[R] regression coefficients
Spencer Graves
spencer.graves at pdf.com
Tue May 20 17:55:27 CEST 2003
Dear Prof. Ripley: Of course, you are correct on both counts. Thanks
for the correction and elaboration. spencer graves
Prof Brian Ripley wrote:
> Why is s assumed known and common to the k groups? I doubt if that is
> what was meant (although it was too imprecise to be at all sure).
>
> If `common' is a viable assumption, you can just fit a model with by-group
> regressions vs one with a common regression (which seems to be what you
> are testing) and use anova().
>
> If not, the case k=2 encompasses the Welch t-test so exact distribution
> theory is not going to be possible, but by fitting a common model and
> three separate models and summing the -2log-lik for the latter you can
> easily get the LT test and refer it to its `standard' (asymptotic)
> Chi-squared distribution.
>
> On Tue, 20 May 2003, Spencer Graves wrote:
>
>
>> I don't know of a simply function to do what you want, but I can give
>>you part of the standard log(likelihood ratio) theory:
>>
>> Suppose b[i]|s ~ N.r(b, s^2*W[i]), i = 1, ..., k. Then the
>>log(likelihood) is a sum of k terms of the following form:
>>
>> l[i] = (-0.5)*(r*log(2*pi*s^2)+log|W[i]|
>> +(s^-2)*t(b[i]-b)%*%solve(W[i]%*%(b[i]-b)
>>
>>By differentiating with respect to b and setting to 0, we get the
>>maximum likelihood estimate for b as follows:
>>
>> b.hat = solve(sum(solve(W[i]))%*%sum(solve(W[i])%*%b[i])
>>
>>In words: b.hat = weighted average with weights inversely proportional
>>to the variances. Then log(likelihood ratio) is as follows:
>>
>> log.LR = sum((s^-2)*t(b[i]-b.hat)%*%solve(W[i])%*%(b[i]-b.hat))
>>
>>This problem should be in most good books on multivariate analysis. I
>>would guess that log.LR probably has an F distribution with numerator
>>degrees of freedom = r*(k-1) and with denominator degrees of freedom =
>>degrees of freedom in the estimate of s. However, I don't remember for
>>sure. It's vaguely possible that this is an "unsolved" problem. In the
>>latter case, you should have all the pieces here to generate a Monte
>>Carlo.
>
>
> You have assumed s is known, in which case it is a Chi-squared
> distribution. If s is unknown, you need to maximize over it to get an LR
> test (separately under the null and the alternative).
>
>
>>lamack lamack wrote:
>>
>>>dear all, How can I compare regression coefficients across three (or
>>>more) groups?
>>
>
More information about the R-help
mailing list