Day 58: Mediation: Direct & Indirect Effects

Theory

Nested potential outcomes

Let $X$ be the treatment, $M$ a mediator and $Y$ the outcome. Write $M(x)$ for the mediator under treatment $x$ , and $Y(x, m)$ for the outcome when $X$ is set to $x$ and $M$ to $m$ . The observed outcome under treatment $x$ is $Y(x) = Y(x, M(x))$ .

Four effects (binary $X$ )

Total effect: $TE = E[Y(1, M(1)) - Y(0, M(0))]$ .
Controlled direct effect at level $m$ : $CDE(m) = E[Y(1, m) - Y(0, m)]$ . Fix the mediator for everyone.
Natural direct effect: $NDE = E[Y(1, M(0)) - Y(0, M(0))]$ . Change the treatment but keep each unit's mediator at the value it would have had without treatment.
Natural indirect effect: $NIE = E[Y(1, M(1)) - Y(1, M(0))]$ . Keep the treatment on and shift only the mediator.

They add up: $TE = NDE + NIE$ . The proportion mediated is $NIE/TE$ .

Identification: the mediation formula

Assume, possibly within levels of pre-treatment covariates $C$ : no unmeasured confounding of (1) $X$ and $Y$ , (2) $X$ and $M$ , (3) $M$ and $Y$ , and (4) no confounder of $M$ and $Y$ that is itself affected by $X$ . Then (Pearl's mediation formula)

NDE = \sum_m \big(E[Y \mid X = 1, m] - E[Y \mid X = 0, m]\big)\,P(m \mid X = 0),

NIE = \sum_m E[Y \mid X = 1, m]\,\big(P(m \mid X = 1) - P(m \mid X = 0)\big).

$CDE(m) = E[Y \mid X = 1, m] - E[Y \mid X = 0, m]$ needs only assumptions (1) and (3) (and can be identified even when (4) fails, by the g-formula).

Linear models with interaction

If $E[M \mid x] = \beta_0 + \beta_1 x$ and $E[Y \mid x, m] = \theta_0 + \theta_1 x + \theta_2 m + \theta_3 xm$ :

CDE(m) = \theta_1 + \theta_3 m, \qquad NDE = \theta_1 + \theta_3\beta_0, \qquad NIE = (\theta_2 + \theta_3)\beta_1.

Without interaction ( $\theta_3 = 0$ ): $NDE = CDE = \theta_1$ and $NIE = \theta_2\beta_1$ , the "product of coefficients" of Baron and Kenny.

The trap

"Regress $Y$ on $X$ and $M$ and call the $X$ coefficient the direct effect" is right only under assumption (3). If something unobserved affects both $M$ and $Y$ , $M$ is a collider on $X \to M \leftarrow U \to Y$ , and the coefficient is biased (Day 53, Exercise 9); it can even flip sign.

Part 1 · Warm-up

Exercise 1 · Read the potential outcomes

In words, what is $Y(1, M(0))$ for a jobseeker in a training programme where $M$ is the number of job applications sent?

Show solution

Step 1. First argument: the jobseeker receives training.

Step 2. Second argument: applications are held at the number they would have sent without training.

Answer: earnings with training but without training's effect on job search, so only the skill (direct) channel operates. It is a cross-world quantity: never observed for anyone.

Exercise 2 · Linear, no interaction

$M = 3 + 0.5X + e_M$ and $Y = 2 + 1.0X + 2.0M + e_Y$ , with no confounding. Find $NDE$ , $NIE$ and $TE$ .

Show solution

Step 1. $NDE = \theta_1 = 1.0$ .

Step 2. $NIE = \theta_2\beta_1 = 2.0 \times 0.5 = 1.0$ .

Step 3. $TE = 1.0 + 1.0 = 2.0$ .

Exercise 3 · Linear with interaction

$E[M \mid x] = 1 + 0.5x$ and $E[Y \mid x, m] = \theta_0 + 1.0x + 2.0m + 0.5xm$ . Compute $CDE(0)$ , $CDE(2)$ , $NDE$ , $NIE$ and $TE$ .

