Law of total expectation

Mathematical statement

Let G be the sigma-algebra generated by a complete partition. The tower identity is E(E(X|G))=E(X). Cells are pairwise disjoint and cover the sample space. Positive-mass cells contribute P(B_i)E[X|B_i]. Conditional expectation is determined only almost surely: a null-cell quotient is undefined, but any integrable version there has zero weighted contribution. Do not divide by zero. No independence is required, and this is not Bayes formula.

Representative formula (exact TeX)

E(E(X\mid G))=E(X)

Assumptions

• The normalized finite law, integrable variable X and complete partition are explicit; G is generated by that partition.
• Do not compute a conditional quotient on zero-mass cells.

Symbol scope

• E — expectation and conditional-expectation operator
• X — integrable random variable
• G — partition-generated sigma-algebra

Intuition, definition, and misconceptions

Check partition completeness and disjointness first.

Compute exact conditional means on positive-mass cells.

Null cells contribute zero and the tower expectation is preserved.

Do not divide on a zero-mass cell or demand independence of partition events.

Adopted derivation guide

The original MKO is retained. The guide has its own append-only revision and verification boundaries, not a new Lean proof.

Prerequisites

• Finite conditional expectation — Cell conditional means are required.
• Expectation in finite models — Use the global mean under the same law.
• Event — Cells are events of the partition.

Verification boundary

This slice uses complete finite models; the general sigma-algebra tower identity needs a separate measure-theoretic proof.

not_formalized

This does not mean the content is false; it means no replayable computational or formal evidence is registered yet.

Next obligation

Add the law of total variance.

Same-source structured data

This page and the interactive view use the same object and existing language edition; this adds no proof or human review.