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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.

ExampleEdit

If X1, ..., Xn are independent normally distributed random variables with mean μ and standard deviation σ 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 σ2), 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 − 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 − 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 σ2, 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 σ2(n − 1)/n.

Both these distributions are proportional to the true but unknown variance σ2; thus their ratio is independent of σ2 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).



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