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Mediation and cross-world assumptions — Examples

1 · Prerequisites

2 · Summary

These finite examples separate the assumptions that a mediation calculation needs. Two treatment-outcome and mediator-outcome models can share the same single-world observed law while assigning different natural effects when the cross-world coupling changes. A mediator-outcome common cause and a treatment-induced mediator-outcome confounder each give a separate exact calculation where the observed conditional-mean functional differs from the true nested counterfactual means. The examples do not treat a fitted regression coefficient as an identified natural effect.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Single-world assumptions do not identify natural effects

Statement refuted

The observed law of $(A, M, Y)$ identifies the natural direct and indirect effects whenever treatment is randomized, consistency and treatment/mediator positivity hold, and the same-world mediator–outcome independences $Y (a, m) ⊥ ⁣ ⁣ ⁣ ⊥ M (a)$ hold for each $a, m$ , even without a cross-world restriction.

Facts & Assumptions

Given: The definitions of the natural effects, the single-world conditions in the statement, and the definition of identifiability.

[F1]

$NDE (a, a^{'}) = E [Y (a, M (a^{'})) - Y (a^{'}, M (a^{'}))]$ and $NIE (a, a^{'}) = E [Y (a, M (a)) - Y (a, M (a^{'}))]$ . Controlled and natural mediation effects

[F2]

Identifiability requires the target to be constant over all full-data laws with the same observed-data law. Identifiability of an estimand

[F3]

The standard formula's additional cross-world restriction is $Y (a, m) ⊥ ⁣ ⁣ ⁣ ⊥ M (a^{'}) ∣ σ (L)$ , which is not one of the single-world assumptions stated above. Assumptions for identifying natural effects

Counterexample

Given: Binary treatment $A$ , mediator $M$ , and outcome $Y$ .

1.1givenconstruct

In model I, let $A, M (0), M (1), V$ be independent fair Bernoulli variables. In model II, let $A, M (1), V$ be independent fair Bernoulli variables and set $M (0) = V$ . In both models define $Y (0, m) = 0$ and $Y (1, m) = 1_{{V = m}}$ for $m \in {0, 1}$ . Potential responses define observed $M = M (A)$ and $Y = Y (A, M (A))$ , so consistency holds; treatment is randomized and both mediator values have probability $1 / 2$ in each treatment arm.

2.1step 1.1algebra

Under $A = 0$ , $Y = 0$ and $M (0)$ is fair in both models. Under $A = 1$ , $M (1)$ is fair and independent of $V$ in both models, so $Y = 1_{{V = M (1)}}$ is fair and independent of $M (1)$ . Thus the conditional observed laws of $(M, Y)$ agree in each treatment arm, and the randomized treatment law is the same; the two models have identical observed laws of $(A, M, Y)$ .

2.2F3step 1.1algebra

The same-world independence holds in both models: for $a = 0$ the outcome response $Y (0, m)$ is constant; for $a = 1$ , $Y (1, m)$ is a function of $V$ and $M (1)$ is independent of $V$ . In model I, $M (0)$ is independent of every $Y (1, m)$ ; in model II, $M (0) = V$ makes $Y (1, m)$ dependent on $M (0)$ , so the cross-world restriction in [F3] differs.

2.3F1step 1.1algebra

In model I, $E [Y (1, M (0))] = 1 / 2$ and $E [Y (1, M (1))] = 1 / 2$ ; in model II, $E [Y (1, M (0))] = 1$ and $E [Y (1, M (1))] = 1 / 2$ . Both models have $E [Y (0, M (0))] = 0$ . By [F1], the natural-effect pairs are $(1 / 2, 0)$ in model I and $(1, - 1 / 2)$ in model II, in the order (NDE, NIE).

3.1F2step 2.1step 2.3∎

The two compatible full-data laws have the same observed law but different natural effects by step 2.3. Therefore the target is not constant on the observational equivalence class in [F2], refuting the stated claim.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Mediator confounding changes the natural-effect functional

Example

Let $A$ be randomized with probabilities $1 / 2, 1 / 2$ , independently of $U \sim Bernoulli (1 / 5)$ . Set $M = A x o r U, Y = A + U .$ The mediator and outcome share the unmeasured cause $U$ . The true nested responses are $M (a) = a x o r U$ and $Y (a, m) = a + U$ ; the outcome does not respond to an intervention on $M$ . Both treatment arms have positive probability for each mediator value.

Facts & Assumptions

Given: The definitions of natural effects and the observed-data functional in the mediation formula.

[F1]

$NDE (a, a^{'}) = E [Y (a, M (a^{'})) - Y (a^{'}, M (a^{'}))]$ and $NIE (a, a^{'}) = E [Y (a, M (a)) - Y (a, M (a^{'}))]$ . Controlled and natural mediation effects

[F2]

The mediation functional uses the observed conditional outcome mean $μ_{a} (m) = E [Y ∣ A = a, M = m]$ averaged over the mediator law at the reference treatment; it identifies the nested mean only under the stated assumptions. The natural-effect mediation formula under independent errors

[F3]

The natural-effect formula requires no unmeasured mediator–outcome confounding and the cross-world restriction, as well as positivity. Assumptions for identifying natural effects

Verification

Given: The full-data construction in the Example and treatment levels $a = 1$ , $a^{'} = 0$ .

