TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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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.

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The first-moment method for avoiding or forcing a finite count of bad events

Statement

Let $X$ be a nonnegative integer-valued random variable on a finite probability space.

If $E [X] < 1$ , some outcome has $X = 0$ .
If $E [X] > 0$ , some outcome has $X > 0$ .
More generally, some outcome satisfies $X \geq E [X]$ and some satisfies $X \leq E [X]$ .

Facts & Assumptions

Given: A nonnegative integer-valued random variable $X$ on a finite probability space.

[L1]

Some outcome has value at least the expectation and some has value at most it (Expectation preserves pointwise order and lies between the minimum and maximum attained values).

[L2]

Markov gives $P (X \geq 1) \leq E [X]$ (Markov's inequality on a finite probability space).

[L3]

An event of positive probability in a finite probability space is nonempty (An event of positive probability in a finite probability space is nonempty).

[L4]

An event and its complement have probabilities summing to $1$ (Normalization, nonnegativity, monotonicity, complements, and differences in a finite probability space).

Proof

technique · direct

1.1

If $E [X] < 1$ , [L2] gives $P (X \geq 1) < 1$ , so [L4] gives $P (X = 0) > 0$ because a nonnegative integer is either zero or at least one.

L2L4algebra

1.2

If $E [X] > 0$ , an outcome with $X \geq E [X] > 0$ exists by [L1].

2.1

By [L3], the event ${X = 0}$ is nonempty.

step 1.1L3

3.1

The two averaging assertions are exactly [L1]; steps 1.1 and 2.1 prove avoidance, and step 1.2 proves forcing.

step 1.1step 1.2step 2.1L1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 33 results over 11 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, Chapter 2 (standard reference, not scraped)
Y. Zhao, MIT 18.218 Probabilistic Method in Combinatorics, Section 2.1 (standard reference, not scraped)