1.1 Sample Spaces and Events
A probability space is a triple (Ω,F,P) where:
- Ω is the sample space (set of all possible outcomes).
- F is a sigma-algebra on Ω.
- P:F→[0,1] is a probability measure.
Definition. A sigma-algebra F on Ω is a collection of subsets satisfying:
- Ω∈F.
- If A∈FThen Ac∈F (closed under complementation).
- If A1,A2,…∈FThen ⋃i=1∞Ai∈F (closed under countable unions).
Definition. A probability measure P satisfies:
- Non-negativity: P(A)≥0 for all A∈F.
- Normalisation: P(Ω)=1.
- Countable additivity: If A1,A2,… are pairwise disjoint, then P(⋃i=1∞Ai)=∑i=1∞P(Ai).
1.2 Basic Properties
Proposition 1.1. For any probability space:
- P(∅)=0.
- P(Ac)=1−P(A).
- If A⊆BThen P(A)≤P(B).
- P(A∪B)=P(A)+P(B)−P(A∩B) (inclusion-exclusion).
- Boole”s inequality: P(⋃i=1nAi)≤∑i=1nP(Ai).
- Bonferroni inequality: P(⋂i=1nAi)≥1−∑i=1n(1−P(Ai)).
Proof. (1) Apply countable additivity to the disjoint union Ω=Ω∪∅∪∅∪⋯: 1=1+P(∅)+P(∅)+⋯So P(∅)=0.
(3) B=A∪(B∖A) is a disjoint union, so P(B)=P(A)+P(B∖A)≥P(A).
(4) P(A∪B)=P(A)+P(B∖A)=P(A)+P(B)−P(A∩B). ■
1.3 Conditional Probability and Independence
Definition. The conditional probability of A given B (with P(B)>0) is
P(A∣B)=P(B)P(A∩B)
Theorem 1.2 (Law of Total Probability). If B1,…,Bn form a partition of Ω with P(Bi)>0 for all iThen
P(A)=∑i=1nP(A∣Bi)P(Bi)
Theorem 1.3 (Bayes’ Theorem). Under the same conditions:
P(Bj∣A)=∑i=1nP(A∣Bi)P(Bi)P(A∣Bj)P(Bj)
Definition. Events A and B are independent if P(A∩B)=P(A)P(B).
Proposition 1.4. If A and B are independent with P(B)>0Then P(A∣B)=P(A).
Proof. P(A∣B)=P(A∩B)/P(B)=P(A)P(B)/P(B)=P(A). ■
Definition. Events A1,…,An are mutually independent if for every subset J⊆{1,…,n}:
P(⋂j∈JAj)=∏j∈JP(Aj)
Pairwise independence does not imply mutual independence.
Worked Example: Pairwise vs Mutual Independence
Solution. Roll two fair dice. Let A = “first die is even”, B = “second die is even”, C = “sum is even”.
P(A)=P(B)=P(C)=1/2.
P(A∩B)=1/4=P(A)P(B). P(A∩C)=P(firsteven,sumeven)=P(secondeven)=1/4=P(A)P(C).
P(B∩C)=1/4=P(B)P(C). So A, B, C are pairwise independent.
But P(A∩B∩C)=P(botheven,sumeven)=P(botheven)=1/4=1/8=P(A)P(B)P(C).
So A, B, C are pairwise independent but not mutually independent. ■