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Probability Spaces

1.1 Sample Spaces and Events

A probability space is a triple (Ω,F,P)(\Omega, \mathcal{F}, P) where:

  • Ω\Omega is the sample space (set of all possible outcomes).
  • F\mathcal{F} is a sigma-algebra on Ω\Omega.
  • P:F[0,1]P : \mathcal{F} \to [0, 1] is a probability measure.

Definition. A sigma-algebra F\mathcal{F} on Ω\Omega is a collection of subsets satisfying:

  1. ΩF\Omega \in \mathcal{F}.
  2. If AFA \in \mathcal{F}Then AcFA^c \in \mathcal{F} (closed under complementation).
  3. If A1,A2,FA_1, A_2, \ldots \in \mathcal{F}Then i=1AiF\bigcup_{i=1}^{\infty} A_i \in \mathcal{F} (closed under countable unions).

Definition. A probability measure PP satisfies:

  1. Non-negativity: P(A)0P(A) \geq 0 for all AFA \in \mathcal{F}.
  2. Normalisation: P(Ω)=1P(\Omega) = 1.
  3. Countable additivity: If A1,A2,A_1, A_2, \ldots are pairwise disjoint, then P(i=1Ai)=i=1P(Ai)P\left(\bigcup_{i=1}^{\infty} A_i\right) = \sum_{i=1}^{\infty} P(A_i).

1.2 Basic Properties

Proposition 1.1. For any probability space:

  1. P()=0P(\emptyset) = 0.
  2. P(Ac)=1P(A)P(A^c) = 1 - P(A).
  3. If ABA \subseteq BThen P(A)P(B)P(A) \leq P(B).
  4. P(AB)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B) - P(A \cap B) (inclusion-exclusion).
  5. Boole”s inequality: P(i=1nAi)i=1nP(Ai)P\left(\bigcup_{i=1}^{n} A_i\right) \leq \sum_{i=1}^{n} P(A_i).
  6. Bonferroni inequality: P(i=1nAi)1i=1n(1P(Ai))P\left(\bigcap_{i=1}^{n} A_i\right) \geq 1 - \sum_{i=1}^{n} (1 - P(A_i)).

Proof. (1) Apply countable additivity to the disjoint union Ω=Ω\Omega = \Omega \cup \emptyset \cup \emptyset \cup \cdots: 1=1+P()+P()+1 = 1 + P(\emptyset) + P(\emptyset) + \cdotsSo P()=0P(\emptyset) = 0.

(3) B=A(BA)B = A \cup (B \setminus A) is a disjoint union, so P(B)=P(A)+P(BA)P(A)P(B) = P(A) + P(B \setminus A) \geq P(A).

(4) P(AB)=P(A)+P(BA)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B \setminus A) = P(A) + P(B) - P(A \cap B). \blacksquare

1.3 Conditional Probability and Independence

Definition. The conditional probability of AA given BB (with P(B)>0P(B) > 0) is

P(AB)=P(AB)P(B)P(A \mid B) = \frac{P(A \cap B)}{P(B)}

Theorem 1.2 (Law of Total Probability). If B1,,BnB_1, \ldots, B_n form a partition of Ω\Omega with P(Bi)>0P(B_i) > 0 for all iiThen

P(A)=i=1nP(ABi)P(Bi)P(A) = \sum_{i=1}^{n} P(A \mid B_i)\, P(B_i)

Theorem 1.3 (Bayes’ Theorem). Under the same conditions:

P(BjA)=P(ABj)P(Bj)i=1nP(ABi)P(Bi)P(B_j \mid A) = \frac{P(A \mid B_j)\, P(B_j)}{\sum_{i=1}^{n} P(A \mid B_i)\, P(B_i)}

Definition. Events AA and BB are independent if P(AB)=P(A)P(B)P(A \cap B) = P(A)\,P(B).

Proposition 1.4. If AA and BB are independent with P(B)>0P(B) > 0Then P(AB)=P(A)P(A \mid B) = P(A).

Proof. P(AB)=P(AB)/P(B)=P(A)P(B)/P(B)=P(A)P(A \mid B) = P(A \cap B)/P(B) = P(A)P(B)/P(B) = P(A). \blacksquare

Definition. Events A1,,AnA_1, \ldots, A_n are mutually independent if for every subset J{1,,n}J \subseteq \{1, \ldots, n\}:

P(jJAj)=jJP(Aj)P\left(\bigcap_{j \in J} A_j\right) = \prod_{j \in J} P(A_j)

Pairwise independence does not imply mutual independence.

Worked Example: Pairwise vs Mutual Independence

Solution. Roll two fair dice. Let AA = “first die is even”, BB = “second die is even”, CC = “sum is even”.

P(A)=P(B)=P(C)=1/2P(A) = P(B) = P(C) = 1/2.

P(AB)=1/4=P(A)P(B)P(A \cap B) = 1/4 = P(A)P(B). P(AC)=P(firsteven,sumeven)=P(secondeven)=1/4=P(A)P(C)P(A \cap C) = P(\text{first} even, sum even) = P(\text{second} even) = 1/4 = P(A)P(C).

P(BC)=1/4=P(B)P(C)P(B \cap C) = 1/4 = P(B)P(C). So AA, BB, CC are pairwise independent.

But P(ABC)=P(botheven,sumeven)=P(botheven)=1/41/8=P(A)P(B)P(C)P(A \cap B \cap C) = P(\text{both} even, sum even) = P(\text{both} even) = 1/4 \neq 1/8 = P(A)P(B)P(C).

So AA, BB, CC are pairwise independent but not mutually independent. \blacksquare