ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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Conditional expectation given a discrete random variable

Example

Assume AC for conditional classes. If Y is a countably valued real random variable and X is real integrable, a version of $E [X ∣ σ (Y)]$ takes value $c_{y} = \int_{{Y = y}} X d P / P (Y = y)$ on each positive-mass fibre and zero on all zero-mass fibres.

Facts & Assumptions

Given: A countably valued real random variable Y and real integrable X on a probability space; AC is the conditional-class convention.

[F1]

The measurable integrable event-identity characterization defines the conditional class. (Conditional expectation as an ae class)

[F2]

Versions are unique almost surely. (Conditional expectation is unique almost surely)

[F3]

Increasing nonnegative partial sums pass through the integral. (Monotone convergence for the integral)

[F4]

Finite weights summing to one define a probability space. (Finite probability spaces are exactly finite full-power-set probability spaces)

Verification

technique · direct

1.1

List the at most countably many fibres $A_{y} = {Y = y}$ . Every union of fibres is measurable as a countable union, and these unions are exactly $σ (Y)$ : each fibre is a preimage of a singleton Borel set, and every preimage is a union of fibres. The proposed function T is thus $σ (Y)$ -measurable. Its absolute integral is $\sum_{y} ∣ c_{y} ∣ P (A_{y}) \leq \sum_{y} \int_{A_{y}} ∣ X ∣ = E ∣ X ∣$ , where [F3] applies to nonnegative finite partial sums. Null fibres have zero X integral and their countable union is null.

2.1

On each positive fibre $\int_{A_{y}} T = c_{y} P (A_{y}) = \int_{A_{y}} X$ , and on null fibres both sides are zero. For every union A of fibres, sum these identities; absolute summability follows from step 1.1 and integrability of X, with [F3] applied to positive and negative parts. Thus $\int_{A} T = \int_{A} X$ . By [F1]–[F2] T is the desired version.

step 1.1F1F2F3

3.1

For a concrete instance take atoms a,b,c with masses $1 / 4, 1 / 4, 1 / 2$ by [F4], with Y values (0,0,2) and X values (2,6,10). The zero fibre has mass $1 / 2$ and X integral $2 / 4 + 6 / 4 = 2$ , so its conditional value is 4. The fibre at 2 has mass $1 / 2$ and X integral 5, giving value 10. Hence T=(4,4,10), with mean $4 / 4 + 4 / 4 + 10 / 2 = 7 = E X$ .

step 2.1F4∎

Source notes

Durrett Example 4.1.5, printed p.208; van der Vaart Example 1.7 and its countable-partition extension, printed p.3. The countable sum is justified using ordinary MCT on positive and negative parts.

Depends on

Used by

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Sources

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