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Probability Spaces | Mathematics

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).

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 B Then 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) + \cdots So 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

Section titled “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 ii Then

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) > 0 Then 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

  • The sigma-algebra must be closed under countable operations, not just finite ones.
  • Independence is a property of the probability measure, not of the sets themselves.
  • Conditional probability satisfies the axioms of probability for a fixed conditioning event.
  • Bayes’ theorem converts between P(AB)P(A|B) and P(BA)P(B|A) using the prior and likelihood.
  • Confusing “mutually exclusive” with “independent.” Mutually exclusive events with positive probability are always dependent.
  • Assuming that P(AB)>0P(A \cap B) > 0 implies dependence. Two events can overlap and still be independent.
  • Forgetting that sigma-algebras must be closed under countable unions, not just finite ones.
  • Applying Bayes’ theorem without verifying that the partition actually covers the sample space.
  • Medical testing: Bayes’ theorem computes the true positive rate from sensitivity and prevalence.
  • Quality control: Independence assumptions simplify the probability of multiple component failures.
  • Finance: Insurance pricing uses conditional probability to model claim frequency given risk factors.
  • Machine learning: Naive Bayes classifiers assume feature independence to compute posterior probabilities efficiently.

1.7 Intuition: What Is a Probability Space?

Section titled “1.7 Intuition: What Is a Probability Space?”

A probability space is the mathematical foundation for reasoning about uncertainty. The sample space Ω\Omega lists every possible outcome of an experiment. The sigma-algebra F\mathcal{F} specifies which collections of outcomes we are allowed to assign probabilities to. The probability measure PP assigns numbers between 0 and 1 to those collections, obeying the rule that disjoint events have additive probabilities.

The sigma-algebra is more than a formality. For infinite sample spaces, not every subset can be assigned a meaningful probability. The sigma-algebra ensures closure under countable operations, which is needed for taking limits of events. Conditional probability, P(AB)=P(AB)/P(B)P(A|B) = P(A \cap B)/P(B), is the mathematical formalisation of “updated belief”: it reassigns probabilities after learning that event BB has occurred. Bayes’ theorem then converts between P(AB)P(A|B) and P(BA)P(B|A), which is the foundation of statistical inference. Independence captures the idea that knowing one event occurred tells you nothing about the other, formalised as P(AB)=P(A)P(B)P(A \cap B) = P(A)P(B).

1.8 Worked Example: Applying Bayes’ Theorem

Section titled “1.8 Worked Example: Applying Bayes’ Theorem”

Problem. A disease affects 1 in 1000 people. A test is 99% sensitive (true positive rate) and 95% specific (true negative rate). If a person tests positive, what is the probability they have the disease?

Solution

Let DD be the event of having the disease and TT be the event of testing positive.

P(D)=0.001P(D) = 0.001, P(TD)=0.99P(T|D) = 0.99, P(TDc)=0.05P(T|D^c) = 0.05.

By the law of total probability: P(T)=P(TD)P(D)+P(TDc)P(Dc)=0.99×0.001+0.05×0.999=0.00099+0.04995=0.05094P(T) = P(T|D)P(D) + P(T|D^c)P(D^c) = 0.99 \times 0.001 + 0.05 \times 0.999 = 0.00099 + 0.04995 = 0.05094.

By Bayes’ theorem: P(DT)=P(TD)P(D)/P(T)=0.00099/0.05094=0.0194P(D|T) = P(T|D)P(D)/P(T) = 0.00099/0.05094 = 0.0194.

So only about 1.94% of positive tests are true positives, despite the high sensitivity. \blacksquare

flowchart TD
A[1_Probability Spaces] --> B[Key Concepts]
A --> C[Core Principles]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
C --> F[Design patterns]
D --> G[Real-world usage]

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

Ensure you have mastered the prerequisite material before attempting this advanced content.