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

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The moment generating function of a finite sum of independent variables is the product of their moment generating functions

Statement

If $(X_{i})_{i \in I}$ is a finite mutually independent family, then for every real $t$ , $M_{\sum_{i \in I} X_{i}} (t) = \prod_{i \in I} M_{X_{i}} (t) .$ For $I = \emptyset$ , both sides equal $1$ .

Facts & Assumptions

Given: A finite mutually independent family $(X_{i})_{i \in I}$ and $t \in R$ .

[L1]

$M_{X} (t) = E [\exp (t X)]$ (The moment generating function $M_{X} (t) = E [e^{t X}]$ on a finite probability space).

[L2]

Expectation factors over finite products of mutually independent random variables (Expectation factors over a finite product of mutually independent random variables).

[L3]

$\exp (x + y) = \exp (x) \exp (y)$ for all reals $x, y$ (The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ ).

[L4]

Mutual independence factors every joint attained-value probability (Pairwise and mutual independence of finite-valued random variables).

[L5]

Probability is additive on finite disjoint unions, and finite sums may be regrouped and interchanged (Probability is additive on every finite pairwise-disjoint family of events, Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule).

Proof

technique · direct

1.1

Iterating [L3] gives $\exp (t \sum_{i} X_{i}) = \prod_{i} \exp (t X_{i})$ pointwise. For any joint values of the transformed variables, each corresponding event is a disjoint union of joint-value events of the $X_{i}$ ; summing the products supplied by [L4] and factoring the finite sums with [L5] proves that the transformed variables remain mutually independent.

L3L4L5algebra

2.1

Apply [L2] to step 1.1 and use [L1] in each factor to obtain the formula.

step 1.1L1L2

3.1

For $I = \emptyset$ , the sum is zero, $M_{0} (t) = \exp (0) = 1$ , and the product is empty and equals $1$ .

step 2.1L1L3∎

Depends on

Used by

For n≥1 independent random signs, ℙ(|S|≥ t)≤2 exp(-t²/(2n)) for t>0 Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 81 results over 18 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

Y. Zhao, MIT 18.218 Probabilistic Method in Combinatorics, Section 4.1 (standard reference, not scraped)