Day 59: Counterfactuals & the Ladder of Causation

Theory

The ladder of causation

Rung	Question	Notation	Needs
1. Association	What do I expect to see if I observe $x$ ?	$P(y \mid x)$	data
2. Intervention	What happens if I do $x$ ?	$P(y \mid \operatorname{do}(x))$	data + causal graph (or an experiment)
3. Counterfactual	Given that $x$ and $y$ happened, what would $y$ have been under $x'$ ?	$P(y_{x'} \mid x, y)$	an SCM (functions, not only a graph)

Each rung needs more assumptions than the one below.

Counterfactuals in an SCM

For unit $u$ (a value of the disturbances $\mathbf U$ ), the counterfactual $Y_x(u)$ is the value of $Y$ in the modified model $\mathcal M_x$ . These are Phase 3's potential outcomes, now derived from structural equations. Consistency: if $X(u) = x$ then $Y_x(u) = Y(u)$ .

Three steps

To compute $P(y_{x'} \mid \text{evidence } e)$ :

Abduction: update the disturbances using the evidence: $P(\mathbf u) \to P(\mathbf u \mid e)$ .
Action: replace $X$ 's equation by $X := x'$ .
Prediction: compute $Y$ in the modified model, using $P(\mathbf u \mid e)$ .

Linear models: the same effect for everyone

If $Y := \beta X + U_Y$ (whatever the dependence between $X$ and $U_Y$ ), then $Y_{x'}(u) - Y_x(u) = \beta(x' - x)$ for every unit. Unit-level counterfactuals are easy; they are hard precisely when effects vary across units.

Attribution: necessity and sufficiency

For binary $X, Y$ (write $y$ for $Y = 1$ , $y'$ for $Y = 0$ ):

Probability of necessity: $PN = P(Y_0 = 0 \mid X = 1, Y = 1)$ : among exposed people who got the disease, the share who would not have without exposure ("but-for" causation in law).
Probability of sufficiency: $PS = P(Y_1 = 1 \mid X = 0, Y = 0)$ .
Probability of necessity and sufficiency: $PNS = P(Y_1 = 1, Y_0 = 0)$ : the share of units who respond.

Identification under extra assumptions

If $X$ is exogenous (randomised) and the effect is monotone (no one is harmed: $Y_1 \ge Y_0$ ), then

PN = \frac{P(y \mid x) - P(y \mid x')}{P(y \mid x)} = 1 - \frac{1}{RR}, \qquad PNS = P(y \mid x) - P(y \mid x').

Without monotonicity, experiments give only bounds, for example

\max\{0,\; P(y_x) - P(y_{x'})\} \le PNS \le \min\{P(y_x),\; P(y'_{x'})\}.

Part 1 · Warm-up

Exercise 1 · Which rung?

Classify: (a) "Patients who take aspirin have fewer heart attacks." (b) "If we give everyone aspirin, heart attacks will fall by 10%." (c) "Would Mr Lee have avoided his heart attack if he had taken aspirin?" (d) "What share of heart attacks among non-users would aspirin have prevented?"

Show solution

Step 1. (a) Rung 1 (association).

Step 2. (b) Rung 2 (intervention).

Step 3. (c) Rung 3: a counterfactual about a specific person, given what happened.

Step 4. (d) Rung 3: a counterfactual conditioned on observed status (non-users who had attacks).

Exercise 2 · Abduction, action, prediction

$Y := 2X + U_Y$ . A unit has $X = 3$ , $Y = 7$ . What would $Y$ have been with $X = 5$ ?

Show solution

Step 1. Abduction: $U_Y = Y - 2X = 7 - 6 = 1$ .

Step 2. Action: set $X := 5$ .

Step 3. Prediction: $Y_5 = 2(5) + 1 = 11$ .

Exercise 3 · Counterfactuals downstream

$X := U_X$ , $M := 0.5X + U_M$ , $Y := M + X + U_Y$ . A unit has $X = 2$ , $M = 1.5$ , $Y = 4$ . Compute $Y_{X = 0}$ and $M_{X = 0}$ .

Show solution

Step 1. Abduction: $U_M = 1.5 - 1 = 0.5$ ; $U_Y = 4 - 1.5 - 2 = 0.5$ .

Step 2. Action: $X := 0$ .

Step 3. Prediction: $M_0 = 0 + 0.5 = 0.5$ ; $Y_0 = 0.5 + 0 + 0.5 = 1$ .

