TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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Product weights normalize, and coordinate events are mutually independent

Statement

The weights in every finite product of finite probability spaces are nonnegative and sum to $1$ . For events $A_{i} \subseteq Ω_{i}$ , the coordinate events $C_{i} : = {ω \in \prod_{j \in I} Ω_{j} : ω_{i} \in A_{i}}$ are mutually independent, and for every $J \subseteq I$ , $P ⁣ (⋂_{j \in J} C_{j}) = \prod_{j \in J} P_{j} (A_{j}) .$

Facts & Assumptions

Given: A finite family $((Ω_{i}, w_{i}))_{i \in I}$ of finite probability spaces and coordinate events $C_{i}$ .

[L1]

Product outcomes and their weights are defined coordinatewise, with the empty product equal to $1$ (The finite product of finite probability spaces).

[L2]

Finite Fubini interchanges and factors iterated finite sums, while finite products of nonnegative reals are nonnegative (Laws of finite sums and finite products, Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule).

[L3]

Mutual independence is the intersection product identity for every nonempty subfamily (Independent events, pairwise independence, and mutual independence of a finite family).

Proof

technique · induction

1.1

For $I = \emptyset$ , the unique product outcome has weight $1$ , so normalization holds.

L1base

1.2

If normalization holds for $I$ and another factor $(Ω_{k}, w_{k})$ is appended, finite Fubini gives $\sum_{(ω, u)} w (ω) w_{k} (u) = (\sum_{ω} w (ω)) (\sum_{u} w_{k} (u)) = 1$ .

L1L2ih

1.3

For $J \subseteq I$ , summing the product weights over $⋂_{j \in J} C_{j}$ factors coordinatewise: a coordinate in $J$ contributes $P_{j} (A_{j})$ and a coordinate outside $J$ contributes $1$ .

L1L2

2.1

Induction proves normalization for every finite product; nonnegativity follows from nonnegativity of all factor weights.

step 1.1step 1.2L2discharge-induction

3.1

Thus $P (⋂_{j \in J} C_{j}) = \prod_{j \in J} P_{j} (A_{j})$ . The empty $J$ gives $1 = 1$ , so [L3] proves mutual independence in all cases.

step 2.1step 1.3L3discharge-induction∎

Depends on

Used by

Cited to discharge well-definedness by The finite product of finite probability spaces.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 59 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

C. M. Grinstead and J. L. Snell, Introduction to Probability, 2nd ed., Section 4.1 (standard reference, not scraped)
H. Pishro-Nik, Introduction to Probability, Statistics, and Random Processes, Section 1.4.1 (standard reference, not scraped)