Answer to Question #149133 in Statistics and Probability for Chembo Balamani

Question

Define and prove the Bayes’ theorem.

Answer 1

Bayes’ theorem states that, if $A_1, A_2, …, A_n$ are mutually exclusive and exhaustive events in a sample space s, such that $P(A_i) > 0$ for $i = 1, 2, …, n$ and E is any event which is subset of $\bigcup_{i=1}^{n}A_i \; with\; P(E) > 0$ , then

$P (A_k|E) = \frac{P(A_k) \times P(E|A_k)}{\sum_{i=1}^{n}P(A_i) \times P(E|A_i)}$ for $k = 1, 2, 3 … n$ .

Proof:

Since $A_1, A_2,...A_n$ are mutually exclusive and exhaustive events in a sample space s, we have $\bigcup_{i=1}^{n}A_i….s$

Since $A_1, A_2, A_3...A_n$ are mutually disjointed, $E \cap A_1, E \cap A_2, E \cap A_3, … E \cap A_n$ are mutually disjointed.

$P(E) = P(E \cap S) \\ = P[E \cap (\bigcup_{i=1}^{n}A_i)] \\ = P(\bigcup_{i=1}^{n}E \cap A_i) \\ = \sum_{i=1}^{n}P(E \cap A_i) \\ = \sum_{i=1}^{n}P(A_i) \times P(E|A_i)$

by multiplication theorem of probability.

Also we have, for k = 1, 2,….,n

$P(E \cap A_k) = P(E) \times P(A_k|E) \\ P(A_k|E) = \frac{P(E \cap A_k)}{P(E)} \\ = \frac{P(A_k) \times P(E|A_k)}{\sum_{i=1}^{n}P(A_i) \times P(E|A_i)}$