CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-04

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.

Linearity, monotonicity, and the modulus bound for expectation

Statement

Let $X, Y$ be integrable real or complex random variables on one probability space.

For scalars $a, b$ , $E [a X + b Y] = a E [X] + b E [Y] .$
If $X$ and $Y$ are real-valued and $X \leq Y$ almost surely, then $E [X] \leq E [Y] .$
$∣ E [X] ∣ \leq E [∣ X ∣] .$

Facts & Assumptions

Given: Integrable random variables $X, Y$ .

[L1]

Expectation is the Lebesgue integral against the probability measure (Expectation of a nonnegative or integrable random variable).

[L2]

The Lebesgue integral is linear on $L^{1}$ , the nonnegative integral is monotone, and the modulus of an integral is bounded by the integral of the modulus (The Lebesgue integral is linear on $L^{1} (μ)$ , Monotonicity and nonnegative homogeneity of the nonnegative integral, The modulus of an integral is bounded by the integral of the modulus).

Proof

technique · direct

1.1

Rewriting expectation as the integral by [L1], linearity in [L2] gives $E [a X + b Y] = a E [X] + b E [Y] .$

L1L2

1.2

Applying the integral triangle inequality from [L2] after [L1] gives $∣ E [X] ∣ = ∣ \int X d P ∣ \leq \int ∣ X ∣ d P = E [∣ X ∣] .$

L1L2

2.1

If $X \leq Y$ almost surely, then $Y - X \geq 0$ almost surely. Hence [L1] and [L2] give $0 \leq E [Y - X] = E [Y] - E [X],$ so $E [X] \leq E [Y]$ .

step 1.1L1L2

3.1

Steps 1.1, 1.2, and 2.1 prove the three assertions.

step 1.1step 1.2step 2.1∎

Depends on

Used by

Variance and covariance identities for random variables Lemma

Dependency tree · two levels

11 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

Rick Durrett, Probability: Theory and Examples, 5th ed., Section 1.6 (standard reference, not scraped)