PARTITION OF SAMPLE SPACE

Tags: #◇Computer Science #◇STT101A #◇Conceptual #◇Partition of Sample Space

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◇Overview

A Partition of a Sample Space $S$ is a collection of non-empty events $\{B_1, B_2, \dots, B_k\}$ that divides $S$ into non-overlapping regions covering the entire space. Every elementary outcome $\omega \in S$ belongs to one and only one event $B_i$ in the partition.

◇Key Definitions

A collection of events $\{B_1, B_2, \dots, B_k\}$ forms a partition of $S$ if and only if it satisfies two conditions simultaneously:

1. Pairwise Mutually Exclusive (Disjoint): $B_i \cap B_j = \emptyset$ for all $i \ne j$.
2. Collectively Exhaustive: $\bigcup_{i=1}^k B_i = B_1 \cup B_2 \cup \cdots \cup B_k = S$ (which implies $\sum_{i=1}^k P(B_i) = 1$).

• Decomposition of Any Event $A$: Any event $A \subseteq S$ is partitioned into $k$ disjoint pieces $A = (A \cap B_1) \cup (A \cap B_2) \cup \cdots \cup (A \cap B_k)$, which is the foundation of the Law of Total Probability and Bayes' Theorem.

◇Worked Example

Problem: For rolling a six-sided die with $S = \{1, 2, 3, 4, 5, 6\}$, verify which of the following collections is a valid partition of $S$:

• $\mathcal{P}_1 = \{\{1, 2\}, \{3, 4\}, \{5, 6\}\}$: Valid (disjoint and union is $S$).
• $\mathcal{P}_2 = \{\{1, 2, 3\}, \{3, 4, 5, 6\}\}$: Invalid (not disjoint, since $3$ is in both).
• $\mathcal{P}_3 = \{\{1, 2\}, \{4, 5\}\}$: Invalid (not exhaustive, since $3$ and $6$ are missing).

◇References & Navigation

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• Related Notes: ◇Bayes's Rules, ◇Independent Events