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As far as I'm aware a probability distribution only requires that some function $p(x)$ is non-negative over the reals, and integrates to unity.

If this is all it takes to define a probability distribution is it valid of me to normalise a likelihood function and call it a probability.

I have looked at the example from here:

What is the reason that a likelihood function is not a pdf?

where the integral of the given Bernoulli likelihood function is given to be 1/2.

If I then take this knowledge and divide this parameter (the 1/2 value) through my previous integral this will ensure that the likelihood function integrates to unity. However since the requirements for a valid probability distribution seem so relaxed (non-negative and integrate to unity), does this not mean I'm now dealing with a justifiably valid probability distribution? Or is there something else that I'm missing?

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    $\begingroup$ If $\int p(x)\,\text{d}x <\infty$, you can turn $p$ into a probability density. Which meaning can you associate with this probabilistic object? In which sense is it justified or valid? $\endgroup$
    – Xi'an
    Feb 14, 2017 at 11:59
  • $\begingroup$ Yeah so is it a valid probabilistic object, but with no meaning? Or is it something else entirely? $\endgroup$
    – tisPrimeTime
    Feb 14, 2017 at 13:07
  • $\begingroup$ And I'm not 100% sure if my question is a duplicate of the link you gave because in my question I refer to an example in that link. $\endgroup$
    – tisPrimeTime
    Feb 14, 2017 at 13:09
  • $\begingroup$ Sure, you're dealing with a probability distribution when you can normalize the likelihood--but whether that has any meaning or relevance for the problem you're trying to solve is another matter altogether. It is tantamount to adopting a specific Bayes prior--and that prior might be altogether contrary to your beliefs, your model, or other evidence you might have. $\endgroup$
    – whuber ♦
    Feb 14, 2017 at 18:27

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It is conceptually awkward: it is the probability of drawing the data that we have drawn given the model and a set of parameters. But we have already drawn the data; the probability of drawing data given that has already been drawn is just 1. We could make it work, but Fisher just solved it by calling it a likelihood rather than a probability.

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  • $\begingroup$ I like where your reasoning is going. Could you possibly expand on it a little bit more? Also would you have a link to Fisher's statement on this topic, where he "solves" it? Thanks. :) $\endgroup$
    – tisPrimeTime
    Feb 14, 2017 at 13:11
  • $\begingroup$ I think to state that "it is the probability of drawing the data that we have drawn given the model and a set of parameters" is incorrect since this is a probability distribution on the parameter space, not on the observation space. $\endgroup$
    – Xi'an
    Feb 14, 2017 at 16:06
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    $\begingroup$ ps_And Fisher murkied the whole thing by creating the fiducial distribution! $\endgroup$
    – Xi'an
    Feb 14, 2017 at 16:08

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