Cochran's theorem

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

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

[\sum_^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 Us. 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.

Example

If X₁, ..., 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}\right)^2+ n\left(\frac-\mu}\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+\overline-\mu)^2]

and expand to give

[\sum(X_i-\overline)^2+\sum(\overline-\mu)^2+2\sum(X_i-\overline)(\overline-\mu).]

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

[\sum(\overline-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\right)^2=\sum\left(\frac}\right)^2+n\left(\frac-\mu}\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-\mu)^2\sim \frac\chi^2_1.]

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

[\widehat^2=\frac\sum\left(X_i-\overline\right)^2. ]

Cochran's theorem shows that

[\widehat^2\sim\frac\chi^2_]

which shows that the expected value of [\widehat^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-\mu\right)^2}\sum\left(X_i-\overline\right)^2}\simF_]

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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