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# Cochran's Theorem

This is a very important theorem which allows us to decompose sums of squares into several quadratic forms and identify their distributions and establish their independence. It can be used to great advantage in Analysis of Variance and Regression. The importance of the terms in the model is assessed via the distributions of their sums of squares.

Theorem 4..7

Given , suppose that is decomposed into quadratic forms,
, , where the rank of is and the are positive semidefinite, then any one of the following conditions implies the other two.

(a)
the ranks of the add to ;
(b)
each ;
(c)
all the are mutually independent.

Proof We can write

That is,

(i)
Given (a) we will prove (b).

Select an arbitrary , say . If we make an orthogonal transformation which diagonalizes , we obtain from

 (4.9)

Since the first and last terms are diagonal, so is the second. Since and therefore , of the leading diagonal elements of are zero. Thus the corresponding elements of are and since by (a) the rank of is , the other elements of its leading diagonal are and the corresponding elements of are . Hence from Theorem 4.4, and is idempotent.

The same result holds for the other and we have established (b) from (a).

(ii)
Given (b) we will prove (c).
 (4.10)

and (b) implies that each is idempotent (with rank ). Choose an arbitrary , say . There is an orthogonal matrix such that

Premultiplying (4.10) by and post-multiplying by , we have

Now each is idempotent and can't have any negative elements on its diagonal. So must have the first leading diagonal elements , and submatrices for rows , columns and for rows , columns must have all elements . So

and thus which can only be so if .
Since was arbitrarily chosen, we have proved (c) from (b).
(iii)
Given (b) we will prove (a).

If (b) holds, has eigenvalues and zero and since , taking traces we have .

(iv)
Given (c) we will prove (b).

If (c) holds, taking powers of , we have for all positive integers . Taking traces we have

This can hold if and only if every eigenvalue of is . That is, if each .

So we have proved (b) from (c).

A more general version of Cochran's Theorem is stated (without proof) in Theorem 4.8.

Theorem 4..8

Given , suppose that is decomposed into quadratic forms, , , when . Then are mutually independent and if and only if .

Example 4..2
We will consider again Example 4.4 from the point of view of Cochran's Theorem. Recall that are iid and

That is,

where is defined in the usual way. Equivalently,

where and are defined in Example 4.1.
We can apply Cochran's Theorem, noting that we can easily show that (a) is true, since , and . So we may conclude that

and that and are independent.

Next: Order Statistics Up: Multivariate Normal Distribution Previous: The role of c.g.f.   Contents
Bob Murison 2000-10-31