# Cochran's theorem

In statistics, **Cochran's theorem**, devised by William G. Cochran,^{[1]} is a theorem used to justify results relating to the probability distributions of statistics that are used in the analysis of variance.^{[2]}

## Contents

## StatementEdit

Suppose *U*_{1}, ..., *U*_{N} are i.i.d. standard normally distributed random variables, and there exist matrices , with . Further suppose that , where *r*_{i} is the rank of . If we write

so that the are quadratic forms, then **Cochran's theorem** states that the *Q*_{i} are independent, and each *Q*_{i} has a chi-squared distribution with *r*_{i} degrees of freedom.^{[1]}

Less formally, it is the number of linear combinations included in the sum of squares defining *Q*_{i}, provided that these linear combinations are linearly independent.

### ProofEdit

We first show that the matrices *B*^{(i)} can be simultaneously diagonalized and that their non-zero eigenvalues are all equal to +1. We then use the vector basis that diagonalize them to simplify their characteristic function and show their independence and distribution.^{[3]}

Each of the matrices *B*^{(i)} has rank *r*_{i} and thus *r*_{i} non-zero eigenvalues. For each *i*, the sum has at most rank . Since , 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)}. In this basis, it is of the form:

Thus the lower rows are zero. Since , it follows that these rows in *C*^{(i)} in this basis contain a right block which is a 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. Moreover, the above analysis can be repeated in the diagonal basis for . In this basis is the identity of an vector space, so it follows that both *B*^{(2)} and are simultaneously diagonalizable in this vector space (and hence also together *B*^{(1)}). By iteration it follows that all *B*-s are simultaneously diagonalizable.

Thus there exists an orthogonal matrix such that for all , is diagonal, where any entry is equal to 1 for and is equal to 0 for any other indices.

Let denote some specific linear combination of all after transformation by . Note that due to the length preservation of the orthogonal matrix S.

The characteristic function of *Q*_{i} is:

This is the Fourier transform of the chi-squared distribution with *r*_{i} degrees of freedom. Therefore this is the distribution of *Q*_{i}.

Moreover, the characteristic function of the joint distribution of all the *Q*_{i}s is:

From this it follows that all the *Q*_{i}s are independent.

## ExamplesEdit

### Sample mean and sample varianceEdit

If *X*_{1}, ..., *X*_{n} are independent normally distributed random variables with mean μ and standard deviation σ
then

is standard normal for each *i*. It is possible to write

(here is the sample mean). To see this identity, multiply throughout by and note that

and expand to give

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

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

and hence

Now the rank of *Q*_{2} is just 1 (it is the square of just one linear combination of the standard normal variables). The rank of *Q*_{1} can be shown to be *n* − 1, and thus the conditions for Cochran's theorem are met.

Cochran's theorem then states that *Q*_{1} and *Q*_{2} 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.^{[4]}

### DistributionsEdit

The result for the distributions is written symbolically as

Both these random variables are proportional to the true but unknown variance σ^{2}. Thus their ratio does not depend on σ^{2} and, because they are statistically independent. The distribution of their ratio is given by

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 varianceEdit

To estimate the variance σ^{2}, one estimator that is sometimes used is the maximum likelihood estimator of the variance of a normal distribution

Cochran's theorem shows that

and the properties of the chi-squared distribution show that

## Alternative formulationEdit

The following version is often seen when considering linear regression.^{[5]} Suppose that is a standard multivariate normal random vector (here denotes the *n*-by-*n* identity matrix), and if are all *n*-by-*n* symmetric matrices with . Then, on defining , any one of the following conditions implies the other two:

- (thus the are positive semidefinite)
- is independent of for

## See alsoEdit

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

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## ReferencesEdit

- ^
^{a}^{b}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*.**30**(2): 178–191. doi:10.1017/S0305004100016595. **^**Bapat, R. B. (2000).*Linear Algebra and Linear Models*(Second ed.). Springer. ISBN 978-0-387-98871-9.**^**Craig A.T. (1938) On The Independence of Certain Estimates of Variances. Ann. Math. Statist. 9, pp. 48–55**^**Geary, R.C. (1936). "The Distribution of the "Student's" Ratio for the Non-Normal Samples".*Supplement to the Journal of the Royal Statistical Society*.**3**(2): 178–184. doi:10.2307/2983669. JFM 63.1090.03. JSTOR 2983669.**^**"Cochran's Theorem (A quick tutorial)" (PDF).