Day 41: Non-Compliance & Intention-to-Treat · Math for Causal Inference

Theory

Assignment is not receipt

In many experiments the researcher randomises an offer $Z$ (a voucher, an invitation, an encouragement letter), but people decide whether to take up the treatment $D$ . Each unit has potential take-up $D(1), D(0)$ and potential outcomes. Four compliance types:

Type	$D(1)$	$D(0)$
Complier	1	0
Always-taker	1	1
Never-taker	0	0
Defier	0	1

Types are defined by potential take-up, so they are fixed before assignment and unaffected by it.

Intention-to-treat

Random assignment of $Z$ identifies two intention-to-treat effects:

\text{ITT}_D = E[D \mid Z = 1] - E[D \mid Z = 0], \qquad \text{ITT}_Y = E[Y \mid Z = 1] - E[Y \mid Z = 0].

$\text{ITT}_Y$ is the effect of offering the programme: often exactly the policy-relevant quantity (governments can offer, not compel).

The effect for compliers

Assume (i) $Z$ is randomised; (ii) exclusion: $Z$ affects $Y$ only through $D$ ; (iii) monotonicity: no defiers; (iv) $\text{ITT}_D > 0$ . Then

\text{CACE} = E[Y(1) - Y(0) \mid \text{complier}] = \frac{\text{ITT}_Y}{\text{ITT}_D}.

This is the Wald (IV) estimator (Days 12 and 19); Day 63 develops it as the LATE. With one-sided non-compliance (no one assigned to control can get treated), there are no always-takers, so compliers are exactly the treated and CACE = ATT (Bloom 1984).

What not to do

As-treated: compare those who took the treatment with those who did not, ignoring $Z$ . This is an observational comparison with self-selection.
Per-protocol: compare those who followed their assignment in each arm. The arms then contain different types (e.g. compliers + always-takers versus compliers + never-takers), so randomisation is lost.

Part 1 · Warm-up

A job-training offer is randomised. Employment is 45% among those offered and 40% among those not offered. Take-up is 60% among the offered and 10% among the non-offered (who found similar training elsewhere).

Exercise 1 · ITT on employment

Compute $\text{ITT}_Y$ .

Show solution

Step 1. $\text{ITT}_Y = 0.45 - 0.40 = 0.05$ .

Answer: offering training raises employment by 5 points.

Exercise 2 · ITT on take-up

Compute $\text{ITT}_D$ .

Show solution

Step 1. $\text{ITT}_D = 0.60 - 0.10 = 0.50$ .

Answer: the offer raises training participation by 50 points.

Exercise 3 · The complier effect

Compute the CACE.

Show solution

Step 1. $\text{CACE} = 0.05/0.50 = 0.10$ .

Answer: training raises employment by 10 points for workers who train only when offered.

Exercise 4 · Type shares

Assuming no defiers, find the shares of always-takers, compliers and never-takers.

Show solution

Step 1. In the control arm, those who train are always-takers: 10%.

Step 2. In the offer arm, those who do not train are never-takers: $1 - 0.60 = 40\%$ .

Step 3. Compliers: $1 - 0.10 - 0.40 = 50\%$ (equal to $\text{ITT}_D$ ).

Answer: 10% always-takers, 50% compliers, 40% never-takers.

Exercise 5 · One-sided non-compliance

A vocational programme can only be accessed with the voucher; 70% of voucher winners enrol. What are $\text{ITT}_D$ and the share of always-takers?

Show solution

Step 1. No one without a voucher can enrol: $E[D \mid Z = 0] = 0$ , so no always-takers.

Step 2. $\text{ITT}_D = 0.70 - 0 = 0.70$ .

Answer: 0.70; always-takers 0%; so CACE = ATT (Exercise 7).

Part 2 · Proofs

Under randomisation, exclusion and no defiers, prove $\text{ITT}_Y = \pi_c\,\text{CACE}$ and $\text{ITT}_D = \pi_c$ , where $\pi_c$ is the complier share.

Show solution

Step 1. By exclusion, write each unit's outcome under assignment $z$ as $Y(D(z))$ . Then $Y_i(Z{=}1) - Y_i(Z{=}0) = Y_i(D_i(1)) - Y_i(D_i(0))$ .

Step 2. By type: always-takers and never-takers have $D(1) = D(0)$ , so their difference is 0. Compliers have $D(1) = 1$ , $D(0) = 0$ , so their difference is $Y_i(1) - Y_i(0)$ . There are no defiers.

Step 3. Randomisation makes $\text{ITT}_Y$ the average of these differences: $\text{ITT}_Y = \pi_c\,E[Y(1) - Y(0) \mid \text{complier}] = \pi_c\,\text{CACE}$ .

Step 4. Same argument for take-up: $D(1) - D(0)$ is 1 for compliers and 0 otherwise, so $\text{ITT}_D = \pi_c$ .