Show solution

Step 1. $CDE(0) = 1.0$ ; $CDE(2) = 1.0 + 0.5(2) = 2.0$ .

Step 2. $NDE = \theta_1 + \theta_3\beta_0 = 1.0 + 0.5(1) = 1.5$ .

Step 3. $NIE = (\theta_2 + \theta_3)\beta_1 = 2.5 \times 0.5 = 1.25$ .

Step 4. $TE = 1.5 + 1.25 = 2.75$ . Check directly: $E[Y \mid \operatorname{do}(1)] - E[Y \mid \operatorname{do}(0)] = [1.0 + 2.5(1.5)] - [2.0(1)] = 4.75 - 2 = 2.75$ .

Exercise 4 · Proportion mediated

In Exercise 2, what share of the effect runs through $M$ ? In Exercise 3?

Show solution

Step 1. Exercise 2: $1.0/2.0 = 50\%$ .

Step 2. Exercise 3: $1.25/2.75 \approx 45\%$ .

Exercise 5 · Which effect answers which question?

Match: (a) "What would the effect of training be if we also guaranteed every trainee exactly 10 applications?" (b) "How much of training's effect is due to its effect on job search?" (c) "Does training work at all?"

Show solution

Step 1. (a) $CDE(10)$ : the mediator is fixed by policy.

Step 2. (b) $NIE$ (or the proportion mediated): a mechanism question.

Step 3. (c) $TE$ .

Part 2 · Proofs

Exercise 6 · The decomposition

Prove $TE = NDE + NIE$ .

Show solution

Step 1. Add and subtract $Y(1, M(0))$ : $Y(1, M(1)) - Y(0, M(0)) = [Y(1, M(1)) - Y(1, M(0))] + [Y(1, M(0)) - Y(0, M(0))]$ .

Step 2. Take expectations: the first bracket is $NIE$ , the second $NDE$ . $\blacksquare$

Exercise 7 · Deriving the mediation formula for the NDE

Assume (a) $Y(x, m) \perp\!\!\!\perp X$ , (b) $M(x) \perp\!\!\!\perp X$ , (c) $Y(x, m) \perp\!\!\!\perp M \mid X$ , and (d) $Y(x, m) \perp\!\!\!\perp M(x')$ (the cross-world assumption). Prove the NDE formula.

Show solution

Step 1. $E[Y(1, M(0))] = \sum_m E[Y(1, m) \mid M(0) = m]\,P(M(0) = m)$ .

Step 2. By (d), $E[Y(1, m) \mid M(0) = m] = E[Y(1, m)]$ .

Step 3. By (a) and (c), $E[Y(1, m)] = E[Y(1, m) \mid X = 1] = E[Y(1, m) \mid X = 1, M = m] = E[Y \mid X = 1, M = m]$ (consistency).

Step 4. By (b), $P(M(0) = m) = P(M(0) = m \mid X = 0) = P(m \mid X = 0)$ .

Step 5. So $E[Y(1, M(0))] = \sum_m E[Y \mid 1, m]P(m \mid 0)$ ; similarly $E[Y(0, M(0))] = \sum_m E[Y \mid 0, m]P(m \mid 0)$ . Subtract. $\blacksquare$

Exercise 8 · Linear formulas

With $E[M \mid x] = \beta_0 + \beta_1 x$ and $E[Y \mid x, m] = \theta_0 + \theta_1 x + \theta_2 m + \theta_3 xm$ , derive $NDE$ and $NIE$ from the mediation formula.

Show solution

Step 1. $E[Y \mid 1, m] - E[Y \mid 0, m] = \theta_1 + \theta_3 m$ .

Step 2. $NDE = \sum_m(\theta_1 + \theta_3 m)P(m \mid 0) = \theta_1 + \theta_3 E[M \mid X = 0] = \theta_1 + \theta_3\beta_0$ .

Step 3. $NIE = \sum_m(\theta_0 + \theta_1 + (\theta_2 + \theta_3)m)\big(P(m \mid 1) - P(m \mid 0)\big)$ .

