Suppose that $X_1,X_2,...,X_n$ is a random sample from normal$(\mu, σ^2)$.

Find the UMVUE for $\mu ^{2}$ by assuming $\sigma ^{2}$ is unknown.

My approach:

The distribution of the sample mean, namely $$\bar X\sim\mathcal N\left(\mu,\frac{\sigma^2}{n}\right)$$

If $\sigma$ is known, a complete sufficient statistic for $\mu$ is $$\sum_{i=1}^n X_i \quad(\text{ and hence }\bar X)$$


\begin{align} \operatorname{Var}(\bar X)&=\frac{\sigma^2}{n} \\\implies E(\bar X^2)&=\frac{\sigma^2}{n}+\mu^2 \end{align}

That is, $$E\left(\bar X^2-\frac{\sigma^2}{n}\right)=\mu^2 $$

By Lehmann-Scheffe, $$\bar X^2-\frac{\sigma^2}{n}$$ is the UMVUE of $\mu^2$ when $\sigma^2$ is known.

My problem is how do I show the case where $\sigma^2$ is unknown?

  • $\begingroup$ You mean sample from $N(\mu,\sigma^2)$? For $\sigma$ unknown, find an unbiased estimator of $\sigma^2$ based on a complete sufficient statistic. $\endgroup$ – StubbornAtom Apr 24 at 16:25
  • $\begingroup$ I made a correction concerning $\mu$. I know $S^2$ is complete sufficient statistic for $\sigma^2$. But what do I do from there? $\endgroup$ – Lady Apr 24 at 16:42
  • $\begingroup$ Yes... I apologize for leaving out essential information. $\endgroup$ – Lady Apr 24 at 16:46
  • $\begingroup$ Then $(\bar X,S^2)$ is complete sufficient and $S^2$ is unbiased for $\sigma^2$. That's all you need. $\endgroup$ – StubbornAtom Apr 24 at 16:48
  • $\begingroup$ No.... I need the UMVUE of $\mu^2$ instead. $\endgroup$ – Lady Apr 24 at 16:49

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