Re: [R] GLM output for deviance and loglikelihood

From: Juliet Hannah <juliet.hannah_at_gmail.com>
Date: Wed, 20 Apr 2011 23:14:41 -0400

As you mentioned, the deviance does not always reduce to:

D = -2(loglikelihood(model))

It does for ungrouped data, such as for binary logistic regression. So let's stick with the original definition. In this case, we need the log-likelihood for the saturated model.

x = rnorm(10)

 y = rpois(10,lam=exp(1 + 2*x))

 test = glm(formula = y ~ x, family = poisson)

sm <- glm(y ~ factor(1:10),family=poisson)

mydev <- as.numeric(2*(logLik(sm)-logLik(test))) mydev
deviance(test)

On Fri, Apr 15, 2011 at 7:00 AM, Jeffrey Pollock <jpollock_at_williamhill.co.uk> wrote:
> It has always been my understanding that deviance for GLMs is defined
> by;
>
>
>
> D =  -2(loglikelihood(model) - loglikelihood(saturated model))
>
>
>
> and this can be calculated by (or at least usually is);
>
>
>
> D = -2(loglikelihood(model))
>
>
>
> As is done so in the code for 'polr' by Brian Ripley (in the package
> 'MASS') where the -loglikehood is minimised using optim;
>
>
>
> res <- optim(s0, fmin, gmin, method = "BFGS", hessian = Hess, ...)
>
> .
>
> .
>
> .
>
> deviance <- 2 * res$value
>
>
>
> If so, why is it that;
>
>
>
>> x = rnorm(10)
>
>> y = rpois(10,lam=exp(1 + 2*x))
>
>> test = glm(formula = y ~ x, family = poisson)
>
>> deviance(test)
>
> [1] 5.483484
>
>> -2*logLik(test)
>
> [1] 36.86335
>
>
>
> I'm clearly not understanding something here, can anyone shed any light?
> Why is;
>
>
>
> -2*logLik(test) =/= deviance(test) ???
>
>
>
> I think this is something that is poorly understood all over the
> internet (at least from my google searches anyway!)
>
>
>
> Thanks,
>
>
>
> Jeff
>
>
>
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