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CorollaryStatement: Literature-sourcedProof: AI-adapted
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Layer-cake formulas for random variables

Statement

Let (Ω,F,P) be a probability space.

  1. If X:Ω→[0,+∞] is measurable, then E[X]=∫0∞P(X>t) dt, where the right-hand side may be +∞.
  2. If X is an integrable real random variable, then E[X]=∫0∞P(X>t) dt−∫0∞P(X<−t) dt.

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]

For a finite-valued measurable real function f, the layer-cake formula with p=1 gives ∫∣f∣ dμ=∫0∞μ({∣f∣>t}) dt (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 L1 (The Lebesgue integral is linear on L1(μ)).

[L4]

Increasing sequences of nonnegative measurable functions have integrals increasing to the integral of their pointwise limit, including infinite values (Monotone convergence for the integral).

Proof

technique · direct
1.1L1L2L4construct

For each integer n≥1, set Xn=min⁡{X,n}. This is a finite-valued nonnegative measurable real function: for a real a, {Xn>a} is Ω if a<0, is {X>a} if 0≤a<n, and is empty if a≥n. Also Xn↑X, including where X=+∞. Apply [L2] to Xn to obtain E[Xn]=∫0∞P(Xn>t) dt. For t≥0, the tail P(Xn>t) equals P(X>t) when t<n and zero when t≥n, so these tails increase pointwise to P(X>t). Every tail is a nonincreasing Borel function of t: each strict upper level set is an initial interval, possibly empty or the whole half-line. Apply [L4] first on Ω and then on (0,∞) to pass to the limits in the identities for Xn. Using [L1] gives E[X]=∫0∞P(X>t) dt, with +∞ allowed.

2.1L1L3step 1.1∎

If X is integrable and real, then X+,X−∈L1 and [L1] gives X=X+−X−. By step 1.1 applied to X+ and X−, E[X+]=∫0∞P(X>t) dt,E[X−]=∫0∞P(X<−t) dt. Subtracting these identities and using [L3] proves the second formula.

Depends on

Used by

Dependency tree · two levels

26 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