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

Suppose *U*_{1}, ..., *U*_{n} 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 *Q*_{i} is a sum of squares of linear combinations of the *U*s. Further suppose that

- $ r_1+\cdots +r_k=n $

where *r*_{i} is the rank of *Q*_{i}. Cochran's theorem states that the *Q*_{i} are independent, and *Q*_{i} has a chi-square distribution with *r*_{i} degrees of freedom.

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

### ExampleEdit

If *X*_{1}, ..., *X*_{n} 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 σ^{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 *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 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 σ^{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 σ^{2}(*n* − 1)/*n*.

Both these distributions are proportional to the true but unknown variance σ^{2}; thus their ratio is independent of σ^{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 *F*_{1,n} is the F-distribution with 1 and *n* degrees of freedom (see also Student's t-distribution).

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