Derivation — The Sampling Distribution of S² and Cochran's Theorem

The previous lesson quoted two facts without proof: that the one-sample $t$-statistic $T=(\bar X-\mu)/(S/\sqrt n)$ exactly follows $t(n-1)$, and that a $\chi^2$-based confidence interval for $\sigma^2$ needs $(n-1)S^2/\sigma^2\sim\chi^2(n-1)$. Both rest on a single, non-obvious fact: under normality, $\bar X$ and $S^2$ are independent random variables — a pure coincidence of the normal distribution's algebra, not a general truth (it fails for most other population distributions). This note supplies the missing proof, in three steps: independence of $\bar X$ from the vector of centred deviations $(X_1-\bar X,\ldots,X_n-\bar X)$, the consequent independence of $\bar X$ and $S^2$, and the exact chi-squared distribution of $(n-1)S^2/\sigma^2$. It closes with Cochran's theorem, the general orthogonal-decomposition result of which this is the two-piece special case, and which reappears whenever a sum of squares is split into independent pieces — the algebraic engine underneath ANOVA and regression's sum-of-squares decompositions later in this unit.

Learning objectives. By the end of this note you should be able to: