Conditional Probability and a Model's Information Structure

Conceptual map

II-01.01Probability mappings
II-01.02Conditional means and variances
II-01.03Forms of independence
II-01.04Integration order and limits
II-01.05Changes in statistical experiments

1. Information enters through sigma-fields

A conditional expectation is indexed by the information revealed by the model. Marginal familiarity does not license independence after selection.

Let $(\Omega,\mathcal{F},P)$ be a probability space, let $$Z$$ be an integrable shock, and let $\mathcal{G}\subseteq \mathcal{F}$ represent observed information. A version of $\mathbb{E}[Z\mid\mathcal{G}]$ is a $\mathcal{G}$ -measurable random variable whose integral agrees with that of $$Z$$ on every event in $\mathcal{G}$ . Conditioning on a discrete variable $$S$$ is conditioning on $\sigma(S)$ .

Proposition 1 · Total expectation and total variance

For square-integrable $$Z$$ ,

\[\begin{aligned}E[Z]=E[E[Z|\mathcal{G}]], \\ \operatorname{Var}(Z)=E[\operatorname{Var}(Z|\mathcal{G})]+\operatorname{Var}(E[Z|\mathcal{G}]).\end{aligned}\](1)

Proof. The first identity follows by applying the defining integral equality to $\Omega$ . For the second, write $Z-\mathbb{E}Z=[Z-\mathbb{E}(Z\mid\mathcal{G})]+[\mathbb{E}(Z\mid\mathcal{G})-\mathbb{E}Z]$ . The cross term has conditional expectation zero. ∎

2. A selected sample changes a symmetric shock

Let $$Z$$ equal $$-1$$ or 1 with probability one half. A sample indicator $$S$$ satisfies $$P(S=1|Z=1)=3/4$$ and $$P(S=1|Z=-1)=1/4$$ . The joint table is

\[P(1,1)=\frac{3}{8}, P(1,0)=\frac{1}{8}, P(-1,1)=\frac{1}{8}, P(-1,0)=\frac{3}{8}.\](2)

Thus $$P(S=1)=1/2$$ , $\mathbb{E}[Z\mid S=1]=1/2$ , and $\mathbb{E}[Z\mid S=0]=-1/2$ . Each conditional variance is 3/4. Equation (1) gives 0=(1/2)(1/2)+(1/2)( $$-1/2$$ ) and 1=3/4+1/4.

The product expectation $\mathbb{E}[ZS]=1/4$ differs from $\mathbb{E}[Z]\mathbb{E}[S]=0$ . The marginal symmetry of $$Z$$ survives; independence between $$Z$$ and selection does not.

3. Restrictions support different algebra

Independence of $$X$$ and $$Y$$ gives $\mathbb{E}[f(X)g(Y)]=\mathbb{E}[f(X)]\mathbb{E}[g(Y)]$ for integrable products. Mean independence $\mathbb{E}[X\mid Y]=\mathbb{E}[X]$ supports $\mathbb{E}[XY]=\mathbb{E}[X]\mathbb{E}[Y]$ when integrability holds, while leaving nonlinear transformations unfactored. Conditional independence given $$H$$ factors conditional laws and conditional product moments, followed by integration over $$H$$ .

Every factorization should name its level. A structural shock independent of treatment assignment before selection can become dependent on treatment inside a selected or post-treatment sample.

4. Zero-probability and endogenous conditioning events

Failure case · Conditional cell treated as an unconditional population

The selected mean in Section 2 equals one half even though the population mean is zero. Substituting the marginal shock law inside the selected cell erases the selection rule that defines that cell.

For an event $$A$$ with $$P(A)=0$$ , the ratio $\mathbb{E}[Z1_{A}]/P(A)$ is undefined. Regular conditional distributions can define versions at continuously indexed values only up to almost-sure equivalence. Claims at support endpoints or off-support conditioning values require a chosen version, a limiting argument, or additional continuity.

5. Implementation, exercises, and sources

Begin with a normalized joint table or density. Derive every marginal by summation or integration, discard no cell before its selection role is recorded, and check probabilities sum to one. For simulated tables, report rare-cell counts and treat empty empirical cells separately from population zero-probability events.

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Exercises

Reproduce all conditional means and variances in Section 2.
Replace the two selection probabilities by $$a$$ and $$b$$ and derive $\mathbb{E}[Z\mid S=1]$ .
Construct mean-independent variables that are dependent and show a nonlinear product factorization that fails.

Partial solutions

1. Within $$S=1$$ , the positive and negative probabilities are 3/4 and 1/4, so the mean is 1/2 and the second moment is one. 2. Provided $$a+b>0$$ , the selected mean is $$(a-b)/(a+b)$$ . 3. Let $$X$$ be symmetric with three or more support points and set $Y=X^{2}$ ; suitable centering can give $\mathbb{E}[X\mid Y]=0$ while dependence remains.

Olav Kallenberg, Foundations of Modern Probability, chapters on conditional expectation and disintegration.Measure-theoretic conditioning and versions.
Rick Durrett, Probability: Theory and Examples, sections on conditional expectation.Tower property and variance decomposition.
James J. Heckman (1979), “Sample Selection Bias as a Specification Error,” Econometrica 47, 153–161.Economic consequences of nonrandom selection.

6. Audit checkpoint

State the probability space, generated information, joint law, support, conditioning event, its probability, selection timing, independence level, integrability, version at continuous conditioning values, and every normalization check.

7. Scope boundary

The chapter covers the conditional-expectation identities used in finite and dominated models. General measure extension and abstract disintegration theory remain outside the computational treatment.

Prerequisites

I-01 · Define the Research Object as a Contract