# 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

${\displaystyle \sum _{i=1}^{n}U_{i}^{2}=Q_{1}+\cdots +Q_{k}}$

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

${\displaystyle r_{1}+\cdots +r_{k}=n}$

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 μ and standard deviation σ then

${\displaystyle U_{i}=(X_{i}-\mu )/\sigma }$

is standard normal for each i.

It is possible to write

${\displaystyle \sum U_{i}^{2}=\sum \left({\frac {X_{i}-{\overline {X}}}{\sigma }}\right)^{2}+n\left({\frac {{\overline {X}}-\mu }{\sigma }}\right)^{2}}$

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

${\displaystyle \sum (X_{i}-\mu )^{2}=\sum (X_{i}-{\overline {X}}+{\overline {X}}-\mu )^{2}}$

and expand to give

${\displaystyle \sum (X_{i}-{\overline {X}})^{2}+\sum ({\overline {X}}-\mu )^{2}+2\sum (X_{i}-{\overline {X}})({\overline {X}}-\mu ).}$

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

${\displaystyle \sum ({\overline {X}}-X_{i}),}$

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

Combining the above results (and dividing by σ2), we have:

${\displaystyle \sum \left({\frac {X_{i}-\mu }{\sigma }}\right)^{2}=\sum \left({\frac {X_{i}-{\overline {X}}}{\sigma }}\right)^{2}+n\left({\frac {{\overline {X}}-\mu }{\sigma }}\right)^{2}=Q_{1}+Q_{2}.}$

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

${\displaystyle ({\overline {X}}-\mu )^{2}\sim {\frac {\sigma ^{2}}{n}}\chi _{1}^{2}.}$

To estimate the variance σ2, one estimator that is often used is

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

Cochran's theorem shows that

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

which shows that the expected value of ${\displaystyle {\widehat {\sigma }}^{2}}$ 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

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

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