Finite conditional expectation: computing E[X|G] for a partition-generated σ-algebra, including a zero-mass block

Goal and scope

On Ω={0,1,2,3,4} with masses 1/8, 3/8, 1/4, 1/4, 0 and X taking the values 4, 0, 2, 6, 10 in order, let G be generated by the partition {0,1}, {2,3}, {4}. From the two defining conditions of conditional expectation (G-measurable, same integral on every event of G), find all versions of E[X|G]: prove they must equal 1 and 4 on the two positive-mass blocks and may take any value on the zero-mass block {4} (determined only almost surely), and explain why pointwise conditioning, checking only the total expectation, and the unweighted mean are all wrong.

Works inside the MKO's scope: the complete finite outcomes, exact rational masses, X values and conditioning partition all belong to one probability space (assumption a1); the main computation uses positive-mass blocks, and versions are unique almost surely, while the zero-mass block retains pointwise freedom without a forced value (assumption a2). It does not treat existence of conditional expectation in general (non-discrete) spaces, filtrations or martingales, and does not change the canonical MKO or its English edition.

Givens and symbols

• prob_axioms
• finite_sum_rules
• set_operations_def
• frac_rules
• mul_inverse_nonzero

• G1 — Ω={0,1,2,3,4}; P({0})=1/8, P({1})=3/8, P({2})=1/4, P({3})=1/4, P({4})=0 (summing to 1); X(0)=4, X(1)=0, X(2)=2, X(3)=6, X(4)=10; all on the same space (MKO assumption a1).
• G2 — The conditioning partition B_1={0,1}, B_2={2,3}, B_3={4}; G=σ(B_1,B_2,B_3).
• G3 — Definition: Y is a version of E[X|G] if (i) Y is G-measurable and (ii) E[Y𝟏_A]=E[X𝟏_A] for every A∈G, where E[Z𝟏_A]=Σ_{ω∈A} Z(ω)P({ω}).

• \Omega,\ P — the sample space and the probability
• B_1,B_2,B_3 — the three blocks {0,1}, {2,3}, {4} of the conditioning partition
• \mathcal{G}=\sigma(B_1,B_2,B_3) — the σ-algebra generated by the partition
• E[Z\mathbf{1}_A] — the integral of Z over A, Σ_{ω∈A} Z(ω)P({ω})
• Y_c — a version of E[X|G] taking the value c on B_3

Rules and sources

• prob_axioms · prerequisite — Probability axioms: P(A)≥0; P(Ω)=1; the probability of a (countable) union of pairwise disjoint events is the sum of their probabilities. Hence P(∅)=0 and P(Aᶜ)=1−P(A).
• finite_sum_rules · prerequisite — Finite sums: Σ_{i=1}^{0} a_i=0 and Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}; Σ(a_i+b_i)=Σa_i+Σb_i; Σ c·a_i=c·Σa_i.
• set_operations_def · prerequisite — Set operations: x∈A∪B ⇔ x∈A or x∈B; x∈A∩B ⇔ x∈A and x∈B; x∈A∖B ⇔ x∈A and x∉B; in a universe U, Aᶜ=U∖A.
• frac_rules · prerequisite — Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0.
• mul_inverse_nonzero · prerequisite — Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined.
• numeric_evaluation · prerequisite — Exact evaluation of an explicit integer or rational expression; independently recomputable.
• image_preimage_def · prerequisite — Image and preimage: f(S)={f(a) ∣ a∈S}; f⁻¹(T)={a∈A ∣ f(a)∈T}. The preimage notation does not presuppose that f has an inverse.
• subset_def · prerequisite — Subset: A⊆B means every element of A is an element of B, i.e. ∀x (x∈A⇒x∈B).
• forall_elim · prerequisite — ∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds.
• distributive · prerequisite — Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu.
• mul_zero · prerequisite — Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0.
• and_intro_elim · prerequisite — Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q.
• iff_intro · prerequisite — Biconditional: P⇔Q holds exactly when both P⇒Q and Q⇒P hold.
• def_unfold · prerequisite — Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true.
• mko_condexp_def · cited — On the same finite probability space, E[X|G] must be G-measurable and have the same integral as X on every G-measurable event. When G is generated by a partition, its value is the weighted mean on each positive-mass block. On zero-mass blocks it is determined only almost surely; division by zero is not performed. · mko-conditional-expectation
• sigma_partition · prerequisite — The σ-algebra σ(𝒫) generated by a partition 𝒫 of a finite set Ω consists exactly of the unions of blocks (including the empty union ∅); it has 2^{|𝒫|} events.
• meas_def · prerequisite — On a finite Ω, a function Y is G-measurable exactly when Y^{-1}(S)∈G for every set S of reals.
• L_meas_const · lemma — Y is σ(𝒫)-measurable exactly when Y is constant on each block of 𝒫.
• L_block_reduce · lemma — For σ(𝒫), condition (ii) holds for every A∈σ(𝒫) exactly when it holds for every block.

