Fundamentals

Axioms:

  1. $ P(A) \geq 0 $ for all $ A $
  2. $ P(\Omega) = 1 $ where $ \Omega $ is the sample space
  3. If $A_i$ are disjoint (or mutually exclusive), then $ P( \bigcup_{i}^{\infty} A_i ) = \sum_{i}^{\infty} P(A_i) $

Interpretations of $P(A)$:

  • Frequentist: long-run frequency
  • Bayesian: degree of belief

Lemmas:

  • Complement: $P(A^c) = 1 - P(A)$
  • Subset: $A \subset B \implies P(A) \leq P(B)$
  • Venn diagram: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Independence:

  • $A,B$ are independent if $ P(A \cap B) = P(A) P(B) $
    • Extends to a set of indepedent events.
  • Either assumed or verified (by the above equation).
    • Trick: if $A,B$ are disjoint, each with positive probability, can they be independent? No, equation is violated: $P(A \cap B) = 0$ (by disjointness) but $P(A)P(B) > 0$ (by positivity), and so $ P(A \cap B) \neq P(A) P(B) $.
    • In general: no way to judge independence based on Venn diagram alone!

Conditional Probability:

  • Define $P(A \mid B) = \frac{P(A,B)}{P(B)}$ if $P(B)>0$.
    • Fraction of times that A occurs among events in which B also occurs
  • Axioms of probability apply to events left of the bar, not the right
    • E.g. if $A,B$ disjoint, then $P(A \cup B \mid C) = P(A \mid C) + P(B \mid C)$, but in general $P(C \mid A \cup B) \neq P(C \mid A) + P(C \mid B).$
  • In general, $P(A \mid B) \neq P(B \mid A)$.
    • E.g. probability of tumor given cancer is 1, but probability of cancer given tumor < 1 (could be benign)

Lemmas:

  • If $A,B$ independent, $P(A \mid B) = \frac{P(A)P(B)}{P(B)} = P(A)$
  • $P(A,B) = P(A \mid B)P(B) = P(B \mid A)P(A)$
    • $\implies P(A \mid B) = \frac{P(B \mid A)P(A)}{P(B)}$ (Bayes’ rule)

Bayes’ Theorem:

  • Law of total probability: if $A_i$ partitions $\Omega$, then for any event $B$, $ P(B) = \sum_{\forall i} P(B \mid A_i)P(A_i) $
    • For conditional probabilities, we have $P(B \mid C) = \sum_{\forall i} P(B \mid C \cap A_i) P(A_i \mid C)$
  • Bayes’ rule: $P(A \mid B) = \frac{P(B \mid A)P(A)}{P(B)} = \frac{P(B \mid A)P(A)}{\sum_{\forall i}P(B \mid A_i)P(A_i)}$