Cochran's theorem

In statistics, Cochran's theorem, devised by William G. Cochran, is a theorem used to justify results relating to the probability distributions of statistics that are used in the analysis of variance.

Statement

Let U₁, ..., U_N be i.i.d. standard normally distributed random variables, and $U=[U_1,...,U_N{]}^T$. Let $B^{(1)},B^{(2)},\dots ,B^{(k)}$be symmetric matrices. Define r_i to be the rank of $B^{(i)}$. Define $Q_i=U^T{B}^{(i)}U$, so that the Q_i are quadratic forms. Further assume $\sum _i{Q}_i=U^{T}U$.

Cochran's theorem states that the following are equivalent:

• $r_1+\cdots +r_k=N$,
• the Q_i are independent
• each Q_i has a chi-squared distribution with r_i degrees of freedom.

Often it's stated as $\sum _i{A}_i=A$, where $A$ is idempotent, and $\sum _i{r}_i=N$ is replaced by $\sum _i{r}_i=rank(A)$. But after an orthogonal transform, $A=diag(I_M,0)$, and so we reduce to the above theorem.

Proof

Claim: Let $X$ be a standard Gaussian in ${\mathbb{ℝ}}^n$, then for any symmetric matrices $Q,Q^{\prime }$, if $X^{T}QX$ and $X^T{Q}^{\prime }X$ have the same distribution, then $Q,Q^{\prime }$ have the same eigenvalues (up to multiplicity).

Proof

Let the eigenvalues of $Q$ be ${\lambda }_1,...,{\lambda }_n$, then calculate the characteristic function of $X^{T}QX$. It comes out to be

$\varphi (t)={(\prod _j(1-2i{\lambda }_{j}t))}^{-1/2}$

(To calculate it, first diagonalize $Q$, change into that frame, then use the fact that the characteristic function of the sum of independent variables is the product of their characteristic functions.)

For $X^{T}QX$ and $X^T{Q}^{\prime }X$ to be equal, their characteristic functions must be equal, so $Q,Q^{\prime }$ have the same eigenvalues (up to multiplicity).

Claim: $I=\sum _i{B}_i$.

Proof

$U^T(I-\sum _i{B}_i)U=0$. Since $(I-\sum _i{B}_i)$ is symmetric, and $U^T(I-\sum _i{B}_i)U{=}^d{U}^{T}0U$, by the previous claim, $(I-\sum _i{B}_i)$ has the same eigenvalues as 0.

Lemma: If $\sum _i{M}_i=I$, all $M_i$ symmetric, and have eigenvalues 0, 1, then they are simultaneously diagonalizable.

Proof

Fix i, and consider the eigenvectors v of $M_i$ such that $M_{i}v=v$. Then we have $v^{T}v=v^{T}Iv=v^{T}v+\sum _{j\ne i}{v}^T{M}_{j}v$, so all $v^T{M}_{j}v=0$. Thus we obtain a split of ${\mathbb{ℝ}}^N$ into $V\oplus V^{\perp }$, such that V is the 1-eigenspace of $M_i$, and in the 0-eigenspaces of all other $M_j$. Now induct by moving into $V^{\perp }$.

Now we prove the original theorem. We prove that the three cases are equivalent by proving that each case implies the next one in a cycle ($1\to 2\to 3\to 1$).

Proof

Case: All $Q_i$ are independent

Fix some $i$, define $C_i=I-B_i=\sum _{j\ne i}{B}_j$, and diagonalize $B_i$ by an orthogonal transform $O$. Then consider $O{C}_i{O}^T=I-O{B}_i{O}^T$. It is diagonalized as well.

Let $W=OU$, then it is also standard Gaussian. Then we have

$Q_i=W^T(O{B}_i{O}^T)W;\sum _{j\ne i}{Q}_j=W^T(I-O{B}_i{O}^T)W$

Inspect their diagonal entries, to see that $Q_i\perp \sum _{j\ne i}{Q}_j$ implies that their nonzero diagonal entries are disjoint.

Thus all eigenvalues of $B_i$ are 0, 1, so $Q_i$ is a ${\chi }^2$ dist with $r_i$ degrees of freedom.

Case: Each $Q_i$ is a ${\chi }^2(r_i)$ distribution.

Fix any $i$, diagonalize it by orthogonal transform $O$, and reindex, so that $O{B}_i{O}^T=diag({\lambda }_1,...,{\lambda }_{r_i},0,...,0)$. Then $Q_i=\sum _j{\lambda }_j{U^{\prime }}_j^2$ for some $U{\text{'}}_j$, a spherical rotation of $U_i$.

