Suppose we have a linear model for a dependent variable $y$ in terms of two independent variables $x_1$ and $x_2$, given by $y_i=x_{i1} \beta_1+x_{i2}\beta_2+\epsilon_i$.

If we were to estimate the parameters $\beta_1$ and $\beta_2$ by ML we would have to specify a distribution for $\epsilon_i$ (assuming $x_1$ and $x_2$ are 'fixed'). Suppose we choose two different density functions $f$ and $g$ for $\epsilon$. Does it make sense to compare the two corresponding maximum likelihood values for the two models to decide on which error distribution is more appropriate?

My intuition would tell me that this is not a correct approach because likelihood values are not absolutely comparable if we go from one distribution to another.

  • 3
    $\begingroup$ You are right, in general it only makes sense to compare the likelihood score when the models are nested. I say in general because I don't know if there exists some edge case where you actually can do this $\endgroup$
    – Repmat
    Commented May 8, 2018 at 10:44

1 Answer 1


To get a sense of the problem, contemplate that density functions used to define likelihood functions are defined with respect to some dominating measure. So if we change the dominating measure, the likelihood function will change.

With more details (but informally) let the statistical model be given as a family of probability measures $P(\cdot; \theta)$ where $\theta$ indexes a family of probability measures. We must assume that all this measures are absolutely continuous with respect to some dominating measure $\mu$. Then we can write $$ P(A;\theta) = \int_A f(x;\theta) \mu(dx) $$ where $f(\cdot;\theta)$ is the Radon-Nikodym derivative of $P(\cdot;\theta)$ with respect to $\mu$. But the dominating measure $\mu$ will not be unique, suppose we change to define densities with respect to some other dominating measure $\lambda$, equivalent to $\mu$ (meaning that they have the same null sets). The likelihood function defined with respect to $\mu$ is $$ f(x;\theta) $$ (viewed as a function of $\theta$ for given $x$). The likelihood function with respect to $\lambda$ becomes $$ f(x;\theta) \frac{\mu}{\lambda}(x) $$ where $\frac{\mu}{\lambda}$ is the Radon-Nikodym derivative of $\mu$ with respect to $\lambda$.

So by changing the dominating measure can we get many different versions of the likelihood functions, but they will all be proportional (as functions of $\theta$), since the factor $\frac{\mu}{\lambda}(x)$ do not depend on $\theta$. See also What does "likelihood is only defined up to a multiplicative constant of proportionality" mean in practice?.

One consequence of this is that to be able to compare likelihoods (and then AIC) for different models, the likelihoods must be defined with respect to the same dominating measure. This also implies that they must be defined for exactly the same data. Sometimes one uses continuous models as approximations for discrete data. If one contemplates both continuous and discrete models, these two kinds of models cannot be compared with AIC, since they use different dominating measures (Lebesgue measure, counting measure).

A point raised in one comment is about nested models. Some theoreticians hold that AIC can only be used to compare nested models. Others disagree. But, if you want to use AIC to compare non-nested model classes, you have to be careful. AIC as implemented in R, for instance, is based on likelihoods where "irrelevant constants" are neglected. That have the effect of making this AIC's noncomparable! So, if you still want to do it, you must program the AIC calculations yourself.

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    $\begingroup$ Why is it the dominating measure that matters instead of the likelihood itself? For instance, what you've written here seems to imply that it is fine to compare AIC between a Poisson and a negative binomial model, but that does not make sense to me. $\endgroup$
    – Dave
    Commented Jul 26, 2022 at 16:08
  • $\begingroup$ @Dave: Why do you think it does not make sense? $\endgroup$ Commented Aug 13, 2022 at 23:13
  • $\begingroup$ It’s a comparison of likelihoods using different definitions of likelihood, so it sounds like a comparison of different loss functions (such as MSE corresponding to Gaussian likelihood and MAE corresponding to Laplace likelihood). $\endgroup$
    – Dave
    Commented Aug 13, 2022 at 23:40
  • $\begingroup$ Interesting explanation but hard to understand the simple answer to OP's question of "Can we compare the LL from two different density functions 𝑓 and 𝑔 for 𝜖". Your answer is in terms of "the likelihoods must be defined with respect to the same dominating measure"... how does the "dominating measure" relate to the "two different densities f and g"? Thanks! $\endgroup$
    – Matifou
    Commented Sep 24, 2023 at 20:47
  • $\begingroup$ @Matifou: In practice, this will mosly mean that both are continuous ("density with respect to Lebesgue measure" or both discrete "density with respect to counting measure" $\endgroup$ Commented Sep 25, 2023 at 1:40

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