Step 5. Divide: $\text{CACE} = \text{ITT}_Y/\text{ITT}_D$ . $\blacksquare$

Exercise 7 · Bloom's result

With one-sided non-compliance, prove CACE = ATT.

Show solution

Step 1. No one assigned to control can be treated, so $D(0) = 0$ for all: there are no always-takers or defiers.

Step 2. The treated ( $D = 1$ ) are exactly those with $Z = 1$ and $D(1) = 1$ : compliers assigned to treatment.

Step 3. Randomisation makes compliers assigned to treatment representative of all compliers, so $E[\tau \mid D = 1] = E[\tau \mid \text{complier}]$ .

Answer: ATT = CACE $= \text{ITT}_Y/\text{ITT}_D$ . $\blacksquare$

Exercise 8 · Per-protocol is biased

In Exercise 4's population, show that the per-protocol comparison (offered who trained versus not-offered who did not train) compares different mixtures of types.

Show solution

Step 1. Offered who trained: compliers and always-takers, in proportions $0.5 : 0.1$ .

Step 2. Not offered who did not train: compliers and never-takers, $0.5 : 0.4$ .

Step 3. If always-takers are more motivated and never-takers less, the first group would have higher employment even without training.

Answer: per-protocol mixes the training effect with differences between types; only the comparison by assignment ( $Z$ ) is protected by randomisation.

Exercise 9 · As-treated is an observational study

Show that the as-treated difference $E[Y \mid D = 1] - E[Y \mid D = 0]$ contains a selection-bias term even when $Z$ is randomised.

Show solution

Step 1. $D$ is chosen by people (types), not assigned: the treated are compliers-with-offer plus always-takers; the untreated are compliers-without-offer plus never-takers.

Step 2. Apply Day 34's decomposition with $D$ as the treatment: naive $=$ ATT $+ \{E[Y(0) \mid D = 1] - E[Y(0) \mid D = 0]\}$ .

Step 3. The bracket compares $Y(0)$ across different type mixtures, so it is generally not zero.

Answer: as-treated analysis throws away the randomisation and reintroduces selection bias.

Exercise 10 · When exclusion fails

Suppose the offer letter itself raises employment by $\delta$ for everyone (e.g. it signals that the government is watching). Show the Wald ratio equals CACE $+ \delta/\pi_c$ .

Show solution

Step 1. Each unit's outcome difference becomes (its training effect if a complier, else 0) $+ \delta$ .

Step 2. $\text{ITT}_Y = \pi_c\,\text{CACE} + \delta$ , while $\text{ITT}_D = \pi_c$ is unchanged.

Step 3. Ratio: $\text{CACE} + \delta/\pi_c$ . $\blacksquare$

A small direct effect is magnified when compliance is low: exclusion matters most for weak instruments.

Part 3 · Applied

Exercise 11 · A training voucher

Training vouchers are randomised; 60% of winners enrol and no losers can. Winners earn $1,000 more per year than losers. Compute the ITT, the CACE, and say which a finance ministry deciding whether to fund vouchers should use.

Show solution

Step 1. $\text{ITT}_Y = 1{,}000$ ; $\text{ITT}_D = 0.6$ .

Step 2. CACE $= 1{,}000/0.6 \approx 1{,}667$ , which equals the ATT here (one-sided non-compliance).

Step 3. The ministry funds vouchers, not compulsory training, and 40% of winners will not use them.

Answer: ITT $1,000 per voucher (for the cost–benefit of the voucher policy); CACE ≈ $1,667 per trainee (for the value of training to those who take it).

Exercise 12 · Flu-shot encouragement

Doctors are randomly encouraged by letter to vaccinate elderly patients. Vaccination rises from 20% to 30%; flu-related hospitalisation falls from 9% to 8%. Compute the CACE and state the key assumption.

Show solution

Step 1. $\text{ITT}_D = 0.30 - 0.20 = 0.10$ ; $\text{ITT}_Y = 0.08 - 0.09 = -0.01$ .

Step 2. CACE $= -0.01/0.10 = -0.10$ .

Step 3. Key assumption: exclusion. The letter must affect hospitalisation only through vaccination, not, for example, by making doctors more attentive in general.

Answer: among patients vaccinated because of the encouragement, vaccination cuts hospitalisation risk by 10 points, if the letter has no other effect.

Exercise 13 · Adherence in a drug trial

In a heart-disease trial, patients who took at least 80% of their placebo pills had 15% mortality versus 28% for poor adherers. What does this imply for per-protocol analyses of the active drug?

Show solution

Step 1. A placebo has no effect, so the 13-point gap is pure selection: good adherers are healthier or more careful.

Step 2. Comparing adherent patients on the drug with all placebo patients (or with non-adherent ones) would credit the drug with this gap.

