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TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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Basic algebra and order properties of conditional expectation

Statement

Assume AC for existence. For real X,YL1(P) and a,bR, E[aX+bYG]=aE[XG]+bE[YG]. Conditional expectation is positive, preserves order and constants, satisfies E(E[XG])=EX, and E[XG]E[XG] almost surely. Also X<Y almost surely implies E[XG]<E[YG] almost surely.

Facts & Assumptions

Given: AC, a probability space, a sub-sigma-algebra G, real integrable X,Y and real scalars a,b; for the strict clause assume X<Y almost surely.

[F1]

Under AC the conditional class exists and each version has the defining event integrals. (Conditional expectation as an ae class)

[F2]

Versions with the same defining data agree almost surely. (Conditional expectation is unique almost surely)

[F3]

Integrability and integrals are preserved by finite linear combinations. (The Lebesgue integral is linear on L1(μ))

[F4]

A nonnegative measurable function has zero integral exactly when it is zero almost everywhere. (A nonnegative measurable function has integral 0 exactly when it vanishes almost everywhere)

[F5]

Linear combinations, absolute values and discrepancy sets are measurable. (Closure properties of measurable functions used by the integral)

Proof

technique · direct
1.1

Choose versions U=E[XG] and V=E[YG]. The function aU+bV is G-measurable and integrable. For every AG, A(aU+bV)=aAX+bAY=A(aX+bY). It is a version of the left side, so uniqueness proves linearity.

F1F2F3F5
2.1

If X0 almost surely, on A={U<0}G we have 0AX=AU0. Thus A(U)=0, and [F4] gives P(A)=0. For XY apply this to YX and use linearity; this proves order preservation.

step 1.1F1F4F5
3.1

The constant function c is integrable, G-measurable, and has its own event integrals, so E[cG]=c by uniqueness. Testing A=Ω in [F1] gives EU=EX. Finally XXX and steps 1.1–2.1 give E[XG]UE[XG]. Hence UE[XG] and EUEX.

step 1.1step 2.1F1F2
4.1

If W=YX>0 almost surely, let T=E[WG]0 almost surely by step 2.1. The event A={T=0} has AW=AT=0, so W1A=0 almost surely by [F4]. Since W>0 off a null set, this forces P(A)=0. Together with P(T<0)=0 this gives T>0 almost surely; linearity identifies T=VU.

step 1.1step 2.1F1F4

Source notes

Durrett Lemma 4.1.1 and Theorem 4.1.9(a)–(b), printed pp.206,210–211; van der Vaart Lemma 1.9(i),(iii),(iv), printed p.4. The strict almost-sure statement is derived by the zero-event argument, not attributed to a counterexample.

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Sources