[Math] UMVU for $\sigma ^ p$ normal distribution.

normal distributionparameter estimationstatistical-inferencestatistics

I am having some trouble with the following problem:
let $X_1, \ldots, X_n$ independent from a Normal distribution with unknown mean $\mu$ and variance $\sigma^2$. Find the UMVU estimator for $\sigma^p$ where $p>0$ is real.
I have found the maximum likelihood estimator but was not able to find and correct its mean, so I do not know how to proceed. (I have also tried to use Rao-Blackwell theorem but I did not know which unbiased estimator to use).
Thank you for your help.

Best Answer

Define $\overline X=\frac{1}{n}\sum\limits_{k=1}^n X_k$ and $S^2=\frac{1}{n-1}\sum\limits_{k=1}^n (X_k-\overline X)^2$.

Then a complete sufficient statistic for $(\mu,\sigma^2)$ is given by $(\overline X, S^2)$.

By Lehmann-Scheffe theorem, any unbiased estimator of $\sigma^p$ based on $(\overline X, S^2)$ will be the UMVUE of $\sigma^p$.

Recall that $$\frac{(n-1)S^2}{\sigma^2}\sim \chi^2_{n-1}$$

So,

\begin{align} E\left[\frac{(n-1)S^2}{\sigma^2}\right]^{p/2}&=\frac{1}{2^{\frac{n-1}{2}}\Gamma\left(\frac{n-1}{2}\right)}\int_0^\infty t^{p/2}\,e^{-t/2}\,t^{\frac{n-1}{2}-1}\,\mathrm{d}t \end{align}

Simplifying both sides of the above equation you will finally arrive at $$E\left[cS^p\right]=\sigma^p$$

for some function $c(.)$ of $n$ and $p$.

Thus the UMVUE of $\sigma^p$ is $c(n,p) S^p$.

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