TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-13

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The finite second-moment bound $P (X \neq 0) \geq E [X]^{2} / E [X^{2}]$ when $E [X^{2}] > 0$

Statement

Let $X$ be a finite real random variable. If $E [X^{2}] > 0$ , then $P (X \neq 0) \geq \frac{E [X]^{2}}{E [X^{2}]} .$ If $E [X^{2}] = 0$ , then $X = 0$ on every positive-weight outcome and $P (X \neq 0) = 0$ .

Facts & Assumptions

Given: A finite real random variable $X$ .

[L1]

$1_{A}$ has expectation $P (A)$ (Indicators turn event probabilities, intersections, and finite counts into expectations and products).

[L2]

Cauchy-Schwarz states $E [U V]^{2} \leq E [U^{2}] E [V^{2}]$ (Cauchy-Schwarz for finite random variables: $E [X Y]^{2} \leq E [X^{2}] E [Y^{2}]$ ).

[L3]

Expectation is the finite sum of values times nonnegative outcome weights (Expectation of a real random variable on a finite probability space).

Proof

technique · cases

1.1

Assume $E [X^{2}] > 0$ . Pointwise, $X 1_{{X \neq 0}} = X$ . Apply [L2] to $U = X$ and $V = 1_{{X \neq 0}}$ ; using [L1] gives $E [X]^{2} \leq E [X^{2}] P (X \neq 0)$ .

assume-case positiveL1L2algebra

1.2

Assume $E [X^{2}] = 0$ . The nonnegative summands $X (ω)^{2} w (ω)$ then force $X = 0$ at every positive-weight outcome, so $P (X \neq 0) = 0$ .

assume-case zeroL3algebra

2.1

Dividing by the positive second moment gives the displayed bound.

step 1.1algebra

3.1

Nonnegativity of $E [X^{2}]$ makes the two cases exhaustive.

step 2.1step 1.2cases-exhaustive∎

Depends on

Used by

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Sources

H. Pishro-Nik, Introduction to Probability, Statistics, and Random Processes, Section 6.2.4 (standard reference, not scraped)
Y. Zhao, MIT 18.218 Probabilistic Method in Combinatorics, Section 3.3 (standard reference, not scraped)

The finite second-moment bound P(X≠0)≥E[X]2/E[X2]P(X=0)≥E[X]2/E[X2] when E[X2]>0E[X2]>0