1. ## Cochran's theorem

Hi,

I found once again a non proved important theorem, Cochran's theorem, in the introduction of ANOVA decomposition of the sum of squares (indipendent chi squares if the degrees of freedom add up to the total degrees of freedom).

Where could I find a proof that isn't too advanced. I have a decent knowledge of linear algebra and a basic knowledge of the multivariate normal distriubution.

I searched the net, but the original paper was not available for free and the free papers were giving advanced extensions of this theorem, with rather complex proofs.

Thank you again for all your support!

2. So you're asking how to prove that the sum of independent chi squares is a chi square with degrees of freedom equal to the sum of the dfs?

There are lots of ways to do that. One way is using the fact that they're independent we know that the MGF (moment generating function) of the sum will be the product of the individual MGFs so then you look at that and recognize it as a the MGF of a chisquare with df equal to the sum of the dfs.

3. Ok but how do we know, say we decompose the original sum of squares SS with r dfs in 2 sum of squares q1 and q2 with dfs r1 and r2

So if we use MGFs we get, say [Chisq(s)]^r is the mgf of a chisquare distribution with r degrees of freedom:

[Chisq(s)]^r=[A(s)]^r1*[B(s)]^r2

But the fact that the product is a chisquare doesn't imply that A(s) and B(s) separately are mgfs of a Chisquare themselves. So how do I proceed from here?

4. Ok I think I got it. There are two important facts:

1-the q are always positive so there is a restriction on the possible distributions they might assume
2-their sum is a chi square with r degrees of freedom, and the chi square is always positive and assumes every possible value from 0 inlcuded to +infinity.

So if we put it this way:

[B(s)]^r2=[Chisq(s)/A(s)]^r1*[Chisq(s)]^r2

Let's call Chisq(s)/A(s)=D(s)

We know that this product is the mgf of a convolution of a chisquare with r2 dfs and an unknown distribution d(x), whose final result must be non negative.

Now, since the second term of the convolution assumes every possible value from 0 to + infinity, then d(x) must be defined only for nonnegative values, otherwise their convolution (sum of the two random variables) could take negative values.

So D(s) is the mgf of the distribution of a non negative value.

Now let's examine D(s)=Chisq(s)/A(s). It means it is the mgf of the distribution of the difference between a random variable distributed like a chisquare with 1 df and another random variable with unknown distribution a(x).

Now, even if a(x) is defined only for nonnegative values, the difference between it and the chisquare could take negative values (since the chisquare could assume any value between 0 and + infinity), UNLESS they coincide exactly, which means their difference is always 0, the distribution is a dirac delta centered in 0 and the mgf is therefore 1.

So A(s)=Chisq(s), and B(s)=Chisq(s), Q.E.D.

Thanks again for the hint!

Btw this proof is more general than one with linear algebra, because we can also assume non-integer dfs!

 Tweet

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts