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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

For a uniform random sign $ε$ , $E [e^{t ε}] \leq e^{t^{2} / 2}$

Statement

If $ε$ is uniform on ${- 1, 1}$ , then for every real $t$ , $E [\exp (t ε)] \leq \exp (t^{2} / 2) .$

Facts & Assumptions

Given: A uniform random sign $ε$ and a real $t$ .

[L1]

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

[L2]

$\exp (x) = \sum_{k \geq 0} x^{k} / k!$ for real $x$ (The real exponential function and the number $e$ by a power series).

[L3]

Factorials are finite products with $0! = 1$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ).

[L4]

Convergent series may be added and scaled, and termwise comparison of nonnegative series passes to their sums (Convergent series add and scale termwise, If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ , A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum).

[L5]

Finite sums obey addition, scaling, and monotonicity (Laws of finite sums and finite products).

Proof

technique · direct

1.1

Direct averaging gives $E [\exp (t ε)] = (\exp (t) + \exp (- t)) / 2$ .

1.2

For every $j \geq 0$ , $(2 j)! = \prod_{r = 1}^{j} (2 r - 1) (2 r) \geq \prod_{r = 1}^{j} 2 r = 2^{j} j!$ , including $j = 0$ .

L3L5algebra

2.1

Expanding both exponentials by [L2] and using [L4], the odd powers cancel and the result is $\sum_{j \geq 0} t^{2 j} / (2 j)!$ .

step 1.1L2L4

3.1

Hence $t^{2 j} / (2 j)! \leq (t^{2} / 2)^{j} / j!$ term by term, and [L4] gives $E [\exp (t ε)] \leq \sum_{j \geq 0} (t^{2} / 2)^{j} / j! = \exp (t^{2} / 2)$ .

step 2.1step 1.2L2L4∎

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: 93 results over 22 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

J. Matousek and J. Vondrak, The Probabilistic Method, Section 7.1 (standard reference, not scraped)
Y. Zhao, MIT 18.218 Probabilistic Method in Combinatorics, Section 4.1 (standard reference, not scraped)

For a uniform random sign εε, E[etε]≤et2/2E[etε]≤et2/2