Answer: $M_0 = 0.5$ and $Y_0 = 1$ .

Exercise 4 · Probability of necessity

In a randomised trial, $P(\text{recover} \mid \text{drug}) = 0.8$ and $P(\text{recover} \mid \text{placebo}) = 0.5$ . Assuming no one is harmed by the drug, among treated patients who recovered, what share recovered because of the drug?

Show solution

Step 1. $PN = \dfrac{0.8 - 0.5}{0.8} = 0.375$ .

Answer: 37.5%; the other 62.5% would have recovered anyway.

Exercise 5 · Effect of treatment on the treated

Write the ATT as a counterfactual quantity and say which rung it belongs to.

Show solution

Step 1. $ATT = E[Y_1 - Y_0 \mid X = 1] = E[Y \mid X = 1] - E[Y_0 \mid X = 1]$ , using consistency.

Step 2. $E[Y_0 \mid X = 1]$ conditions a counterfactual on the actual treatment.

Answer: rung 3 in form, though it is identified by the same assumptions as rung-2 quantities (back-door, Exercise 14).

Part 2 · Proofs

Exercise 6 · Consistency from the SCM

Prove that if $X(u) = x$ then $Y_x(u) = Y(u)$ .

Show solution

Step 1. $\mathcal M_x$ differs from $\mathcal M$ only in $X$ 's equation.

Step 2. For unit $u$ , the original equation already produces $X = x$ , so replacing it with $X := x$ changes no value.

Step 3. All other equations and disturbances are identical, so every variable, including $Y$ , takes the same value. $\blacksquare$

Exercise 7 · Linear counterfactuals

Let $Y := \beta X + U_Y$ , with $U_Y$ arbitrarily dependent on $X$ . Prove $E[Y_{x'} \mid X = x, Y = y] = y + \beta(x' - x)$ .

Show solution

Step 1. Abduction: given $X = x$ , $Y = y$ , we know $U_Y = y - \beta x$ exactly.

Step 2. Action and prediction: $Y_{x'} = \beta x' + U_Y = \beta x' + y - \beta x$ .

Step 3. This is non-random given the evidence, so its expectation is itself. $\blacksquare$ (Rung 3 collapses to $\beta$ , a rung-2 quantity, in linear additive models.)

Exercise 8 · Probability of necessity under exogeneity and monotonicity

Assume $X \perp\!\!\!\perp (Y_0, Y_1)$ and $Y_1 \ge Y_0$ . Prove $PN = [P(y \mid x) - P(y \mid x')]/P(y \mid x)$ .

Show solution

Step 1. $PN = \dfrac{P(Y_0 = 0, X = 1, Y = 1)}{P(X = 1, Y = 1)} = \dfrac{P(Y_0 = 0, Y_1 = 1, X = 1)}{P(Y_1 = 1, X = 1)}$ (consistency).

Step 2. By exogeneity, $X$ factors out: $PN = P(Y_0 = 0, Y_1 = 1)/P(Y_1 = 1)$ .

Step 3. Monotonicity: $Y_1 = 1$ whenever $Y_0 = 1$ , so $P(Y_0 = 0, Y_1 = 1) = P(Y_1 = 1) - P(Y_0 = 1)$ .

Step 4. By exogeneity $P(Y_x = 1) = P(y \mid x)$ . Substitute. $\blacksquare$

Exercise 9 · Bounds on PNS

From experimental data alone, prove $\max\{0, P(y_x) - P(y_{x'})\} \le PNS \le \min\{P(y_x), P(y'_{x'})\}$ .

Show solution

Step 1. Four response types: $(Y_1, Y_0) \in \{(1, 1), (1, 0), (0, 1), (0, 0)\}$ with probabilities $p_{11}, p_{10}, p_{01}, p_{00}$ ; $PNS = p_{10}$ .

Step 2. Experiments give $P(y_x) = p_{11} + p_{10}$ and $P(y_{x'}) = p_{11} + p_{01}$ .

Step 3. Upper: $p_{10} \le p_{11} + p_{10} = P(y_x)$ and $p_{10} \le p_{10} + p_{00} = 1 - P(y_{x'}) = P(y'_{x'})$ .

Step 4. Lower: $p_{10} = P(y_x) - P(y_{x'}) + p_{01} \ge P(y_x) - P(y_{x'})$ , and $p_{10} \ge 0$ . $\blacksquare$

Exercise 10 · Same experiments, different counterfactuals

Construct two SCMs for binary $X, Y$ with the same $P(y \mid \operatorname{do}(x))$ for both $x$ but different $PNS$ .