Since $Q_i\sim {\chi }^2(r_i)$, we get all ${\lambda }_j=1$. So all $B_i⪰0$, and have eigenvalues $0,1$.

So diagonalize them simultaneously, add them up, to find $\sum _i{r}_i=N$.

Case: $r_1+\cdots +r_k=N$.

We first show that the matrices B⁽ⁱ⁾ can be simultaneously diagonalized by an orthogonal matrix and that their non-zero eigenvalues are all equal to +1. Once that's shown, take this orthogonal transform to this simultaneous eigenbasis, in which the random vector $[U_1,...,U_N{]}^T$ becomes $[U{\text{'}}_1,...,U{\text{'}}_N{]}^T$, but all ${U_i}^{\prime }$ are still independent and standard Gaussian. Then the result follows.

Each of the matrices B⁽ⁱ⁾ has rank r_i and thus r_i non-zero eigenvalues. For each i, the sum $C^{(i)}\equiv \sum _{j\ne i}{B}^{(j)}$ has at most rank $\sum _{j\ne i}{r}_j=N-r_i$. Since $B^{(i)}+C^{(i)}=I_{N\times N}$, it follows that C⁽ⁱ⁾ has exactly rank N − r_i.

Therefore B⁽ⁱ⁾ and C⁽ⁱ⁾ can be simultaneously diagonalized. This can be shown by first diagonalizing B⁽ⁱ⁾, by the spectral theorem. In this basis, it is of the form:

$\left[\begin{array}{ccccccc}{\lambda }_1& 0& 0& \cdots & \cdots & & 0\\ 0& {\lambda }_2& 0& \cdots & \cdots & & 0\\ 0& 0& \ddots & & & & ⋮\\ ⋮& ⋮& & {\lambda }_{r_i}& & \\ ⋮& ⋮& & & 0& \\ 0& ⋮& & & & \ddots \\ 0& 0& \dots & & & & 0\end{array}\right].$

Thus the lower $(N-r_i)$ rows are zero. Since $C^{(i)}=I-B^{(i)}$, it follows that these rows in C⁽ⁱ⁾ in this basis contain a right block which is a $(N-r_i)\times (N-r_i)$ unit matrix, with zeros in the rest of these rows. But since C⁽ⁱ⁾ has rank N − r_i, it must be zero elsewhere. Thus it is diagonal in this basis as well. It follows that all the non-zero eigenvalues of both B⁽ⁱ⁾ and C⁽ⁱ⁾ are +1. This argument applies for all i, thus all B⁽ⁱ⁾ are positive semidefinite.

Moreover, the above analysis can be repeated in the diagonal basis for $C^{(1)}=B^{(2)}+\sum _{j>2}{B}^{(j)}$. In this basis $C^{(1)}$ is the identity of an $(N-r_1)\times (N-r_1)$ vector space, so it follows that both B⁽²⁾ and $\sum _{j>2}{B}^{(j)}$ are simultaneously diagonalizable in this vector space (and hence also together with B⁽¹⁾). By iteration it follows that all B-s are simultaneously diagonalizable.

Thus there exists an orthogonal matrix $S$ such that for all $i$, $S^{\mathrm{T}}{B}^{(i)}S\equiv B^{(i)\prime }$ is diagonal, where any entry $B_{x,y}^{(i)\prime }$ with indices $x=y$, $\sum _{j=1}^{i-1}{r}_j<x=y\le \sum _{j=1}^i{r}_j$, is equal to 1, while any entry with other indices is equal to 0.

Examples

Sample mean and sample variance

If X₁, ..., X_n are independent normally distributed random variables with mean μ and standard deviation σ then

$U_i=\frac{X_i-\mu }{\sigma }$

is standard normal for each i. Note that the total Q is equal to sum of squared Us as shown here:

$\sum _i{Q}_i=\sum _{jik}{U}_j{B}_{jk}^{(i)}{U}_k=\sum _{jk}{U}_j{U}_k\sum _i{B}_{jk}^{(i)}=\sum _{jk}{U}_j{U}_k{\delta }_{jk}=\sum _j{U}_j^2$

which stems from the original assumption that $B_1+B_2\dots =I$. So instead we will calculate this quantity and later separate it into Q_i's. It is possible to write

$\sum _{i=1}^n{U}_i^2=\sum _{i=1}^n{\left(\frac{X_i-\bar{X}}{\sigma }\right)}^2+n{\left(\frac{\bar{X}-\mu }{\sigma }\right)}^2$

(here $\bar{X}$ is the sample mean). To see this identity, multiply throughout by ${\sigma }^2$ and note that