Step 4. Constants sum to zero against a difference of probabilities; the $m$ term gives $(\theta_2 + \theta_3)(E[M \mid 1] - E[M \mid 0]) = (\theta_2 + \theta_3)\beta_1$ . $\blacksquare$

Exercise 9 · No interaction, one direct effect

Show that if $E[Y \mid x, m]$ has no $x \times m$ interaction, then $CDE(m) = NDE$ for every $m$ .

Show solution

Step 1. No interaction: $E[Y \mid 1, m] - E[Y \mid 0, m] = c$ , a constant.

Step 2. $CDE(m) = c$ for all $m$ .

Step 3. $NDE = \sum_m c\,P(m \mid 0) = c$ . $\blacksquare$

Exercise 10 · A treatment-induced mediator–outcome confounder

Let $X \to L \to M \to Y$ , $L \to Y$ , $X \to M$ , $X \to Y$ (all observed). Explain why $CDE(m)$ is identified but the $NDE$ is not, without further assumptions.

Show solution

Step 1. $CDE(m) = E[Y \mid \operatorname{do}(x, m)]$ is a joint intervention; with $X$ unconfounded and $\{X, L\}$ blocking the back-door paths of $M$ , the g-formula gives $E[Y \mid \operatorname{do}(x, m)] = \sum_l E[Y \mid x, l, m]P(l \mid x)$ (Day 55, Exercise 13).

Step 2. The NDE needs $E[Y(1, M(0))]$ , which involves $L(1)$ (through the direct path $X \to L \to Y$ ) jointly with $M(0)$ , which depends on $L(0)$ .

Step 3. The joint distribution of $L(1)$ and $L(0)$ for the same unit is never observed and is not fixed by any observable distribution (a "recanting witness").

Answer: CDEs are identified; natural effects need extra assumptions or are reported with bounds. $\blacksquare$

Part 3 · Applied

Exercise 11 · Training, job finding and earnings

A training programme $X$ may raise monthly earnings $Y$ ($100s) directly (skills) and by speeding up job finding $M$ (found a job within 3 months). With no confounding: $P(M = 1 \mid X = 0) = 0.4$ , $P(M = 1 \mid X = 1) = 0.6$ ; $E[Y \mid X, M]$ : $(0, 0)$ : 10, $(0, 1)$ : 20, $(1, 0)$ : 12, $(1, 1)$ : 24. Compute $TE$ , $NDE$ , $NIE$ , $CDE(0)$ , $CDE(1)$ and the proportion mediated.

Show solution

Step 1. $E[Y \mid X = 1] = 0.4(12) + 0.6(24) = 19.2$ ; $E[Y \mid X = 0] = 0.6(10) + 0.4(20) = 14$ . $TE = 5.2$ .

Step 2. $NDE = (12 - 10)(0.6) + (24 - 20)(0.4) = 1.2 + 1.6 = 2.8$ .

Step 3. $NIE = 24(0.6 - 0.4) + 12(0.4 - 0.6) = 4.8 - 2.4 = 2.4$ .

Step 4. Check: $2.8 + 2.4 = 5.2$ . $CDE(0) = 2$ , $CDE(1) = 4$ .

Answer: about 46% of the effect ( $2.4/5.2$ ) runs through faster job finding. The direct effect depends on the mediator level (2 or 4), so $NDE$ is a weighted average of the CDEs.

Exercise 12 · Cash transfers, food spending and child growth

A cash transfer $X$ raises monthly food spending $M$ ($): $E[M \mid x] = 50 + 20x$ . Child height-for-age z-scores follow $E[Y \mid x, m] = \theta_0 + 0.10x + 0.004m + 0.002xm$ . Assume no unmeasured confounding. Compute the effects and the proportion mediated by food spending.

Show solution

Step 1. $NDE = 0.10 + 0.002 \times 50 = 0.20$ .

Step 2. $NIE = (0.004 + 0.002) \times 20 = 0.12$ .

Step 3. $TE = 0.32$ ; proportion mediated $0.12/0.32 = 37.5\%$ .