Show solution

Step 1. Model A: $U \sim \text{Bernoulli}(0.5)$ , $Y := U$ . Then $P(Y = 1 \mid \operatorname{do}(x)) = 0.5$ for both $x$ , and nobody responds: $PNS = 0$ .

Step 2. Model B: $Y := X \oplus U$ (exclusive or). Again $P(Y = 1 \mid \operatorname{do}(x)) = 0.5$ for both $x$ .

Step 3. In Model B, units with $U = 0$ have $Y_1 = 1, Y_0 = 0$ (helped) and units with $U = 1$ have $Y_1 = 0, Y_0 = 1$ (harmed): $PNS = 0.5$ .

Answer: identical at rungs 1 and 2, different at rung 3. $\blacksquare$ A zero average effect can hide large offsetting individual effects.

Part 3 · Applied

Exercise 11 · A worker's counterfactual wage

In Day 50's SCM, $\log W := 1.5 + 0.08S + 0.10A + U_W$ , a worker has 12 years of schooling and $\log W = 2.6$ . What would her log wage have been with 16 years? Does the answer depend on her (unobserved) ability?

Show solution

Step 1. Abduction: $0.10A + U_W = 2.6 - 1.5 - 0.08(12) = 0.14$ . We cannot separate $A$ from $U_W$ , but we do not need to.

Step 2. Action: $S := 16$ . Ability is upstream of schooling, so it is unchanged.

Step 3. Prediction: $\log W_{16} = 1.5 + 0.08(16) + 0.14 = 2.92$ .

Answer: 2.92, about 38% higher wages ( $e^{0.32} - 1$ ). Her ability does not matter because the model is linear and additive (Exercise 7); with heterogeneous returns it would.

Exercise 12 · But-for causation in court

A worker exposed to asbestos develops mesothelioma and sues. A cohort study (exposure as good as random given the covariates) gives a lifetime risk of 2.0% for exposed and 0.5% for comparable unexposed workers. Assuming asbestos never prevents the disease, is it "more likely than not" that the exposure caused this worker's disease?

Show solution

Step 1. Exogeneity and monotonicity give $PN = (0.020 - 0.005)/0.020 = 0.75$ .

Step 2. Equivalently, $RR = 4$ and $PN = 1 - 1/4 = 0.75$ .

Answer: 75% > 50%: the civil standard is met. The rule "RR > 2 implies PN > 0.5" holds under these assumptions; without monotonicity only bounds are available.

Exercise 13 · How many people does a vaccine protect?

A trial finds infection risks of 2% with vaccine and 10% with placebo. Let $y$ = "not infected". Bound the share of people who are protected (not infected with the vaccine, infected without), without assuming monotonicity. What does monotonicity add?

Show solution

Step 1. $P(y_1) = 0.98$ and $P(y_0) = 0.90$ ; $P(y'_0) = 0.10$ .

Step 2. Lower bound: $\max\{0, 0.98 - 0.90\} = 0.08$ .

Step 3. Upper bound: $\min\{0.98, 0.10\} = 0.10$ .

Step 4. With monotonicity (the vaccine never causes infection), $PNS = 0.98 - 0.90 = 0.08$ exactly.

Answer: between 8% and 10% of people are protected; at most 2% could be harmed (infected only because vaccinated), which is what widens the interval.

Exercise 14 · Effect of training on the trained

Use Day 52's training data: half previously employed ( $Z = 1$ ); training rates 30% ( $Z = 1$ ) and 70% ( $Z = 0$ ); employment $P(E = 1 \mid T, Z)$ : $Z = 1$ : 0.60 untrained, 0.70 trained; $Z = 0$ : 0.20 untrained, 0.35 trained. Compute $E[E_0 \mid T = 1]$ and the ATT.

Show solution

Step 1. Composition of the trained: $P(Z = 1 \mid T = 1) = \dfrac{0.5(0.3)}{0.5} = 0.3$ .

Step 2. With $Z$ a back-door set, $E[E_0 \mid T = 1] = \sum_z P(E = 1 \mid T = 0, z)P(z \mid T = 1) = 0.3(0.60) + 0.7(0.20) = 0.32$ .

Step 3. $E[E \mid T = 1] = 0.3(0.70) + 0.7(0.35) = 0.455$ .

Step 4. $ATT = 0.455 - 0.32 = 0.135$ .

