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Appendix 2.1: Derivation of Bayes’ Theorem

We know that the following formulas are correct:

P (A | B) = \frac{P (A B)}{P (B)}

$P (A | B) = \frac{P (A B)}{P (B)}$ (1)

P (B | A) = \frac{P (A B)}{P (A)}

$P (B | A) = \frac{P (A B)}{P (A)}$ (2)

[which can be rewritten as P (A B) = P (B | A) P (A)] and

$[which can be rewritten as P (A B) = P (B | A) P (A)] and$

P (B | A') = \frac{P (A' B)}{P (A')}

$P (B | A') = \frac{P (A' B)}{P (A')}$

[which can be rewritten as P (A' B) = P (B | A') P (A')] .

$[which can be rewritten as P (A' B) = P (B | A') P (A')] .$ (3)

Furthermore, by definition, we know that

\begin{array}{l} P (B) & = & P (A B) + P (A^{'} B) \\ = & P (B | A) P (A) + P (B | A^{'}) P (A^{'}) \\ ↖ from (2) from (3) ↗ \end{array}

$\begin{array}{l} P (B) & = & P (A B) + P (A^{'} B) \\ = & P (B | A) P (A) + P (B | A^{'}) P (A^{'}) \\ ↖ from (2) from (3) ↗ \end{array}$ (4)

Substituting Equations 2 and 4 into Equation 1, we have

\begin{array}{l} \begin{array}{l} P (A | B) & = & \frac{P (A B)}{P (B)} ↙ from (2) \\ = & \frac{P (B | A) P (A)}{\underset{}{\underset{︸}{P (B | A) P (A) + P (B | A^{'}) P (A^{'})}}} \end{array} \\ from (4) ↗ \end{array}

$\begin{array}{l} \begin{array}{l} P (A | B) & = & \frac{P (A B)}{P (B)} ↙ from (2) \\ = & \frac{P (B | A) P (A)}{\underset{}{\underset{︸}{P (B | A) P (A) + P (B | A^{'}) P (A^{'})}}} \end{array} \\ from (4) ↗ \end{array}$ (5)

This is the general form of Bayes’ Theorem, shown as Equation 2-5 in this chapter.

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Table of Contents for Appendix 2.1: Derivation of Bayes&#8217; TheoremDerivation of Bayes’ Theorem

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Table of Contents for
Appendix 2.1: Derivation of Bayes’ TheoremDerivation of Bayes’ Theorem