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I was looking at AIC, which is given by AIC = 2K - ln(L).

However, to my understanding, and observation, L = Log-Likelihood can be negative.

So in the case where L is negative, is AIC not applicable ?

Otherwise, ln(L) would give complex number, and I don't see complex number would be a useful value for this case.

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  • $\begingroup$ You seem to be confused by the definitions. If L is the likelihood (a product of probabilities, so to speak) it will be always positive. Then ln(L) is the loglikelihood, which can (and actially will, since 0$\leq$prob$\leq$1) be negative. And then it doesn't really matter what are the values of AIC themselves, only the differences matter. $\endgroup$ – corey979 Apr 1 at 10:40
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Log-likelihood can be negative, indeed it will be if the likelihood is essentially a product of probabilities less than $1$. Thus $-\ln(L)$ will be positive, and so too will be $2k-\ln(L)$.

But the likelihood will be more than $0$, so the log-likelihood will be a real number.

So you do not need to worry.

Likelihoods are usually stated as being proportional up to a multiplicative constant, so log-likelihoods are really stated up to an additive constant. You do not need to worry about this either, so long as you use consistent calculations for comparisons of likelihoods .

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