Main derivation

\mathcal{G}=\sigma(B_1,B_2,B_3)

\mathcal{G}=\{\varnothing,B_1,B_2,B_3,B_1\cup B_2,B_1\cup B_3,B_2\cup B_3,\Omega\}

Y\ \mathcal{G}\text{-measurable}

Y=c_1\ \text{on}\ B_1,\ Y=c_2\ \text{on}\ B_2,\ Y=c_3\ \text{on}\ B_3

\forall A\in\mathcal{G}:\ E[Y\mathbf{1}_A]=E[X\mathbf{1}_A]

\forall j:\ E[Y\mathbf{1}_{B_j}]=E[X\mathbf{1}_{B_j}]

E[Y\mathbf{1}_{B}]=E[X\mathbf{1}_{B}],\ Y=c\ \text{on}\ B,\ P(B)>0

c=\frac{E[X\mathbf{1}_B]}{P(B)}

B_1=\{0,1\}

P(B_1)=\tfrac12,\ E[X\mathbf{1}_{B_1}]=\tfrac12,\ c_1=1

B_2=\{2,3\}

P(B_2)=\tfrac12,\ E[X\mathbf{1}_{B_2}]=2,\ c_2=4

B_3=\{4\},\ P(B_3)=0

c_3\cdot 0=10\cdot 0\ \text{for every}\ c_3\in\mathbb{R}

c_1=1,\ c_2=4,\ c_3\ \text{arbitrary}

E[X|\mathcal{G}]=Y_c,\ Y_c=\begin{cases}1&\text{on }B_1\\4&\text{on }B_2\\c&\text{on }B_3\end{cases},\ c\in\mathbb{R}

A=B_1\cup B_2,\ A=\Omega

E[Y_c\mathbf{1}_A]=E[X\mathbf{1}_A]=\tfrac52

Z=X

Z\ \text{not}\ \mathcal{G}\text{-measurable}

W\equiv\tfrac52

E[W]=E[X]\ \text{but}\ E[W\mathbf{1}_{B_1}]=\tfrac54\neq\tfrac12

c_1'=\tfrac{4+0}{2}=2

E[c_1'\mathbf{1}_{B_1}]=1\neq\tfrac12

\text{S08, S10 to S12}

E[X|\mathcal{G}]=1\ \text{on}\ B_1,\ 4\ \text{on}\ B_2,\ \text{arbitrary on the null block}\ B_3

Substeps and lemmas

Y=c_j\ \text{on}\ B_j\ (j=1,2,3)

Y^{-1}(S)=\bigcup_{j:\,c_j\in S}B_j\in\mathcal{G}

Y\ \mathcal{G}\text{-measurable},\ \omega\in B_j,\ y=Y(\omega)

Y=y\ \text{on all of}\ B_j

\forall A\in\mathcal{G}:\ E[Y\mathbf{1}_A]=E[X\mathbf{1}_A]

\forall j:\ E[Y\mathbf{1}_{B_j}]=E[X\mathbf{1}_{B_j}]

\forall j:\ E[Y\mathbf{1}_{B_j}]=E[X\mathbf{1}_{B_j}]

\forall A\in\mathcal{G}:\ E[Y\mathbf{1}_A]=E[X\mathbf{1}_A]

