Is this Bayes Theorem?

I'm reading this blog and it says this is Bayes Theorem. I thought Bayes Theorem just had two probabilities divided by one probability.

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1 Answer 1


Yes, it's just Bayes theorem with some additional probability rules applied (and then some additional manipulations).

e.g. one such rule being used is $p(z,\theta) = p(z|\theta)\, p(\theta)$, which follows from the definition of conditional probability. See

You might start with $p(z|y,\theta)=p(y|z,\theta)\, p(z,\theta)/p(y,\theta)$ and then apply common probability rules.

If it is not immediately clear that this expression is just Bayes, begin with: $p(z|y,\theta)\, p(y,\theta) = p(z,y,\theta) = p(y|z,\theta)\, p(z,\theta)$ (which follows from conditional probability axioms) and divide through by $p(y,\theta)$, in similar fashion to deriving the most basic form of Bayes' theorem.

Then proceed to apply simple manipulations using basic facts as needed.

  • $\begingroup$ Thank you. I know I can infer it from the equations, but what does the comma mean here exactly? How is it different from the | symbol? $p(z|y,\theta)$ $\endgroup$
    – Renoldus
    Commented Apr 27, 2022 at 0:26
  • $\begingroup$ When you're dealing with the joint distribution of more than one variable you put commas between the variables -- see en.wikipedia.org/wiki/… and en.wikipedia.org/wiki/…, compare that with the use of the vertical bar in conditional distributions, with the thing after $|$ being what is conditioned on -- see en.wikipedia.org/wiki/… $\endgroup$
    – Glen_b
    Commented Apr 27, 2022 at 0:39

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