# When can likelihood be interpreted as probability function? [duplicate]

I'm interested in likelihood $L(\theta | x) = P( x | \theta)$ seen as the function of $\theta$ for a given $x$.

I have been reading in books and other posts that $L$ is generally not a probability function (with respect to $\theta$) because it may not integrate to 1 and, moreover, there my be even no $\sigma$-field defined over the $\theta$ space.

In the Bayesian approach, we define a prior probability $P(\theta)$ and so the $\sigma$-field had to be defined.

What other properties we need to be able to interpret $L(\theta|x)$ as the conditional probability $P(\theta|x)$? Perhaps normalization to integrate to 1? What else? (whuber mentions some regularity conditions in this post Wikipedia entry on likelihood seems ambiguous. What would these be?)

# EDIT

After some searching, I believe the answer is in Fisher's fiducial distribution. See Casella&Berger(2002, sec. 6.3.1), the original Fisher's papers (1933,1935) or this question: What is the fiducial argument and why has it not been accepted?

# EDIT2

Now that I know the key-word is fiducial, there is a fairly good explanation in the answer to this question: What does "fiducial" mean (in the context of statistics)?

## marked as duplicate by Xi'an, kjetil b halvorsen, Peter Flom♦May 18 '17 at 11:34

• @Tim - Thanks. Yes, I've seen this discussion before. It indeed explains well the standard view of likelihood function (not a prob distrib of $\theta|x$) and the bayesian interpretation. I will read a bit more on the fiducial argument of Fisher who was known to be an anti-Bayesian yet, I believe, in this notion (the fiducial distribution) wanted to capture the possibility of treating theta as r.v. I'll update my post accordingly if I find a neat answer to my question. – mag Jun 21 '17 at 11:04