In statistics, Cochran's theorem, devised by William G. Cochran,[1] is a theorem used to justify results relating to the probability distributions of statistics that are used in the analysis of variance.
[2] If X1, ..., Xn are independent normally distributed random variables with mean μ and standard deviation σ then is standard normal for each i.
Note that the total Q is equal to sum of squared Us as shown here: which stems from the original assumption that
So instead we will calculate this quantity and later separate it into Qi's.
and note that and expand to give The third term is zero because it is equal to a constant times and the second term has just n identical terms added together.
as the addition of all its rows is equal to zero.
Thus the conditions for Cochran's theorem are met.
Cochran's theorem then states that Q1 and Q2 are independent, with chi-squared distributions with n − 1 and 1 degree of freedom respectively.
This can also be shown by Basu's theorem, and in fact this property characterizes the normal distribution – for no other distribution are the sample mean and sample variance independent.
[3] The result for the distributions is written symbolically as Both these random variables are proportional to the true but unknown variance σ2.
Thus their ratio does not depend on σ2 and, because they are statistically independent.
The distribution of their ratio is given by where F1,n − 1 is the F-distribution with 1 and n − 1 degrees of freedom (see also Student's t-distribution).
The final step here is effectively the definition of a random variable having the F-distribution.
To estimate the variance σ2, one estimator that is sometimes used is the maximum likelihood estimator of the variance of a normal distribution Cochran's theorem shows that and the properties of the chi-squared distribution show that The following version is often seen when considering linear regression.
is a standard multivariate normal random vector (here
denotes the n-by-n identity matrix), and if
standard normally distributed random variables, and
, change into that frame, then use the fact that the characteristic function of the sum of independent variables is the product of their characteristic functions.)
symmetric, and have eigenvalues 0, 1, then they are simultaneously diagonalizable.
We prove that the three cases are equivalent by proving that each case implies the next one in a cycle (
implies that their nonzero diagonal entries are disjoint.
So diagonalize them simultaneously, add them up, to find
We first show that the matrices B(i) can be simultaneously diagonalized by an orthogonal matrix and that their non-zero eigenvalues are all equal to +1.
Once that's shown, take this orthogonal transform to this simultaneous eigenbasis, in which the random vector
This can be shown by first diagonalizing B(i), by the spectral theorem.
unit matrix, with zeros in the rest of these rows.
This argument applies for all i, thus all B(i) are positive semidefinite.
Moreover, the above analysis can be repeated in the diagonal basis for
are simultaneously diagonalizable in this vector space (and hence also together with B(1)).
By iteration it follows that all B-s are simultaneously diagonalizable.