Answer: training raised the employment of those trained by 13.5 points, slightly more than the ATE of 12.5 points, because the trained are mostly people for whom training helps more.

Exercise 15 · A patient's counterfactual

A patient refused a drug and died. Her family asks whether she would have survived with it. A trial shows survival of 70% with the drug and 60% without. What can be said?

Show solution

Step 1. The question is rung 3: $P(\text{survive}_1 \mid X = 0, \text{died})$ , the probability of sufficiency.

Step 2. Patients who refuse may differ from trial participants (refusal is not random), so even population-level quantities require care.

Step 3. Treating her as exchangeable with trial patients and assuming monotonicity: $PS = \dfrac{P(y \mid x) - P(y \mid x')}{1 - P(y \mid x')} = \dfrac{0.70 - 0.60}{0.40} = 0.25$ .

Answer: about a 25% chance she would have survived, under strong assumptions. Without monotonicity, only bounds are available; the honest answer is a range.

Verify in Python

import numpy as np

# Exercise 3: abduction-action-prediction
X, M, Y = 2, 1.5, 4
uM, uY = M - .5 * X, Y - M - X
M0 = .5 * 0 + uM; Y0 = M0 + 0 + uY
print("M_0 =", M0, " Y_0 =", Y0)

# Exercise 11: counterfactual log wage
resid = 2.6 - 1.5 - .08 * 12
print("log W_16 =", round(1.5 + .08 * 16 + resid, 3), " wage gain", f"{np.exp(.32) - 1:.1%}")

# Exercises 4, 12, 13: necessity and sufficiency
PN = lambda p1, p0: (p1 - p0) / p1
print("PN drug:", round(PN(.8, .5), 3), " PN asbestos:", round(PN(.02, .005), 3))
p1, p0 = .98, .90
print("PNS bounds:", (round(max(0, p1 - p0), 3), round(min(p1, 1 - p0), 3)))

# Exercise 10: two models, same experiments, different counterfactuals
rng = np.random.default_rng(59); U = rng.random(1_000_000) < .5
for name, f in [("A: Y = U", lambda x: U), ("B: Y = X xor U", lambda x: np.logical_xor(x, U))]:
    Y1, Y0 = f(1), f(0)
    print(f"{name:15s} P(Y=1|do1) {Y1.mean():.3f}  P(Y=1|do0) {Y0.mean():.3f}  PNS {(Y1 & ~Y0).mean():.3f}")

# Exercise 14: ATT
pT = {1: .3, 0: .7}; pE = {(1, 0): .60, (1, 1): .70, (0, 0): .20, (0, 1): .35}    # key (z, t)
pz_t1 = {z: .5 * pT[z] / .5 for z in (0, 1)}
EY0_T1 = sum(pE[z, 0] * pz_t1[z] for z in (0, 1)); EY_T1 = sum(pE[z, 1] * pz_t1[z] for z in (0, 1))
print(f"E[E_0|T=1] {EY0_T1:.3f}  E[E|T=1] {EY_T1:.3f}  ATT {EY_T1 - EY0_T1:.3f}")

Key takeaways

Rung 3 conditions counterfactuals on what actually happened; computing them takes abduction, action and prediction in an SCM.
In linear additive models every unit has the same effect; heterogeneity is what makes counterfactuals hard.
Probabilities of necessity and sufficiency answer attribution questions; they are identified under exogeneity and monotonicity and otherwise bounded.

Day 59. Counterfactuals & the Ladder of Causation

Theory

The ladder of causation

Counterfactuals in an SCM

Three steps

Linear models: the same effect for everyone

Attribution: necessity and sufficiency

Identification under extra assumptions

Part 1 · Warm-up

Exercise 1 · Which rung?

Exercise 2 · Abduction, action, prediction

Exercise 3 · Counterfactuals downstream

Exercise 4 · Probability of necessity

Exercise 5 · Effect of treatment on the treated

Part 2 · Proofs

Exercise 6 · Consistency from the SCM

Exercise 7 · Linear counterfactuals

Exercise 8 · Probability of necessity under exogeneity and monotonicity

Exercise 9 · Bounds on PNS

Exercise 10 · Same experiments, different counterfactuals

Part 3 · Applied

Exercise 11 · A worker's counterfactual wage

Exercise 12 · But-for causation in court

Exercise 13 · How many people does a vaccine protect?

Exercise 14 · Effect of training on the trained

Exercise 15 · A patient's counterfactual

Verify in Python

Key takeaways

Simulation Lab: Counterfactuals & Attribution