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

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 $$\sum_{i=1}^n X_i \quad(\text{ and hence }\bar X)$$

Now,

\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?

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