Answer: most of the effect runs through other channels (health care, schooling, reduced stress); the interaction means that food spending matters more for treated households.

Exercise 13 · Smoking, birth weight and infant mortality

Return to Day 48's paradox in mediation form. In a linear model, $M$ = birth weight (standardised), $U$ = birth defects (unobserved): $M = -X - U + e_M$ , $Y = 0.2X - 0.5M + 1.0U + e_Y$ , with $\operatorname{Var}(U) = \operatorname{Var}(e_M) = 1$ . What is the true direct effect of smoking, and what does the regression of mortality on smoking and birth weight report?

Show solution

Step 1. True direct effect: $\theta_1 = 0.2$ (harmful).

Step 2. Given $X$ and $M$ , $-U + e_M = M + X$ is known, so $E[U \mid X, M] = -k(M + X)$ with $k = 1/(1 + 1) = 0.5$ .

Step 3. $E[Y \mid X, M] = 0.2X - 0.5M + 1.0 \times (-0.5)(M + X) = -0.3X - 1.0M$ .

Answer: the regression reports $-0.3$ : smoking looks protective holding birth weight fixed, because birth weight is a collider of smoking and defects. The simulation confirms −0.3.

Exercise 14 · Occupation and the gender pay gap

A study regresses wages on gender, "controlling for occupation", and calls the remaining gap "discrimination" (a direct effect). Draw the DAG with gender $G$ , occupation $O$ , wage $W$ and unobserved ability or preferences $A$ that affect both occupation and wages. What is wrong, and which quantity could be defended?

Show solution

Step 1. $G \to O \to W$ , $G \to W$ , $A \to O$ , $A \to W$ .

Step 2. Occupation is a mediator and a collider ( $G \to O \leftarrow A$ ). Conditioning on it opens $G \to O \leftarrow A \to W$ , so the "within-occupation gap" mixes discrimination with ability differences induced by selection into occupations.

Step 3. The total gap $TE$ is identified (gender is not caused by the other variables); the CDE given occupation needs $A$ measured.

Answer: report the total gap, and treat within-occupation gaps as descriptive unless the occupation–wage confounders are measured. Note also that occupational sorting can itself reflect discrimination, so the "indirect" part is not necessarily benign.

Exercise 15 · Obesity, hypertension and stroke

A health ministry can fund either a weight-loss programme or a hypertension-control programme. Researchers estimate the effect of obesity $X$ on stroke $Y$ and the part mediated by hypertension $M$ . Which mediation quantity answers "how many strokes would remain among obese people if everyone's blood pressure were controlled?" What assumptions does it need?

Show solution

Step 1. Controlling blood pressure fixes the mediator at a normal level $m_0$ for everyone: $CDE(m_0)$ is the effect of obesity that remains.

Step 2. Strokes prevented by hypertension control among the obese: $E[Y(1)] - E[Y(1, m_0)]$ , which needs the same assumptions.

Step 3. Assumptions: no unmeasured confounding of obesity and stroke, and of hypertension and stroke given obesity and covariates (e.g. diet, smoking, kidney function); no cross-world assumption is needed.

Answer: the CDE answers the policy question; natural effects answer the mechanism question and need stronger assumptions.

Verify in Python

import numpy as np

# Exercise 11: mediation formula with a binary mediator
pM = {0: .4, 1: .6}; EY = {(0, 0): 10, (0, 1): 20, (1, 0): 12, (1, 1): 24}
P = lambda m, x: pM[x] if m else 1 - pM[x]
TE = sum(EY[1, m] * P(m, 1) for m in (0, 1)) - sum(EY[0, m] * P(m, 0) for m in (0, 1))
NDE = sum((EY[1, m] - EY[0, m]) * P(m, 0) for m in (0, 1))
NIE = sum(EY[1, m] * (P(m, 1) - P(m, 0)) for m in (0, 1))
print(f"TE {TE:.2f}  NDE {NDE:.2f}  NIE {NIE:.2f}  proportion mediated {NIE / TE:.2f}")

