TheoremStatement: AI-adaptedProof: AI-adaptedverified 2026-09-26 (gpt-6-sol)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A finite family of events is mutually independent exactly when its indicators are mutually independent

Statement

A finite family of events is mutually independent if and only if its indicator random variables are mutually independent. The same equivalence holds with pairwise independence in place of mutual independence.

Facts & Assumptions

Given: A finite family of events $(A_{i})_{i \in I}$ .

[L1]

Mutual independence is preserved when events are replaced by complements (Mutual independence is inherited by subfamilies and by replacing events with complements).

[L2]

The event ${1_{A} = 1}$ is $A$ , and ${1_{A} = 0}$ is $A^{c}$ (The indicator random variable of an event).

[L3]

Random variables are mutually independent exactly when all finite joint attained-value probabilities factor (Pairwise and mutual independence of finite-valued random variables).

Proof

technique · direct

1.1

Suppose the events are mutually independent. Every joint assignment $1_{A_{j}} = b_{j}$ , with $b_{j} \in {0, 1}$ , is an intersection of events $A_{j}$ and complements $A_{j}^{c}$ , whose probability factors by [L1].

L1L2

1.2

Conversely, suppose the indicators are mutually independent and fix a nonempty subfamily. If one of its events is empty, both the probability of the intersection and the product of the individual probabilities are zero. Otherwise $1$ is an attained value of every indicator in the subfamily, so specialize the joint-value identity of [L3] to $b_{j} = 1$ for every chosen index. By [L2] this is exactly the event-intersection product identity.

L2L3algebra

2.1

Hence the indicators are mutually independent by [L3].

step 1.1L3

3.1

Step 1.1 together with step 2.1 proves the forward direction, and step 1.2 proves the reverse direction. Restricting the same arguments to two indices proves the pairwise equivalence.

step 1.1step 1.2step 2.1∎

Depends on

Used by

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Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

H. Pishro-Nik, Introduction to Probability, Statistics, and Random Processes, Sections 1.4.1 and 3.1.5 (standard reference, not scraped)
J. Matousek and J. Vondrak, The Probabilistic Method, Sections 1.1 and 3.1 (standard reference, not scraped)