Cochran's theorem

In statistics, Cochran's theorem is used in the analysis of variance.

Suppose U1, ..., Un are independent standard normally distributed random variables, and an identity of the form



\sum_{i=1}^n U_i^2=Q_1+\cdots + Q_k $$

can be written where each Qi is a sum of squares of linear combinations of the Us. Further suppose that



r_1+\cdots +r_k=n $$

where ri is the rank of Qi. Cochran's theorem states that the Qi are independent, and Qi has a  chi-square distribution with ri degrees of freedom.

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

Example
If X1, ..., Xn are independent normally distributed random variables with mean &mu; and standard deviation &sigma; then


 * $$U_i=(X_i-\mu)/\sigma$$

is standard normal for each i.

It is possible to write



\sum U_i^2=\sum\left(\frac{X_i-\overline{X}}{\sigma}\right)^2 + n\left(\frac{\overline{X}-\mu}{\sigma}\right)^2 $$

(here, summation is from 1 to n, that is over the observations). To see this identity, multiply throughout by $$\sigma$$ and note that



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

and expand to give



\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


 * $$\sum(\overline{X}-X_i),$$

and the second term is just n identical terms added together.

Combining the above results (and dividing by &sigma;2), we have:



\sum\left(\frac{X_i-\mu}{\sigma}\right)^2= \sum\left(\frac{X_i-\overline{X}}{\sigma}\right)^2 +n\left(\frac{\overline{X}-\mu}{\sigma}\right)^2 =Q_1+Q_2. $$

Now the rank of Q2 is just 1 (it is the square of just one linear combination of the standard normal variables). The rank of Q1 can be shown to be n &minus; 1, and thus the conditions for Cochran's theorem are met.

Cochran's theorem then states that Q1 and Q2 are independent, with Chi-squared distribution with n &minus; 1 and 1 degree of freedom respectively.

This shows that the sample mean and sample variance are independent; also



(\overline{X}-\mu)^2\sim \frac{\sigma^2}{n}\chi^2_1. $$ To estimate the variance &sigma;2, one estimator that is often used is



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

Cochran's theorem shows that



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

which shows that the expected value of $$\widehat{\sigma}^2$$ is &sigma;2(n &minus; 1)/n.

Both these distributions are proportional to the true but unknown variance &sigma;2; thus their ratio is independent of &sigma;2 and because they are independent we have



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

where F1,n is the F-distribution with 1 and n degrees of freedom (see also Student's t-distribution).

twierdzenie Cochrana Teorema di Cochran