How is the "proportional" calculated in Bayes theorem?

Question

I am struggling with the Wikipedia entry on Likelihood.

In an example it mentions $L(P_H = 0.5 |HH) = 0.25$

It mentions that

Bayes' theorem implies that the posterior probability is proportional to the likelihood times the prior probability.

I am trying to understand, in our scenario, what the prior and the post should be. I thought of

post = prior * likelihood = 0.5 * 0.25 = 0.125

This seems way too small. How is the "proportional" calculated?

@Sycorax thanks I am reading https://en.wikipedia.org/wiki/Bayes%27_theorem — Kirsten, Jan 25 '22 at 04:04
In $P(A|B) = P(B|A) P(A)/P(B)$ the denominator $P(B)$ is the normalizing constant — Tim, Jan 25 '22 at 08:02

score 2 · Accepted Answer · answered Jan 25 '22 at 06:31

2

When you multiply a prior probability distribution by the likelihood function you can get a distribution that has an integral of more or less than one. It is not a probability distribution until you scale it to get that integral back to one.

answered Jan 25 '22 at 06:31

Michael Lew

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How is the "proportional" calculated in Bayes theorem?

1 Answers1