TITLE: Independence of sample mean and variance through covariance
QUESTION [5 upvotes]: I have seen the text book derivation where the independence is established through factoring the joint distribution. But has anyone tried to prove that the covariance is zero?.
Let $Z_{i}$ come from a standard normal distribution.
Let $X = \bar{Z}$ be the sample mean and let $$Y = \sum_i\frac{1}{n-1} (Z_{i} -\bar{Z})^2$$ be the sample variance.
Prove that $\operatorname{Cov}(X,Y) = 0$ .
I am getting terms like E($\bar{Z}^3$) in the expansion which is making it very cumbersome to handle...

REPLY [3 votes]: First show that
$$
\operatorname{cov} (\bar{Z}, Z_i - \bar{Z}) = 0.
$$
These two random variables are not just normally distributed, but jointly normally distributed, since both are linear combinations of the same sequence of independent normals.  Therefore, they are uncorrelated only if they're independent.
Hence the sample variance, which is a function of $Z_i-\bar{Z}$ for $i=1,\ldots,n$, is independent of the sample mean.
Later addendum: I should add that the whole vector $(Z_1- \bar Z,\ldots,Z_n-\bar Z)$ is independent of $\bar Z$, since the covariance between $\bar Z$ and that vector is a matrix whose every entry is $0$ and we have joint normality.  If $(Z_1- \bar Z,\ldots,Z_n-\bar Z)$ is independent of $\bar Z$, then $((Z_1- \bar Z)^2+\cdots+(Z_n-\bar Z)^2$ is independent of $\bar Z$.