# Exercises 3 and 12: linear formulas, checked by simulating nested potential outcomes
def effects(b0, b1, t1, t2, t3, n=1_000_000, seed=58):
    rng = np.random.default_rng(seed)
    eM = rng.normal(size=n); eY = rng.normal(size=n)
    M = lambda x: b0 + b1 * x + eM
    Y = lambda x, m: t1 * x + t2 * m + t3 * x * m + eY
    return (Y(1, M(0)) - Y(0, M(0))).mean(), (Y(1, M(1)) - Y(1, M(0))).mean()
for args in [(1, .5, 1, 2, .5), (50, 20, .10, .004, .002)]:
    b0, b1, t1, t2, t3 = args
    nde, nie = effects(*args)
    print(f"simulated NDE {nde:.3f} (formula {t1 + t3 * b0:.3f}), NIE {nie:.3f} (formula {(t2 + t3) * b1:.3f})")

# Exercise 13: regression 'direct effect' with a confounded mediator
rng = np.random.default_rng(13); n = 1_000_000
X = rng.normal(size=n); U = rng.normal(size=n)
M = -X - U + rng.normal(size=n); Y = .2 * X - .5 * M + 1.0 * U + rng.normal(size=n)
coef = np.linalg.lstsq(np.column_stack([np.ones(n), X, M]), Y, rcond=None)[0]
print(f"coefficient on smoking given birth weight: {coef[1]:.3f} (true direct effect 0.2)")

Key takeaways

$TE = NDE + NIE$ ; controlled direct effects fix the mediator by policy, natural effects describe mechanisms.
The mediation formula identifies natural effects under no confounding of $X$ – $Y$ , $X$ – $M$ and $M$ – $Y$ , plus a cross-world assumption that fails when treatment affects a mediator–outcome confounder.
"Controlling for the mediator" gives a direct effect only if the mediator is unconfounded with the outcome; otherwise it is collider bias.

Day 58. Mediation: Direct & Indirect Effects

Theory

Nested potential outcomes

Four effects (binary $X$ )

Identification: the mediation formula

Linear models with interaction

The trap

Part 1 · Warm-up

Exercise 1 · Read the potential outcomes

Exercise 2 · Linear, no interaction

Exercise 3 · Linear with interaction

Exercise 4 · Proportion mediated

Exercise 5 · Which effect answers which question?

Part 2 · Proofs

Exercise 6 · The decomposition

Exercise 7 · Deriving the mediation formula for the NDE

Exercise 8 · Linear formulas

Exercise 9 · No interaction, one direct effect

Exercise 10 · A treatment-induced mediator–outcome confounder

Part 3 · Applied

Exercise 11 · Training, job finding and earnings

Exercise 12 · Cash transfers, food spending and child growth

Exercise 13 · Smoking, birth weight and infant mortality

Exercise 14 · Occupation and the gender pay gap

Exercise 15 · Obesity, hypertension and stroke

Verify in Python

Key takeaways

Simulation Lab: Direct & Indirect Effects

Day 58. Mediation: Direct & Indirect Effects

Theory

Nested potential outcomes

Four effects (binary XXX)

Identification: the mediation formula

Linear models with interaction

The trap

Part 1 · Warm-up

Exercise 1 · Read the potential outcomes

Exercise 2 · Linear, no interaction

Exercise 3 · Linear with interaction

Exercise 4 · Proportion mediated

Exercise 5 · Which effect answers which question?

Part 2 · Proofs

Exercise 6 · The decomposition

Exercise 7 · Deriving the mediation formula for the NDE

Exercise 8 · Linear formulas

Exercise 9 · No interaction, one direct effect

Exercise 10 · A treatment-induced mediator–outcome confounder

Part 3 · Applied

Exercise 11 · Training, job finding and earnings

Exercise 12 · Cash transfers, food spending and child growth

Exercise 13 · Smoking, birth weight and infant mortality

Exercise 14 · Occupation and the gender pay gap

Exercise 15 · Obesity, hypertension and stroke

Verify in Python

Key takeaways

Simulation Lab: Direct & Indirect Effects

Four effects (binary $X$ )