Alphabeta Math
CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Markov's conclusion can fail without nonnegativity

Statement refuted

The conclusion of Markov's inequality remains valid when the hypothesis X≥0 is removed.

Facts & Assumptions

Given: The uniform two-point probability space and a random variable X taking values 1 and −3.

[L1]

Each point in a uniform two-point space has probability 1/2 (The uniform probability space on a nonempty finite set).

[L3]

Markov's theorem assumes that X is nonnegative (Markov's inequality on a finite probability space).

Counterexample

technique · constructive
1.1

For the constructed two-valued variable, direct calculation gives E[X]=(1−3)/2=−1.

givenL1L2algebraconstruct
2.1

At threshold a=1, P(X≥1)=1/2, while E[X]/a=−1.

step 1.1L1algebra
3.1

The purported conclusion would be 1/2≤−1, which is false. This refutes removal of the nonnegativity hypothesis, not [L3].

step 2.1L3discharge-construct∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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