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

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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.

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

Layer-cake formulas for random variables

Statement

Let $(Ω, F, P)$ be a probability space.

If $X : Ω \to [0, + \infty]$ is measurable, then $E [X] = \int_{0}^{\infty} P (X > t) d t,$ where the right-hand side may be $+ \infty$ .
If $X$ is an integrable real random variable, then $E [X] = \int_{0}^{\infty} P (X > t) d t - \int_{0}^{\infty} P (X < - t) d t .$

Facts & Assumptions

Given: A probability space and a random variable $X$ in the relevant clause.

[L1]

Expectation is integration against $P$ , and $X = X^{+} - X^{-}$ with $∣ X ∣ = X^{+} + X^{-}$ for real $X$ (Expectation of a nonnegative or integrable random variable, The positive and negative parts of a function).

[L2]

The layer-cake formula with $p = 1$ gives $\int ∣ f ∣ d μ = \int_{0}^{\infty} μ ({∣ f ∣ > t}) d t$ for measurable $f$ (For 0 < p < infinity, the layer-cake formula computes the integral of |f|^p from the distribution function).

[L3]

The Lebesgue integral is linear on $L^{1}$ (The Lebesgue integral is linear on $L^{1} (μ)$ ).

Proof

technique · direct

1.1

Apply [L2] with $f = X$ and $p = 1$ . Because $X \geq 0$ , one has $∣ X ∣ = X$ and ${∣ X ∣ > t} = {X > t}$ , so [L1] gives $E [X] = \int_{0}^{\infty} P (X > t) d t .$

L1L2

2.1

If $X$ is integrable and real, then $X^{+}, X^{-} \in L^{1}$ and [L1] gives $X = X^{+} - X^{-}$ . By step 1.1 applied to $X^{+}$ and $X^{-}$ , $E [X^{+}] = \int_{0}^{\infty} P (X > t) d t, E [X^{-}] = \int_{0}^{\infty} P (X < - t) d t .$ Subtracting these identities and using [L3] proves the second formula.

L1L3step 1.1∎

Depends on

Used by

Recovering an exponential law from its tail Example
The uniform random variable on [0,1] Example

Dependency tree · two levels

17 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)
J. R. Norris, Probability and Measure, Section 4.2 (standard reference, not scraped)