# Basu's theorem for normal sample mean and variance

I'm working on the following problem: Suppose that $X \sim N(\mu,\sigma^2)$. Let the sample mean and variance, $\overline{X}$ and $S^2$ be defined as usual so that $\mathbb{E} S^2 = \sigma^2$. Prove that the sample mean is independent of the sample variance. Given that both $\mu$ and $\sigma^2$ are unknown, find the MVUE for $\mu \sigma^2$.

The second part is simple. Given that $\overline{X}$ and $S^2$ are independent, and that $(\overline{X}, S^2)$ is a jointly complete and sufficient statistic for $(\mu, \sigma^2)$, we have that $\phi ( \overline{X}, S^2) = \overline{X} S^2$ is unbiased for $\mu \sigma^2$. By Lehmann-Schefee, $\phi$ is the MVUE for $\mu \sigma^2$.

To prove independence, I would like to implore Basu's Theorem. This says, briefly, that any boundedly (which I will ignore) complete sufficient statistic is independent of any ancillary statistic. It is not hard to show that the normal distribution is exponential class. Moreover, given known variance, we get that $\sum X_i$ is a complete and sufficient statistic for $\mu$. It is clear that $S^2$ is ancillary, since (accept my abuse of notation) for any $a \in \mathbb{R}$, $$S^2+a = \frac{1}{n-1} \sum_{i=1}^n \Big( X_i + a - \frac{1}{n}\sum_{i=1}^n \big( X_i + a \big) \Big)^2 = \frac{1}{n-1} \sum_{i=1}^n \Big( X_i - \overline{X}\Big)^2$$ so that $S^2$ is location invariant, and hence ancillary.

So, if we were only interested in estimating $\mu$, we would be done. My question is, how do I deal with applying Basu's theorem when I have a jointly complete and sufficient statistic. Does it suffice to only show that $\sum X$ is complete and sufficient statistic for $\mu$ even though I'm estimating a function of $\mu$ and $\sigma^2$?

Note: there is a similar (mostly unanswered), but not identical question here: UMVUE using complete and sufficient statistic