Basu's theorem

Remarks

Proof idea: fix a measurable set $B$ and let $q = P_\theta(V \in B)$, which does not depend on $\theta$ since $V$ is ancillary. By sufficiency, the conditional probability $k(T) = P_\theta(V \in B \mid T)$ can be chosen free of $\theta$. Then $k(T) - q$ is bounded and $E_\theta[k(T) - q] = 0$ for all $\theta$, so bounded completeness gives $k(T) = q$ almost surely; that is, $V \in B$ is independent of $T$. Example: for $X_1, \dots, X_n$ i.i.d. $\mathcal{N}(\mu, \sigma^2)$ with $\sigma^2$ known and $\mu \in \R$, $\bar{X}$ is complete and sufficient for $\mu$, and the sample variance $S^2$ is ancillary; so $\bar{X}$ and $S^2$ are independent. Since $\sigma^2$ was arbitrary, this proves the independence of $\bar{X}$ and $S^2$ for every normal sample. Completeness cannot be dropped: for $n \ge 2$ the whole sample $T = (X_1, \dots, X_n)$ is sufficient, and $V = X_1 - X_2$ is ancillary, but $V$ is a non-constant function of $T$, so they are not independent.

Going deeper

For a student

The best-known use is the independence of the sample mean and the sample variance of a normal sample, which underlies the $t$-distribution of $\sqrt{n}(\bar{X} - \mu)/S$. More generally, in an exponential family of full rank the natural sufficient statistic is complete, so it is independent of every ancillary statistic. The theorem also allows one to compute $E_\theta[V \mid T]$ as $E_\theta[V]$ for an integrable ancillary $V$, which simplifies Rao–Blackwell computations.