Bayes v. Frequentist — Bayes’s Rule: Derivation

© 26 January 2023 by Michael A. Kohn

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Introduction | Coin Problem | Bayes’s Rule: History | Bayes’s Rule: Derivation | Billiards Problem | Endnotes | References

Bayes’s Rule : Derivation

Whether or not Bayes wrote it, there is nothing controversial about his rule. Everybody accepts that the probability of both A and B is the probability of A given B times the probability of B:

$$P(A \& B) = P(A|B) \times P(B),$$

that the probability of both B and A is the probability of B given A times the probability of A,

$$P(B \& A) = P(B|A) \times P(A),$$

and that the probability of A and B equals the the probability of B and A:

$$P(A \& B) = P(B \& A).$$

So,

\begin{align*} P(A|B) \times P(B) &= P(B|A) \times P(A)\\ P(A|B) &= \frac{P(B|A) \times P(A)}{P(B)}\\ \end{align*}

Nobody can argue with this one-step derivation. The identity was likely known before Bayes and is not the focus of his essay.

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Introduction | Coin Problem | Bayes’s Rule: History | Bayes’s Rule: Derivation | Billiards Problem | Endnotes | References