Ben Bolker --
>> Now we can apply Fisher's Likelihood Ratio Test of Fisher as well
>> as the AIC. The LRT tells us that the expectation
>
> ... under the null hypothesis where M0 (the simpler model) is true?
Yup. I'm considering that case to see how the AIC fits in with the LRT.
Of course the AIC is proposed for a much wider range of cases.
>> of
>> 2 log(L1) - 2 log(L0) is d1 - d0 (because it is distributed as a
>> Chi-Square with that number of degrees of freedom).
>>
>> But the AIC tells us that the expectation is 2(d1 - d0).
>
> Maybe I'm missing something, but I don't see how the AIC tells us
> something about the expectation of 2 log(L1) - 2 log(L0) ? It gives us
> the expectation of the Kullback-Leibler distance, which is something
> like sum(p(i) log(p(i)/q(i)) where p(i) is the true distribution of
> outcomes and q(i) is the predicted distribution of outcomes ... so it's
> something more like a marginal log-likelihood difference rather than a
> maximum log-likelihood difference ...
Well, the AIC ends up with comparing -2 log(L) + 2d for the two
hypotheses. The difference of these for models M1 and M0
is just (the negative of) 2 log(L1/L0) - 2(d1-d0). Or have I
missed something here? So the expectation of the difference
is log likelihood *is* described by the AIC, right? And isn't it
(in view of Fisher's distribution) wrong too? That is what
disturbs me and makes me feel there is something I don't
understand about the AIC argument.
Joe
----
Joe Felsenstein j...@gs.washington.edu
Department of Genome Sciences and Department of Biology,
University of Washington, Box 355065, Seattle, WA 98195-5065 USA
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