Index: The Book of Statistical ProofsGeneral Theorems ▷ Bayesian statistics ▷ Bayesian inference ▷ Bayes' rule

Theorem: Let $A_1$, $A_2$ and $B$ be arbitrary statements about random variables where $A_1$ and $A_2$ are mutually exclusive. Then, Bayes’ rule states that the posterior odds are equal to the Bayes factor times the prior odds, i.e.

$\label{eq:bayes-rule} \frac{p(A_1|B)}{p(A_2|B)} = \frac{p(B|A_1)}{p(B|A_2)} \cdot \frac{p(A_1)}{p(A_2)} \; .$

Proof: Using Bayes’ theorem, the conditional probabilities on the left are given by

$\label{eq:bayes-th-A1} p(A_1|B) = \frac{p(B|A_1) \cdot p(A_1)}{p(B)}$ $\label{eq:bayes-th-A2} p(A_2|B) = \frac{p(B|A_2) \cdot p(A_2)}{p(B)} \; .$

Dividing the two conditional probabilities by each other

$\label{eq:bayes-rule-qed} \begin{split} \frac{p(A_1|B)}{p(A_2|B)} &= \frac{p(B|A_1) \cdot p(A_1) / p(B)}{p(B|A_2) \cdot p(A_2) / p(B)} \\ &= \frac{p(B|A_1)}{p(B|A_2)} \cdot \frac{p(A_1)}{p(A_2)} \; , \end{split}$

one obtains the posterior odds ratio as given by the theorem.

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Metadata: ID: P12 | shortcut: bayes-rule | author: JoramSoch | date: 2020-01-06, 20:55.