \frac{\sum_{i=1}^{7} X_i^2 – \frac{(\sum_{i=1}^{7} X_i)^2}{7}}{6}$. This implies $\fbox{$\frac{\overline{X}}{AS^2}$}$ is a UMVUE of $\frac{\mu}{\sigma^2}$
(b) Thanks to the work by NCH below, we obtain UMVUE of $\frac{\mu}{\sigma^2}$ is $\frac{2}{3} \frac{\overline{X}}{S^2}$. Thus, to compute $Var(\frac{\overline{X}}{S^2})$, we need to compute $E(\frac{\overline{X}^2}{S^4})$. First, since $\overline{X}$ and $\frac{1}{S^2}$ are independent, $\overline{X}^2$ and $\frac{1}{S^4}$ are independent. So $E(\frac{\overline{X}^2}{S^4}) = E(\overline{X}^2)E(\frac{1}{S^4})$. Now, since $Var(\overline{X}) = \frac{Var(X_1)}{7} = \frac{\sigma^2}{7}$, $E(\overline{X}) = \frac{\sigma^2}{7} + \mu^2$
My question: Could anyone please help me determine the constant $A$ above to complete this proof? I feel so shameful not to be able to compute $E(\frac{1}{S^2})$.
Best Answer
By Cochran's theorem, $$\frac{6S^2}{\sigma^2} =\sum_{i=1}^{7}\left(\frac{X_i-\overline{X}}{\sigma}\right)^2 \sim \chi^2_6$$
Denote this r.v. by $Z$.
The p.d.f. of $\chi^2_6$ equals to $$ f_Z(x)=\frac{1}{2^3\Gamma(3)}x^{2} e^{-x/2}=\frac{1}{16}x^{2} e^{-x/2}, \ x>0 $$
Calculate $$ E\left(\frac1{S^2}\right)=E\left(\frac{\sigma^2}{6S^2}\cdot\frac{6}{\sigma^2}\right)=\frac{6}{\sigma^2}E\left(\frac{1}{Z}\right). $$ $$ E\left(\frac{1}{Z}\right) = \int\limits_0^\infty \frac1x \frac{1}{16}x^{2} e^{-x/2}\, dx = \frac{1}{16}\int\limits_0^\infty x e^{-x/2}\, dx = \frac{4}{16}=\frac14. $$ Finally, $$E\left(\frac1{S^2}\right)= \frac{6}{4\sigma^2}$$ and the constant $A=6/4=1.5$.