If I'm not wrong, likelihood functions are sensitive to the size of the sample, i.e. the larger the sample, the lower the likelihood value. Given a sample $x$ of a random variable $X \sim f(\theta)$, and a parameter estimate $\hat\theta$, suppose I want to test the hypothesis that the likelihoods of different subsamples of $x$, let's call them $x_a$ and $x_b$ are equal.

The problem is $x_a$ and $x_b$ have a different number of elements, say $n_a$ and $n_b$ respectively, and $n_a \neq n_b$, so I assume the likelihoods must be normalized in some way. Is it enough to divide by the sample size? For instance, can I use a test statistic such as

$T = \frac{\ell(\hat\theta|x_a)}{n_a} - \frac{\ell(\hat\theta|x_b)}{n_b}$

and then test the hypothesis that $T = 0$. A related question would be how do I find the probability distribution of the above statistic, but that is another story.

  • $\begingroup$ I am not sure whether you are using likelihood in its common meaning. $\endgroup$
    – Henry
    Oct 31, 2012 at 21:50
  • $\begingroup$ I am using it in the sense of probability of observing $x$ given a parameter value $\theta$. I think this is the common meaning. $\endgroup$
    – Ernest A
    Oct 31, 2012 at 22:08

1 Answer 1


Likelihoods are strange things and often are defined only up to a constant multiple or an equivalence class. So for instance if you wanted to do a plot to compare both, you might normalize both to have a maximum of one.

The usual way to do what I think you want to do, is to calculate the likelihood for all the data using common parameters and then a mutiple of two likelihoods for the separate data using parameters that are not all common and then do a likelihood ratio test.

Almost any statistical package will do this for assuming standard probability distributions.

But if you are realy interested in looking under the hood, you might want to read http://andrewgelman.com/movabletype/mlm/plot13.pdf

And remember - it is holloween.

  • 1
    $\begingroup$ Considering that likelihoods are probabilities, wouldn't they always have a maximum of one? $\endgroup$
    – Ernest A
    Nov 1, 2012 at 11:12
  • $\begingroup$ @ErnestA thats another strange thing, the answer is no if one is considering the outcome as continuous and yes if one is considering the observed outcome which is discrete. And remember it is based on the outcomes observed in the sample; the most probably ones may not be in the sample. But if you are really interested, I think have to do some reading and that pdf will suffice. $\endgroup$
    – phaneron
    Nov 1, 2012 at 13:36
  • $\begingroup$ This paper is a little dense for me given that I'm not into Bayesian statistics, and the frequentist methods it discusses are of little help. I think I can simply evaluate the likelihood function with fixed parameters individually for every observation and work something out from there. $\endgroup$
    – Ernest A
    Nov 1, 2012 at 17:00
  • $\begingroup$ Always a good idea to just try and do it. The likelihood function with fixed parameters is no longer a function but a point and loses all of it good properties for inference. Also the likelihood as a function is seldom directly studied in Bayesian statistics or most intermediary or evn advanced Statistics texts - but unfortunately just in books on higher order asymptotics. The book "In All Likelihood" by Pawitan is the most accessible source I am aware of. $\endgroup$
    – phaneron
    Nov 1, 2012 at 17:31
  • $\begingroup$ @ErnestA, no, the likelihood is not a probability. It's a function of the parameters given data. It is easy to verify that the integral (over all the parameters) won't give you 1 (which is the requirement for a probability). $\endgroup$
    – Mayou36
    Feb 8, 2022 at 12:24

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