Answer: per-protocol and as-treated analyses can show large "effects" of pills that do nothing; analyse by assignment (ITT) and, if needed, estimate the CACE with assignment as the instrument. (This pattern was famously documented in the Coronary Drug Project.)

Exercise 14 · Free bednets

Free insecticide-treated bednets are offered to randomly chosen households; 65% use them. A health minister asks: "What will happen to malaria if we offer free nets nationwide?" Which estimand answers this, and what extra assumption is needed?

Show solution

Step 1. The policy is the offer, with realistic take-up: the ITT is the answer.

Step 2. Extrapolating nationwide assumes similar take-up and effects elsewhere (external validity), and no change in spillovers (nets protect neighbours, Day 32) at full scale.

Answer: the ITT, provided take-up and spillovers in the study resemble national conditions.

Exercise 15 · Low take-up

An encouragement design raises gym attendance by only 2 points; the ITT on blood pressure is −0.1 mmHg (SE 0.2). Compute the CACE and its approximate SE, and comment.

Show solution

Step 1. CACE $= -0.1/0.02 = -5$ mmHg.

Step 2. Ignoring uncertainty in $\text{ITT}_D$ , SE $\approx 0.2/0.02 = 10$ mmHg (delta method, Day 12).

Step 3. 95% CI ≈ $-5 \pm 19.6$ : from −25 to +15 mmHg.

Answer: with such a weak first stage, the complier effect is uninformative, and any small exclusion violation would be magnified 50 times (Exercise 10).

Verify in Python

import numpy as np

rng = np.random.default_rng(41)
n = 400_000
types = rng.choice(["complier", "always", "never"], n, p=[0.5, 0.1, 0.4])
Z = rng.random(n) < 0.5
D = np.where(types == "always", 1, np.where(types == "never", 0, Z.astype(int)))
base = np.select([types == "always", types == "never"], [0.55, 0.30], 0.40)   # different Y(0) by type
effect = np.where(types == "complier", 0.10, 0.05)
Y = rng.random(n) < base + effect * D
itt_y = Y[Z].mean() - Y[~Z].mean(); itt_d = D[Z].mean() - D[~Z].mean()
print(f"ITT_Y {itt_y:.3f}, ITT_D {itt_d:.3f}, CACE {itt_y / itt_d:.3f} (truth 0.10)")
print(f"as-treated {Y[D == 1].mean() - Y[D == 0].mean():.3f}")
print(f"per-protocol {Y[Z & (D == 1)].mean() - Y[~Z & (D == 0)].mean():.3f}")
print("flu-shot CACE:", round((0.08 - 0.09) / (0.30 - 0.20), 3))

Key takeaways

Randomising an offer identifies intention-to-treat effects, often the policy-relevant quantity.
Under exclusion and monotonicity, the effect for compliers is ITT_Y / ITT_D; with one-sided non-compliance it equals the ATT.
As-treated and per-protocol comparisons discard the randomisation and reintroduce selection bias.

Day 41. Non-Compliance & Intention-to-Treat

Theory

Assignment is not receipt

Intention-to-treat

The effect for compliers

What not to do

Part 1 · Warm-up

Exercise 1 · ITT on employment

Exercise 2 · ITT on take-up

Exercise 3 · The complier effect

Exercise 4 · Type shares

Exercise 5 · One-sided non-compliance

Part 2 · Proofs

Exercise 7 · Bloom's result

Exercise 8 · Per-protocol is biased

Exercise 9 · As-treated is an observational study

Exercise 10 · When exclusion fails

Part 3 · Applied

Exercise 11 · A training voucher

Exercise 12 · Flu-shot encouragement

Exercise 13 · Adherence in a drug trial

Exercise 14 · Free bednets

Exercise 15 · Low take-up

Verify in Python

Key takeaways

Simulation Lab: Non-Compliance: ITT & CACE

Day 41. Non-Compliance & Intention-to-Treat

Theory

Assignment is not receipt

Intention-to-treat

The effect for compliers

What not to do

Part 1 · Warm-up

Exercise 1 · ITT on employment

Exercise 2 · ITT on take-up

Exercise 3 · The complier effect

Exercise 4 · Type shares

Exercise 5 · One-sided non-compliance

Part 2 · Proofs

Exercise 6 · ITT = complier share × CACE

Exercise 7 · Bloom's result

Exercise 8 · Per-protocol is biased

Exercise 9 · As-treated is an observational study

Exercise 10 · When exclusion fails

Part 3 · Applied

Exercise 11 · A training voucher

Exercise 12 · Flu-shot encouragement

Exercise 13 · Adherence in a drug trial

Exercise 14 · Free bednets

Exercise 15 · Low take-up

Verify in Python

Key takeaways

Simulation Lab: Non-Compliance: ITT & CACE