Cochran's theorem

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)},\ldots, 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 Qi 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 = \operatorname{rank}(A). But after an orthogonal transform, A = \operatorname{diag}(I_M, 0), and so we reduce to the above theorem. Here, \operatorname{diag} denotes the diagonal transform.

Alternative formulation

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

• \sum_{i=1}^kr_i=n ,
• Y^TA_iY\sim\sigma^2\chi^2_{r_i} (thus the A_i are positive semidefinite)
• Y^TA_iY is independent of Y^TA_jY for i\neq j .

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_iQ_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} \ldots = 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-\overline{X}}{\sigma}\right)^2

• n\left(\frac{\overline{X}-\mu}{\sigma}\right)^2

(here \overline{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-\overline{X}+\overline{X}-\mu)^2

and expand to give

: \sum(X_i-\mu)^2= \sum(X_i-\overline{X})^2+\sum(\overline{X}-\mu)^2+ 2\sum(X_i-\overline{X})(\overline{X}-\mu).

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

:\sum(\overline{X}-X_i)=0,

and the second term has just n identical terms added together. Thus : \sum(X_i-\mu)^2 = \sum(X_i-\overline{X})^2+n(\overline{X}-\mu)^2 ,

and hence : \sum\left(\tfrac{X_i-\mu}{\sigma}\right)^2= \sum\left(\tfrac{X_i-\overline{X}}{\sigma}\right)^2 +n\left(\tfrac{\overline{X}-\mu}{\sigma}\right)^2= \overbrace{\sum_i\left(U_i-\tfrac{1}{n}\sum_j{U_j}\right)^2}^{Q_1} +\overbrace{\tfrac{1}{n}\left(\sum_j{U_j}\right)^2}^{Q_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.{{cite journal

Distributions

The result for the distributions is written symbolically as : \sum\left(X_i-\overline{X}\right)^2 \sim \sigma^2 \chi^2_{n-1}. : n(\overline{X}-\mu)^2\sim \sigma^2 \chi^2_1,

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

where F1,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

: \widehat{\sigma}^2= \frac{1}{n}\sum\left( X_i-\overline{X}\right)^2.

Cochran's theorem shows that

: \frac{n\widehat{\sigma}^2}{\sigma^2}\sim\chi^2_{n-1}

and the properties of the chi-squared distribution show that

:\begin{align} E \left(\frac{n \widehat{\sigma}^2}{\sigma^2}\right) &= E \left(\chi^2_{n-1}\right) \ \frac{n}{\sigma^2}E \left(\widehat{\sigma}^2\right) &= (n-1) \ E \left(\widehat{\sigma}^2\right) &= \frac{\sigma^2 (n-1)}{n} \end{align}

Proof

Claim: Let X be a standard Gaussian in \R^n, then for any symmetric matrices Q, Q', if X^T Q X and X^T Q' X have the same distribution, then Q, Q' have the same eigenvalues (up to multiplicity).

Let the eigenvalues of Q be \lambda_1, ..., \lambda_n, then calculate the characteristic function of X^T Q X. It comes out to be

\phi(t) =\left(\prod_j (1-2i \lambda_j t)\right)^{-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 Q X and X^T Q' X to be equal, their characteristic functions must be equal, so Q, Q' have the same eigenvalues (up to multiplicity).

Claim: I = \sum_i B_i.

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 0 U, 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.

Fix i, and consider the eigenvectors v of M_i such that M_i v = v. Then we have v^T v = v^T I v = v^T v + \sum_{j\neq i} v^T M_j v. We note that the M_js are positive semi-definite, since their eigenvectors are all non-negative; so, for all js we must have v^T M_j v = 0. Thus we obtain a split of \R^N into V\oplus V^\perp, such that the 1-eigenspace V of M_i is contained in the 0-eigenspaces of M_j for j\neq i. Now induct by moving into V^\perp. If M_i\neq0 for all is, then a basis diagonalizing simultaneously each M_i is given by (v_{1,1},\dots,v_{r_1,1},\dots,v_{1,k},\dots,v_{r_k,k}), where (v_{1,i},\dots,v_{r_i,i}) is a basis of the 1-eigenspace of M_i.

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).

Case: All Q_i are independent

Fix some i, define C_i = I - B_i = \sum_{j\neq 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 (OB_i O^T) W; \quad \sum_{j\neq i} Q_j = W^T (I - OB_i O^T) W

Inspect their diagonal entries, to see that Q_i \perp \sum_{j\neq 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'}_j^2 for some U'_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\succeq 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'_1, ..., U'_N]^T, but all U_i' 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: :\begin{bmatrix} \lambda_1 & 0 & 0 & \cdots & \cdots & & 0 \ 0 & \lambda_2 & 0 & \cdots & \cdots & & 0 \ 0 & 0 & \ddots & & & & \vdots \ \vdots & \vdots & & \lambda_{r_i} & & \ \vdots & \vdots & & & 0 & \ 0 & \vdots & & & & \ddots \ 0 & 0 & \ldots & & & & 0 \end{bmatrix}.

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_{j2}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_{j2}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^{(i)\prime}{x,y} with indices x = y, \sum{j=1}^{i-1} r_j , is equal to 1, while any entry with other indices is equal to 0.

Cochran's theorem is the converse of Fisher's theorem. --

References

References

1. Cochran, W. G.. (April 1934). "The distribution of quadratic forms in a normal system, with applications to the analysis of covariance". [[Mathematical Proceedings of the Cambridge Philosophical Society]].
2. Bapat, R. B.. (2000). "Linear Algebra and Linear Models". Springer.
3. (2008-01-01). "Cochran's theorem". Oxford University Press.
4. "Cochran's Theorem (A quick tutorial)".