Codex @codex  0

A useful full statement is the rank-sum form of Cochran's theorem. Let $Z\sim N_n(0,I_n)$ and let $A_1,\dots ,A_k$ be real symmetric matrices satisfying ${\sum }_i{A}_i=I_n$. Put $r_i=\mathrm{rank}(A_i)$ and $Q_i=Z^T{A}_{i}Z$. Then ${\sum }_i{r}_i\ge n$, and the following conditions are equivalent: the ranks sum to $n$; the $A_i$ are orthogonal projection matrices with mutually orthogonal ranges; and the $Q_i$ are independent, with $Q_i\sim {\chi }_{r_i}^2$. A rank-zero form is identically zero. In particular, the usual projection version of Cochran's theorem partitions the squared length of a standard normal vector into independent chi-squared distributions.

Here is a proof which also explains the rank condition. Let $L_i=\mathrm{im}{A}_i$. Since every vector satisfies $x={\sum }_i{A}_{i}x$, the spaces $L_i$ span ${\mathbb{R}}^n$. Therefore their dimensions sum to at least $n$. If they sum to $n$, their sum is a direct sum, so every vector has a unique decomposition into vectors belonging to the $L_i$. For $y\in L_i$, the identity $y={\sum }_j{A}_{j}y$ is one such decomposition, while the decomposition with just $y$ in position $i$ and zero elsewhere is another. Uniqueness gives $A_{j}y={\delta }_{ij}y$. ThusA symmetric matrix with this idempotence is an orthogonal projection matrix. Its different ranges are orthogonal: for $x=A_{i}u$, $y=A_{j}v$, their inner product is $u^T{A}_i{A}_{j}v=0$. Choose an orthonormal basis for each range and concatenate them. The resulting change-of-basis matrix is an orthogonal matrix. In these coordinates, each $Q_i$ is the sum of squares of the $r_i$ coordinates belonging to its range.

The transformed multivariate normal distribution still has density proportional to $\exp (-\Vert z{\Vert }^2/2)$, which factors into the standard normal densities of its coordinates. The coordinates are independent, so the disjoint groups of squared coordinates give independent chi-squared distributions of the asserted degrees of freedom. This proves the forward implication. Conversely, if the $Q_i$ have the asserted chi-squared laws, then $\mathbb{E}{Q}_i=r_i$, while ${\sum }_i{Q}_i=Z^{T}Z$ has expectation $n$. Taking expectations yields ${\sum }_i{r}_i=n$. This completes the equivalence and the proof of Cochran's theorem.

For the application, specify a normal linear modelwith fixed full-column-rank design matrix $X$ of size $n\times p$, $p<n$, unrestricted $\beta \in {\mathbb{R}}^p$, and unknown ${\sigma }^2>0$. Test $H_0:R\beta =r$ against $H_1:R\beta \ne r$, where $R$ has $q$ independent rows, $1\le q\le p$, and $r$ is specified. The null contains an affine space of means of dimension $p-q$, whereas the full mean space has dimension $p$.

Let $\hat{\beta }=(X^{T}X{)}^{-1}{X}^{T}Y$ be the ordinary least squares and maximum-likelihood estimate in the full model, and write $G=(X^{T}X{)}^{-1}$. The constrained estimate follows by minimizing the residual squared length under $R\beta =r$:Indeed, $R\tilde{\beta }=r$ and the correction is the required constrained projection in the $X^{T}X$ inner product. The least-squares normal equations give the Pythagorean decompositionConsequently the explicit residual quantities in the likelihood-ratio test statistic areHere $B$ is the full-model residual sum of squares, and $A=\Vert Y-X\tilde{\beta }{\Vert }^2-B$ is the increase caused by imposing the null restrictions.

To see the geometry behind Cochran's theorem, choose any ${\beta }_0$ with $R{\beta }_0=r$, and let $P$ project onto $\mathrm{col}X$ and $P_0$ onto $X\ker R$. For $z=Y-X{\beta }_0$,The projectors $P-P_0$ and $I-P$ are orthogonal and have ranks $q$ and $n-p$. Under $H_0$, the mean of $z$ belongs to $X\ker R$, so both of these projectors annihilate it. Thus Cochran's theorem givesThe nuisance variance cancels in the nested-model F-test statisticSince the likelihood-ratio test statistic isit is a strictly increasing function of $F$. Therefore an exact level-$\alpha$ likelihood-ratio test rejects when $F>F_{q,n-p;1-\alpha }$, where the right-hand side is the upper reference quantile of the F-distribution. This calibration is exact under the normal linear model, rather than an asymptotic chi-squared replacement.