3
$\begingroup$

In a 2012 answer to a question on model selection, a user proposed the following (log-base-2) likelihood ratio with "AIC correction":

#compute the AIC-corrected log-base-2 likelihood ratio (a.k.a. "bits" of evidence)
(AIC(mod1)-AIC(mod2))*log2(exp(1))

I found the approach interesting, but could not find any documentation for it, and not being a statistician myself, I feel unsure in using it.

So my questions are:

  1. Can anybody provide a source for this approach? Edit: This article provides a very similar equation: exp(-1/2 * (AIC(mod1)-AIC(mod2))) Maybe someone with a bit more calculus than me could explain how to get from this formula to the original one in the question?

  2. Does it provides an advantage on simple log likelihood ratio? (e.g. weighting for model complexity?) The same article defines it as evidence ratio, thus exactly how many times mod2 is more evident than mod1.

  3. What does the *log2(exp(1)) bit means?

Note: the referred article is "AIC model selection and multimodel inference in behavioral ecology: some background, observations, and comparisons" by Burnham, Anderson & Huyvaert, 2011.

$\endgroup$
1
$\begingroup$

For your question 2, it looks like they are not the same formula so you can't go from one to the other since

log2(exp(1))=1.44269504089

and

exp(-1/2)=0.60653065971
$\endgroup$
1
  • 1
    $\begingroup$ Now that you point it out, it seems obvious! In the first equation, it is a simple multiplication for a constant. In the second equation instead the delta is exponential. So the interpretation will hardly be comparable. Anyway my conclusion on this topic was that the formula posted in the original question was most probably meaningless, as "bits of evidence" are usually defined as logs of probabilities. eg: lesswrong.com/lw/jn/how_much_evidence_does_it_take $\endgroup$ – Johann Nov 19 '17 at 18:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.