Independence of sample mean and sample variance for normal samples

Remarks

Put $Z_i = (X_i - \mu)/\sigma$, which are independent standard normal. Cochran's theorem with $A_1 = \frac{1}{n} J$ and $A_2 = I_n - \frac{1}{n} J$, where $J$ is the matrix of ones, shows that $n \bar{Z}^2$ and $\sum_i (Z_i - \bar{Z})^2 = (n-1)S^2/\sigma^2$ are independent, with $\chi^2$ laws with $1$ and $n - 1$ degrees of freedom. Independence of $\bar{X}$ itself follows because $\bar{Z}$ is a function of the projection $A_1 Z$, which is independent of $A_2 Z$. Another proof uses an orthogonal matrix whose first row is $(1, \dots, 1)/\sqrt{n}$.

At work: the statistic $T = \sqrt{n}(\bar{X} - \mu)/S$ is a standard normal variable divided by $\sqrt{V/(n-1)}$, with $V \sim \chi^2_{n-1}$ independent of it. So $T$ has the Student t-distribution with $n - 1$ degrees of freedom. This gives the one-sample $t$-test and the $t$ confidence interval for $\mu$.

The normality hypothesis is necessary: for i.i.d. samples from a nondegenerate distribution with finite variance and $n \ge 2$, independence of $\bar{X}$ and $S^2$ holds only if the distribution is normal. For example, for Bernoulli variables, $\bar{X} = 0$ forces $S^2 = 0$, while $S^2 > 0$ is possible when $0 < \bar{X} < 1$.

Going deeper

For a student

The result is the case of two pieces of Cochran's theorem. It underlies the exact $t$-test, the $\chi^2$ confidence interval for $\sigma^2$, and through them much of classical small-sample inference. The analogous result for multivariate normal samples says that the sample mean vector and the sample covariance matrix are independent, the latter with a Wishart distribution.