Solution to Problem 5.

(5) [15 points] Let $X_1$, $X_2$, $\ldots$ be i.i.d. $N(\mu,\sigma^2)$. Define the sample mean and variance

$$\begin{eqnarray*} \bar{X}_n & \; = \; & \frac{1}{n} \sum_{i=1}^n X_i \\ S_n^2... ...& \frac{1}{n-1} \sum_{i=1}^n \left( X_i - \bar{X}_n \right)^2 . \end{eqnarray*}$$

Assume $n > 1$. (a) Show that $E[S_n^p]$ $=$ $2^p \sigma^p \Gamma((p + n-1)/2)/\Gamma((n-1)/2)$ provided $p > -(n-1)$.

Solution: We know that

$$\frac{(n-1) S_n^2}{\sigma^2} \; \sim \; \chi^2_{n-1} \; = \; \mbox{gamma}((n-1)/2,2).$$

Hence,

$$S_n^2 \; \sim \; \mbox{gamma}((n-1)/2,2 \sigma^2/(n-1)).$$

Now, if $X$ $\sim$ gamma($\alpha,\beta$) distribution, we have

$$\begin{eqnarray*} E[X^p] & \; = \; & \int_0^{\infty} \, \frac{1}{\Gamma(\alpha... ... & \; = \; & \frac{\Gamma(p+\alpha) \beta^p}{\Gamma(\alpha)} . \end{eqnarray*}$$

We need $p+\alpha > 0$, i.e. $p > - \alpha$. Hence,

$$E[S_n^p] \; = \; E[(S_n^2)^{p/2}] \; = \; \frac{\Gamma((p+n-1)/2) 2^{p/2} \sigma^p}{\Gamma((n-1)/2)(n-1)^{p/2}},$$

provided $p > -(n-1)$. (b) Find the UMVUE of $\sigma^p$ for $p > -(n-1)/2$.

Solution: We know that $(\bar{X}_n,S_n^2)$ is complete and sufficient for $(\mu,\sigma^2)$. So we just need to find a function of $(\bar{X},S^2)$ whose expectation is $\sigma^p$, and which has finite variance. From the result above, our answer is

$$\frac{\Gamma((n-1)/2)(n-1)^{p/2}}{2^{p/2} \Gamma((p+n-1)/2)} S^p_n .$$

The restriction that $p > -(n-1)/2$ guarantees that this has finite variance.


(c) What is $E[\bar{X}_n/S_n ]$? Find the UMVUE of $\mu/\sigma$.

Solution: We know that $\bar{X}_n$ and $S_n^2$ are independent, so

$$\begin{eqnarray*} E[\bar{X}_n/S_n] & \; = \; & E[\bar{X}_n] E[ S_n^{-1} ] \\ ... ...Gamma((n-2)/2) 2^{-1/2} \sigma^-1}{\Gamma((n-1)/2)(n-1)^{-1/2}} \end{eqnarray*}$$

It follows that the UMVUE of $\mu \sigma^{-1}$ is $C \bar{X}_n$ where

$$C \; = \; {\Gamma((n-1)/2)(n-1)^{-1/2}}{\Gamma((n-2)/2)} .$$