From a paper I am reading, I do not think the meanings of the parameters matters here but I can edit if that's useful to know.

$L(\phi, \theta \mid n, \omega) = Pr(n \mid \phi,\theta)Pr(\omega \mid n, \phi, \theta)$

So the likelihood has been factored into the product of the marginal of $n$ and the conditional distribution of $\omega$ given $n$.

I haven't had much formal stats training so I'm thinking there is some pretty standard knowledge that makes it obvious why you can do this but I don't know what to search for.

I've tried messing around with the chain rule and the definition of conditional probability $Pr(A | B) = Pr(A,B) / Pr(B)$ but I can't seem to make it work out. Can I do it just using these or is there something else I need to know about?

  • $\begingroup$ Rewrite the likelihood as a probability and then use $P(a,b) = P(a)P(b|a)$ $\endgroup$
    – Neil G
    Dec 6, 2017 at 10:53
  • $\begingroup$ Got it, thanks. My mistake was I forgot to switch the variables and the parameters around when writing the likelihood as a probability. $\endgroup$
    – ASeaton
    Dec 6, 2017 at 11:00

1 Answer 1


Thanks to Neil G for his comment.

My mistake was not swapping the parameters and the variables when converting the likelihood to a probability.

$L(\phi, \theta \mid n, \omega) = Pr(n, \omega \mid \phi, \theta) = Pr(n \mid \phi, \theta)Pr(\omega \mid n, \phi, \theta)$

using $Pr(A,B) = Pr(A)Pr(B|A)$


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