# If you normalise a likelihood function, does it become a valid and probability distribution. [duplicate]

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?

• 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? – Xi'an Feb 14 '17 at 11:59
• Yeah so is it a valid probabilistic object, but with no meaning? Or is it something else entirely? – tisPrimeTime Feb 14 '17 at 13:07
• 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. – tisPrimeTime Feb 14 '17 at 13:09
• 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. – whuber Feb 14 '17 at 18:27