0
$\begingroup$

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.

enter image description here

$\endgroup$

1 Answer 1

2
$\begingroup$

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
https://en.wikipedia.org/wiki/Conditional_probability#As_an_axiom_of_probability

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.