$\sum (X_i-\mu {)}^2=\sum (X_i-\bar{X}+\bar{X}-\mu {)}^2$

and expand to give

$\sum (X_i-\mu {)}^2=\sum (X_i-\bar{X}{)}^2+\sum (\bar{X}-\mu {)}^2+2\sum (X_i-\bar{X})(\bar{X}-\mu ).$

The third term is zero because it is equal to a constant times

$\sum (\bar{X}-X_i)=0,$

and the second term has just n identical terms added together. Thus

$\sum (X_i-\mu {)}^2=\sum (X_i-\bar{X}{)}^2+n(\bar{X}-\mu {)}^2,$

and hence

$\sum {\left(\frac{X_i-\mu }{\sigma }\right)}^2=\sum {\left(\frac{X_i-\bar{X}}{\sigma }\right)}^2+n{\left(\frac{\bar{X}-\mu }{\sigma }\right)}^2=\stackrel{Q_1}{\overbrace{\sum _i{(U_i-\frac{1}{n}\sum _j{U}_j)}^2}}+\stackrel{Q_2}{\overbrace{\frac{1}{n}{\left(\sum _j{U}_j\right)}^2}}=Q_1+Q_2.$

Now $B^{(2)}=\frac{J_n}{n}$ with $J_n$ the matrix of ones which has rank 1. In turn $B^{(1)}=I_n-\frac{J_n}{n}$ given that $I_n=B^{(1)}+B^{(2)}$. This expression can be also obtained by expanding $Q_1$ in matrix notation. It can be shown that the rank of $B^{(1)}$ is $n-1$ as the addition of all its rows is equal to zero. Thus the conditions for Cochran's theorem are met.

Cochran's theorem then states that Q₁ and Q₂ are independent, with chi-squared distributions with n − 1 and 1 degree of freedom respectively. This shows that the sample mean and sample variance are independent. This can also be shown by Basu's theorem, and in fact this property characterizes the normal distribution – for no other distribution are the sample mean and sample variance independent.

Distributions

The result for the distributions is written symbolically as

$\sum {(X_i-\bar{X})}^2\sim {\sigma }^2{\chi }_{n-1}^2.$

$n(\bar{X}-\mu {)}^2\sim {\sigma }^2{\chi }_1^2,$

Both these random variables are proportional to the true but unknown variance σ². Thus their ratio does not depend on σ² and, because they are statistically independent. The distribution of their ratio is given by

$\frac{n{(\bar{X}-\mu )}^2}{\frac{1}{n-1}\sum {(X_i-\bar{X})}^2}\sim \frac{{\chi }_1^2}{\frac{1}{n-1}{\chi }_{n-1}^2}\sim F_{1,n-1}$

where F_1,n − 1 is the F-distribution with 1 and n − 1 degrees of freedom (see also Student's t-distribution). The final step here is effectively the definition of a random variable having the F-distribution.

Estimation of variance

To estimate the variance σ², one estimator that is sometimes used is the maximum likelihood estimator of the variance of a normal distribution

${\hat{\sigma }}^2=\frac{1}{n}\sum {(X_i-\bar{X})}^2.$

Cochran's theorem shows that

$\frac{n{\hat{\sigma }}^2}{{\sigma }^2}\sim {\chi }_{n-1}^2$

and the properties of the chi-squared distribution show that

$\begin{array}{rl}E\left(\frac{n{\hat{\sigma }}^2}{{\sigma }^2}\right)& =E\left({\chi }_{n-1}^2\right)\\ \frac{n}{{\sigma }^2}E\left({\hat{\sigma }}^2\right)& =(n-1)\\ E\left({\hat{\sigma }}^2\right)& =\frac{{\sigma }^2(n-1)}{n}\end{array}$

Alternative formulation

The following version is often seen when considering linear regression. Suppose that $Y\sim N_n(0,{\sigma }^2{I}_n)$ is a standard multivariate normal random vector (here $I_n$ denotes the n-by-n identity matrix), and if $A_1,\dots ,A_k$ are all n-by-n symmetric matrices with $\sum _{i=1}^k{A}_i=I_n$. Then, on defining $r_i=\mathrm{Rank}(A_i)$, any one of the following conditions implies the other two:

• $\sum _{i=1}^k{r}_i=n,$
• $Y^T{A}_{i}Y\sim {\sigma }^2{\chi }_{r_i}^2$ (thus the $A_i$ are positive semidefinite)
• $Y^T{A}_{i}Y$ is independent of $Y^T{A}_{j}Y$ for $i\ne j.$

See also

• Cramér's theorem, on decomposing normal distribution
• Infinite divisibility (probability)