1.1givenF3algebra

Consistency holds by construction. For each $a$ , $M (a)$ takes both values with positive probability because $0 < P (U = 1) < 1$ ; treatment is randomized and both treatment levels have positive probability. However, $U$ affects both $M (a)$ and $Y (a, m)$ , so the no-unmeasured mediator–outcome-confounding condition in [F3] fails.

1.2F1algebra

The true nested means are $E [Y (1, M (0))] = E [1 + U] = 6 / 5$ , $E [Y (1, M (1))] = 6 / 5$ , and $E [Y (0, M (0))] = E [U] = 1 / 5$ . Thus [F1] gives $NDE (1, 0) = 1$ and $NIE (1, 0) = 0$ .

1.3F2givenalgebra

The observed mediator probabilities are $Q_{0} (1) = 1 / 5$ and $Q_{0} (0) = 4 / 5$ . At $A = 0$ , $M = U$ and $Y = U$ , so $μ_{0} (0) = 0$ , $μ_{0} (1) = 1$ . At $A = 1$ , $M = 1 - U$ and $Y = 1 + U$ , so $μ_{1} (0) = 2$ , $μ_{1} (1) = 1$ . The observed-data functional therefore gives $\tilde{θ} (1, 0) = (4 / 5) 2 + (1 / 5) 1 = 9 / 5$ , while $\tilde{θ} (0, 0) = 1 / 5$ and $\tilde{θ} (1, 1) = 6 / 5$ .

2.1F2step 1.1step 1.2step 1.3algebra∎

Its resulting contrasts are $\overset{}{NDE} (1, 0) = 8 / 5$ and $\overset{}{NIE} (1, 0) = - 3 / 5$ , different from the true values in step 1.2. This computes the formula's observed-data functional but does not apply the identification assertion, whose assumptions fail by step 1.1.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A treatment-induced mediator–outcome confounder breaks the formula

Example

Let $A \sim Bernoulli (1 / 2)$ , $U \sim Bernoulli (1 / 2)$ , and $W \sim Bernoulli (1 / 5)$ be mutually independent. Define the post-treatment variable $Z = A U$ , mediator $M = Z x o r W$ , and outcome $Y = Z + M$ . Thus $Z$ is caused by treatment and causes both $M$ and $Y$ ; it is a treatment-induced mediator–outcome confounder. Treatment and mediator positivity hold for both treatment levels and mediator values.

Facts & Assumptions

Given: The definitions of natural effects and the observed-data functional in the mediation formula.

[F1]

$NDE (a, a^{'}) = E [Y (a, M (a^{'})) - Y (a^{'}, M (a^{'}))]$ and $NIE (a, a^{'}) = E [Y (a, M (a)) - Y (a, M (a^{'}))]$ . Controlled and natural mediation effects

[F2]

The mediation functional averages the observed conditional outcome mean under one treatment over the mediator law under the reference treatment. The natural-effect mediation formula under independent errors

[F3]

The standard formula excludes a treatment-induced post-treatment common cause of mediator and outcome. Assumptions for identifying natural effects

Verification

Given: The full-data construction in the Example and treatment levels $a = 1$ , $a^{'} = 0$ .

1.1F1givenalgebra

Under interventions, $Z (a) = a U$ , $M (a) = a U x o r W$ , and $Y (a, m) = a U + m$ . Hence the true nested means are $E [Y (1, M (0))] = E [U + W] = 7 / 10$ , $E [Y (1, M (1))] = E [U] + E [M (1)] = 1$ , and $E [Y (0, M (0))] = E [W] = 1 / 5$ . By [F1], the true natural effects are $NDE (1, 0) = 1 / 2$ and $NIE (1, 0) = 3 / 10$ .

1.2givenalgebra

At $A = 0$ , $M = W$ and $Y = M$ , so $Q_{0} (1) = 1 / 5$ and $μ_{0} (m) = m$ . At $A = 1$ , $M = U x o r W$ is fair. The joint probabilities $P (U = 1, M = 1 ∣ A = 1) = 2 / 5$ and $P (U = 1, M = 0 ∣ A = 1) = 1 / 10$ give $μ_{1} (1) = 9 / 5$ and $μ_{1} (0) = 1 / 5$ . Thus $Q_{1} (0) = Q_{1} (1) = 1 / 2$ .

2.1F2F3step 1.1step 1.2algebra∎

The observed-data functional gives $\tilde{θ} (1, 0) = (1 / 5) (9 / 5) + (4 / 5) (1 / 5) = 13 / 25$ , $\tilde{θ} (1, 1) = (1 / 2) (9 / 5) + (1 / 2) (1 / 5) = 1$ , and $\tilde{θ} (0, 0) = 1 / 5$ . Its direct and indirect contrasts are therefore $8 / 25$ and $12 / 25$ , not the true $1 / 2$ and $3 / 10$ from step 1.1. The functional is computable, but the identification theorem does not apply because the post-treatment common cause $Z$ violates [F3].