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Why deviance returned by GLMFIT is not = -2*LogLikelihood?

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Francisco de Castro
Francisco de Castro 2012 年 3 月 6 日
コメント済み: David Nielsen 2017 年 3 月 4 日
I'm working with GLM models using glmfit. After fitting the model I need to calculate the LogLikelihood (which is not returned directly by glmfit). I've seen in many sources that deviance (which IS returned by glmfit) is equal to -2*LogLikelihood. However, if I calculate the LogLikelihood separately (see example below with binomial distribution) I get totally different answers. Any idea what I'm doing wrong? I took the example data from MATLAB doc. on glmfit
x = [2100 2300 2500 2700 2900 3100 3300 3500 3700 3900 4100 4300]';
n = [48 42 31 34 31 21 23 23 21 16 17 21]';
y = [1 2 0 3 8 8 14 17 19 15 17 21]';
[b,dev,stats]= glmfit(x,[y n],'binomial');
yfit= glmval(b, x,'logit','size',n);
dev/-2
logLikelihood= nansum(log( binopdf( y, n, yfit)))

採用された回答

Tom Lane
Tom Lane 2012 年 3 月 6 日
Two things. First, the last argument to binopdf should be the fitted probability, not the fitted counts. Second, the deviance is defined with respect to a "full" model that has a separate fitted value for every observation. So -dev/2 reproduces this value:
sum(log(binopdf(y,n,yfit./n))) - sum(log(binopdf(y,n,y./n)))
If you need the log likelihood value, your way of computing it is fine, once you correct the binopdf input.
  2 件のコメント
Marris Atwood
Marris Atwood 2015 年 6 月 9 日
So you mean the saturated model's log likelihood, i.e., sum(log(binopdf(y,n,y./n))), is not necessarily always 0, right? That is acceptable, but how I should know if that is 0?
David Nielsen
David Nielsen 2017 年 3 月 4 日
Would this not be the log likelihood ratio, i.e. the ratio of the log likelihood of the fitted model and the saturated model?

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