Proof of Bayes' rule: Probability form
Let \(\mathbf H\) be a variable in \(\mathbb P\) for the true hypothesis, and let \(H_k\) be the possible values of \(\mathbf H,\) such that \(H_k\) is mutually exclusive and exhaustive. Then, Bayes’ theorem states:
with a proof that runs as follows. By the definition of conditional probability,
By the law of marginal probability:
By the definition of conditional probability again:
Done.
Note that this proof of Bayes’ rule is less general than the proof of the odds form of Bayes’ rule.
Example
Using the Diseasitis example problem, this proof runs as follows:
Numerically:
Using red for sick, blue for healthy, and + signs for positive test results, the proof above can be visually depicted as follows:
Parents:
- Proof of Bayes' rule
Proofs of Bayes’ rule, with graphics
- Bayes' rule: Probability form
The original formulation of Bayes’ rule.