Re: GLM and interactions

[Jason Stover]

On Sat, Jul 09, 2011 at 08:12:27AM +0000, John Darrington wrote:
> 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

Yes, that one is fine.