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Re: GLM and interactions
From: |
John Darrington |
Subject: |
Re: GLM and interactions |
Date: |
Sat, 9 Jul 2011 08:12:27 +0000 |
User-agent: |
Mutt/1.5.18 (2008-05-17) |
On Fri, Jul 08, 2011 at 03:09:24PM -0400, Jason Stover wrote:
category (5 categories --> 4 degrees of freedom):
a --> 0 0 0 0
b --> 0 0 0 1
c --> 0 0 1 0
d --> 0 1 0 0
e --> 1 0 0 0
drug (3 categories --> 2 degrees of freedom):
1 --> 0 0
2 --> 1 0
3 --> 0 1
drug * category ((5 - 1) * (3 - 1) = 8 degrees of freedom):
a1 --> 0 0 0 0 0 0 0 0
a2 --> 0 0 0 0 0 0 0 0
a3 --> 0 0 0 0 0 0 0 0
b1 --> 0 0 0 0 0 0 0 0
b2 --> 0 0 0 1 0 0 0 0
b3 --> 0 0 0 0 0 0 0 1
c1 --> 0 0 0 0 0 0 0 0
c2 --> 0 0 1 0 0 0 0 0
c3 --> 0 0 0 0 0 0 1 0
d1 --> 0 0 0 0 0 0 0 0
d2 --> 0 1 0 0 0 0 0 0
d3 --> 0 0 0 0 0 1 0 0
e1 --> 0 0 0 0 0 0 0 0
e2 --> 1 0 0 0 0 0 0 0
e3 --> 0 0 0 0 1 0 0 0
This is not the only valid encoding, but it's the one that occurred to
me first. Many different encodings could be considered as being
correct. The only constraint is that we need to estimate the mean of
each factor/level combination by summing the coefficients available.
And we do not want any more coefficients than necessary, lest we lose
degrees of freedom for error (and hence our ability to estimate the
variability).
So there needs to be no particular relationship between the encoding of the
interaction
and the encoding of its composite variables? The only rules are:
1. Exactly N non-zero rows must be present, where N is the degrees of freedom.
2. In any row there may be no more than one non-zero elements.
Hence, would the following also be a valid encoding?
drug * category ((5 - 1) * (3 - 1) = 8 degrees of freedom):
a1 --> 0 0 0 0 0 0 0 1
a2 --> 0 0 0 0 0 0 1 0
a3 --> 0 0 0 0 0 1 0 0
b1 --> 0 0 0 0 1 0 0 0
b2 --> 0 0 0 1 0 0 0 0
b3 --> 0 0 1 0 0 0 0 0
c1 --> 0 1 0 0 0 0 0 0
c2 --> 1 0 0 0 0 0 0 0
c3 --> 0 0 0 0 0 0 0 0
d1 --> 0 0 0 0 0 0 0 0
d2 --> 0 0 0 0 0 0 0 0
d3 --> 0 0 0 0 0 0 0 0
e1 --> 0 0 0 0 0 0 0 0
e2 --> 0 0 0 0 0 0 0 0
e3 --> 0 0 0 0 0 0 0 0
J'
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