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Basic algebra and order properties of conditional expectation

Statement

Assume AC for existence. For real $X, Y \in L^{1} (P)$ and $a, b \in R$ , $E [a X + b Y ∣ G] = a E [X ∣ G] + b E [Y ∣ G]$ . Conditional expectation is positive, preserves order and constants, satisfies $E (E [X ∣ G]) = E X$ , and $∣ E [X ∣ G] ∣ \leq E [∣ X ∣ ∣ G]$ almost surely. Also $X < Y$ almost surely implies $E [X ∣ G] < E [Y ∣ G]$ 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 $L^{1} (μ)$ )

[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, positive and negative parts, absolute values and discrepancy sets are measurable; restricting a nonnegative measurable function to a measurable event preserves measurability. (Closure properties of measurable functions used by the integral)

[F6]

Almost-everywhere equal integrable functions have equal integrals on every measurable event. (Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree)

Proof

technique · direct

1.1F1F2F3F5

Choose versions $U = E [X ∣ G]$ and $V = E [Y ∣ G]$ . The function $a U + b V$ is $G$ -measurable and integrable. For every $A \in G$ , $\int_{A} (a U + b V) = a \int_{A} X + b \int_{A} Y = \int_{A} (a X + b Y)$ . It is a version of the left side, so uniqueness proves linearity.

2.1step 1.1F1F3F4F5F6

If $X \geq 0$ almost surely, its positive part $X^{+}$ is measurable by [F5], integrable because $X^{+} \leq ∣ X ∣$ , and equals $X$ almost surely. Set $A = {U < 0} \in G$ . By [F6] and [F1], $0 \leq \int_{A} X^{+} = \int_{A} X = \int_{A} U$ . The measurable integrable function $U^{-} = \max {- U, 0}$ equals $- U$ on $A$ and zero off $A$ , so [F3] gives $\int_{A} U = - \int U^{-} \leq 0$ . Thus $\int U^{-} = 0$ ; [F4] gives $U^{-} = 0$ almost surely and hence $P (A) = 0$ . For $X \leq Y$ almost surely apply this positivity result to $Y - X$ and use step 1.1, proving order preservation.

3.1step 1.1step 2.1F1F2

The constant function $c$ is integrable, $G$ -measurable, and has its own event integrals, so $E [c ∣ G] = c$ by uniqueness. Testing $A = Ω$ in [F1] gives $E U = E X$ . Finally $- ∣ X ∣ \leq X \leq ∣ X ∣$ and steps 1.1–2.1 give $- E [∣ X ∣ ∣ G] \leq U \leq E [∣ X ∣ ∣ G]$ . Hence $∣ U ∣ \leq E [∣ X ∣ ∣ G]$ and $E ∣ U ∣ \leq E ∣ X ∣$ .

4.1step 1.1step 2.1F1F4F5F6∎

If $W = Y - X > 0$ almost surely, let $T = E [W ∣ G] \geq 0$ almost surely by step 2.1. On $A = {T = 0} \in G$ , [F1] gives $\int_{A} W = \int_{A} T = 0$ . The positive part $W^{+}$ is measurable and integrable, and $W^{+} = W$ almost surely; [F6] therefore gives $\int W^{+} 1_{A} = \int_{A} W = 0$ . The integrand is pointwise nonnegative and measurable by [F5], so [F4] gives $W^{+} 1_{A} = 0$ almost surely. Since $W^{+} > 0$ off a null set, this forces $P (A) = 0$ . Together with $P (T < 0) = 0$ this gives $T > 0$ almost surely; step 1.1 identifies $T = V - U$ almost surely.

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.

Depends on

Used by

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Conditional treatment-indicator factorization Lemma
Conditional variance is well-defined and has the second-moment formula Lemma
Conditional-independence factorization for integrable outcomes Lemma
Conditioning a known state and independent noise Lemma
Convex SCAFFOLD client-drift bound Lemma
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Doob upcrossing inequality Lemma
Finite-horizon Brownian change of drift and relative entropy Lemma
Kernel conditional expectations for square-integrable Markov observables Lemma
Martingale differences are orthogonal in l2 Lemma

…and 68 more results.

Dependency tree · two levels

24 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

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