BAYES'S RULES

Tags: #◇Computer Science #◇STT101A #◇Conceptual #◇Bayes's Rules

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◇Overview

Bayes' Rule (Bayes' Theorem) inverts conditional probabilities, allowing us to compute the posterior probability $P(B_k \mid A)$ of an unobservable cause or hypothesis $B_k$ given observed evidence $A$, using the prior probabilities $P(B_i)$ and likelihoods $P(A \mid B_i)$ across a partition of the sample space.

◇Key Definitions

• Two-Hypothesis Form: $P(B \mid A) = \frac{P(A \mid B)P(B)}{P(A \mid B)P(B) + P(A \mid B^c)P(B^c)}$.
• General Partition Form: For a partition $\{B_1, B_2, \dots, B_n\}$ of $S$:
  $P(B_k \mid A) = \frac{P(A \mid B_k)P(B_k)}{\sum_{i=1}^n P(A \mid B_i)P(B_i)}$
• Posterior Odds Form: $\frac{P(B \mid A)}{P(B^c \mid A)} = \frac{P(A \mid B)}{P(A \mid B^c)} \cdot \frac{P(B)}{P(B^c)} = \text{Likelihood Ratio} \times \text{Prior Odds}$.

◇Worked Example

Problem: Factory machines $M_1$ and $M_2$ produce $60\%$ and $40\%$ of chips, respectively. Defective rates are $P(D \mid M_1) = 0.02$ and $P(D \mid M_2) = 0.05$. If a randomly selected chip is defective ($D$), what is the probability it came from $M_2$?

Solution:
$P(M_2 \mid D) = \frac{P(D \mid M_2)P(M_2)}{P(D \mid M_1)P(M_1) + P(D \mid M_2)P(M_2)} = \frac{(0.05)(0.40)}{(0.02)(0.60) + (0.05)(0.40)} = \frac{0.020}{0.012 + 0.020} = \frac{0.020}{0.032} = 0.625$

◇References & Navigation

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• Related Notes: ◇Random Variable, ◇Partition of Sample Space