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Mathematics · Ch 12 — Introduction to Probability Theory

Bayes' Theorem

12.8

Bayes' Theorem

Thomas Bayes and the idea of reversing a conditional probability. Thomas Bayes (1702-1761) was an English statistician, philosopher, and Presbyterian minister, remembered for a specific and hugely influential result. Bayesian methods link a prior probability (belief held BEFORE running the experiment) and a conditional probability (the likelihood) to a posterior probability (belief held AFTER observing the outcome), via Bayes' rule. Bayesian probability treats probability more broadly, as a strength of belief or epistemic confidence given the available evidence -- not only as a long-run frequency.

Theorem 12.11 -- Bayes' Theorem. If A1,A2,…,AnA_1,A_2,\ldots,A_n are mutually exclusive and exhaustive events (a partition of SS) with P(Ai)>0P(A_i)>0 for every i=1,2,…,ni=1,2,\ldots,n, and BB is any event with P(B)>0P(B)>0, then for each ii:

P(Ai/B)=P(Ai) P(B/Ai)P(A1) P(B/A1)+P(A2) P(B/A2)+⋯+P(An) P(B/An)=P(Ai) P(B/Ai)∑j=1nP(Aj) P(B/Aj).P(A_i/B)=\frac{P(A_i)\,P(B/A_i)}{P(A_1)\,P(B/A_1)+P(A_2)\,P(B/A_2)+\cdots+P(A_n)\,P(B/A_n)}=\frac{P(A_i)\,P(B/A_i)}{\displaystyle\sum_{j=1}^{n}P(A_j)\,P(B/A_j)}.

Proof. By the Total Probability Theorem, P(B)=∑jP(Aj)P(B/Aj)P(B)=\sum_j P(A_j)P(B/A_j) -- exactly the denominator above. By the Multiplication Theorem, P(Ai∩B)=P(B/Ai) P(Ai)P(A_i\cap B)=P(B/A_i)\,P(A_i). By the definition of conditional probability, P(Ai/B)=P(Ai∩B)P(B)P(A_i/B)=\dfrac{P(A_i\cap B)}{P(B)}; substituting the two previous expressions for numerator and denominator gives the stated formula, linking P(Ai/B)P(A_i/B) back to the (usually easier to state) P(B/Ai)P(B/A_i).

Total Probability versus Bayes' Theorem -- two directions of the same setup. Given the same partition A1,…,AnA_1,\ldots,A_n and event BB: Total Probability answers 'GIVEN each possible cause, what is the overall chance of the effect BB?' -- summing forward from causes to effect. Bayes' Theorem answers the REVERSE question -- 'GIVEN that the effect BB was actually observed, what is the chance it came from cause AiA_i?' -- dividing 'this one cause's contribution to BB' by 'every cause's total contribution to BB' (which is exactly P(B)P(B) from Total Probability). This is why Bayes' Theorem always needs the Total Probability computation as its denominator. …