Reverse audits

• S13 · sufficient_goal_reduction · S08, S10, S11, S12 — Goal G: find all versions and refute the three wrong methods. Sufficient condition H: S08 together with S10 to S12; H⇒G by conjunction. H is proved forwards from G1 to G3; G is never used as a premise, and no division is done on the zero-mass block.
• S12 · premise_audit · S04, S05, numeric_evaluation — On a block B of positive probability, the weighted and plain means agree for every real-valued X on B if and only if each conditional point weight P({ω})/P(B) equals 1/|B|. Necessity follows by taking the point indicators, and sufficiency by substitution. Unequal weights may still give agreement for particular X; on B₁ here the plain mean is 2 and the weighted mean is 1.
• S11 · premise_audit · S03, S05, numeric_evaluation — One failing block suffices to refute; this shows that every block is still needed after the reduction of S03.
• S10 · premise_audit · S02, L_meas_const — Two different values inside one block suffice to refute measurability.
• S09 · premise_audit · S08, S03, numeric_evaluation — The numbers can be recomputed; this step does not claim to replace the argument of S03 for all 8 events.
• S08 · inverse_equivalence · S02, S03, S04, S07 — ⇒: every version must be some Y_c (S02, S04). ⇐: every Y_c satisfies (i) and (ii) (S02, S03, S05 to S07). Both directions have reasons.
• S07 · premise_audit · S02, mul_zero, mul_inverse_nonzero — The equivalence of S04 needs P(B)≠0, which fails here, so S04's formula cannot be reused; only u·0=0 is used.
• S06 · premise_audit · S04, G1, numeric_evaluation — Backward check 4·(1/2)=2=E[X𝟏_{B_2}].
• S05 · premise_audit · S04, G1, numeric_evaluation — Recomputable; backward check 1·(1/2)=1/2=E[X𝟏_{B_1}].
• S04 · inverse_equivalence · mul_inverse_nonzero, distributive — Multiplying by 1/P(B) and by P(B) undo each other, given P(B)≠0; so on a positive-mass block c is unique.
• S03.2 · premise_audit · S01, finite_sum_rules, set_operations_def — It needs the blocks to be disjoint, so that the sum splits without double counting.
• S03.1 · premise_audit · S01, forall_elim — It needs only that the blocks lie in G.
• S03 · inverse_equivalence · L_block_reduce — Both directions are proved in the sub-steps.
• S02.2 · premise_audit · meas_def, sigma_partition — The key is that blocks are disjoint: ω lies in exactly one block, so a union of blocks containing ω must contain all of B_j.
• S02.1 · premise_audit · meas_def, sigma_partition — Only one direction is proved here; the other is S02.2.
• S02 · inverse_equivalence · L_meas_const — Both directions of the lemma are proved in the sub-steps.
• S01 · inverse_equivalence · sigma_partition, set_operations_def — The 8 listed sets are exactly all unions of blocks, and sigma_partition says G consists of exactly these; the two descriptions determine each other.

Error controls and negative controls

Pointwise conditioning and division by zero mass

One MKO misconception is pointwise conditioning: the answer X itself takes the values 4 and 0 on B_1 and is not G-measurable (S10). Another is dividing on a zero-mass block: E[X𝟏_{B_3}]/P(B_3)=0/0 is undefined; the correct conclusion is that any constant works and versions are unique only almost surely (S07, S08).

Checking the total expectation is not enough

Another MKO misconception: declaring a full conditional expectation after checking only E[Y]=E[X]. The constant 5/2 matches the total expectation but integrates to 5/4≠1/2 over B_1 (S11). Condition (ii) must hold on every event of G, which after reduction still means every block (S03).

Weight by mass

On a block of positive probability, the value is E[X𝟏_B]/P(B); it cannot in general be replaced by the plain average. Agreement for every real-valued X on that block holds if and only if all conditional point weights equal 1/|B|. For a particular X the means may coincide even with unequal weights. On B₁ here the plain mean is 2 and the correct value is 1 (S12). No division is made on a zero-probability block, whose versions remain as in S07 and S08.

• Every value of X is known, so E[X|G]=X → reject — X is not constant on the block B_1, so it is not G-measurable (S10).
• On B_3, E[X|G]=E[X𝟏_{B_3}]/P(B_3)=0/0=0 → reject — 0/0 is undefined; every constant satisfies the condition, and the value is determined only almost surely (S07).
• W≡5/2 satisfies E[W]=E[X], so W=E[X|G] → reject — E[W𝟏_{B_1}]=5/4≠1/2 (S11).
• On B_1, E[X|G]=(4+0)/2=2 → reject — Masses must be used: 2·(1/2)=1≠1/2; the correct value is 1 (S05, S12).
• Conditional expectation is unique pointwise, so its value at outcome 4 must be X(4)=10 → reject — Y_0 (taking 0 on B_3) is also a version; versions are unique only almost surely (S08).

Result and limitations

The versions of E[X|G] are exactly the Y_c: on B_1={0,1} the mass-weighted mean (4·1/8+0·3/8)/(1/2)=1, on B_2={2,3} the value (2·1/4+6·1/4)/(1/2)=4, and on the zero-mass block B_3={4} any constant c, with no 0/0 division; all versions agree almost surely, and the total expectation 5/2 is preserved. The pointwise candidate Y=X is not G-measurable, although its integral equality itself holds. The other two candidates, W≡5/2 based only on the total expectation and the unweighted block mean, violate the integral condition on B_1.

• This guide treats only partition-generated σ-algebras; existence of conditional expectation on general probability spaces (for example via Radon–Nikodym) and filtrations and martingales (the MKO's next step) are outside its scope.

Verification and source binding

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• source commit: b28a5edbd46279b1db2a3f3f48bb3f447e18899f
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