Probability Spaces

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

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

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

1. $\Omega \in \mathcal{F}$.
2. If $A \in \mathcal{F}$Then $A^c \in \mathcal{F}$ (closed under complementation).
3. If $A_1, A_2, \ldots \in \mathcal{F}$Then $\bigcup_{i=1}^{\infty} A_i \in \mathcal{F}$ (closed under countable unions).

Definition. A probability measure $P$ satisfies:

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

Proposition 1.1. For any probability space:

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

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

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

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

Definition. The conditional probability of $A$ given $B$ (with $P(B) > 0$) is

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

Theorem 1.2 (Law of Total Probability). If $B_1, \ldots, B_n$ form a partition of $\Omega$ with $P(B_i) > 0$ for all $i$Then

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

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

$P(B_j \mid A) = \frac{P(A \mid B_j)\, P(B_j)}{\sum_{i=1}^{n} P(A \mid B_i)\, P(B_i)}$

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

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

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

Definition. Events $A_1, \ldots, A_n$ are mutually independent if for every subset $J \subseteq \{1, \ldots, n\}$:

$P\left(\bigcap_{j \in J} A_j\right) = \prod_{j \in J} P(A_j)$

Pairwise independence does not imply mutual independence.

• 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 $P(A|B)$ and $P(B|A)$ using the prior and likelihood.

• Confusing “mutually exclusive” with “independent.” Mutually exclusive events with positive probability are always dependent.
• Assuming that $P(A \cap B) > 0$ 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.

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?