Day 63: LATE & Compliers (Imbens–Angrist) · Math for Causal Inference

Theory

Potential treatments

Let $Z \in \{0, 1\}$ be the instrument and $D(z)$ the treatment a unit would take if $Z = z$ . Four types:

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

We observe only $D = D(Z)$ , so types are not observed individually.

Assumptions

Independence: $Z$ is independent of all potential outcomes $Y(d, z)$ and potential treatments $D(z)$ .
Exclusion: $Y(d, z) = Y(d)$ : the instrument matters only through treatment.
Relevance: $P(D = 1 \mid Z = 1) \ne P(D = 1 \mid Z = 0)$ .
Monotonicity: $D(1) \ge D(0)$ for everyone: no defiers.

The LATE theorem (Imbens and Angrist, 1994)

Under 1–4,

\frac{E[Y \mid Z = 1] - E[Y \mid Z = 0]}{E[D \mid Z = 1] - E[D \mid Z = 0]} = E[Y(1) - Y(0) \mid \text{complier}] \equiv LATE.

Always-takers and never-takers have the same treatment whatever $Z$ , so they contribute nothing to the reduced form; the ratio rescales the compliers' effect by their share.

What is identified

Shares: $\pi_a = P(D = 1 \mid Z = 0)$ , $\pi_n = P(D = 0 \mid Z = 1)$ , $\pi_c = 1 - \pi_a - \pi_n$ (the first stage).
Complier outcome means:

E[Y(1) \mid c] = \frac{E[YD \mid Z = 1] - E[YD \mid Z = 0]}{\pi_c}, \qquad E[Y(0) \mid c] = \frac{E[Y(1 - D) \mid Z = 0] - E[Y(1 - D) \mid Z = 1]}{\pi_c}.

Complier characteristics: for any pre-treatment $X$ , $E[X \mid c] = \big(E[XD \mid Z = 1] - E[XD \mid Z = 0]\big)/\pi_c$ .

Interpretation

LATE is the effect for a subpopulation defined by the instrument. Another instrument has other compliers and, with heterogeneous effects, another LATE. Whether a LATE is policy-relevant depends on whether the policy would change treatment for similar people. Profiling compliers (Exercise 12) helps judge this.

Part 1 · Warm-up

Exercise 1 · Classify the types

A lottery offers a place in a charter school. Classify: (a) a family that would enrol only if it wins; (b) a family that will enrol through a sibling preference whatever happens; (c) a family that would never move its child; (d) a family that would enrol only if it loses (out of spite).

Show solution

Step 1. (a) Complier: $D(1) = 1$ , $D(0) = 0$ .

Step 2. (b) Always-taker.

Step 3. (c) Never-taker.

Step 4. (d) Defier, ruled out by monotonicity.

Exercise 2 · Type shares

$P(D = 1 \mid Z = 1) = 0.7$ and $P(D = 1 \mid Z = 0) = 0.2$ . Under monotonicity, find the type shares.

Show solution

Step 1. Always-takers: treated even with $Z = 0$ : $\pi_a = 0.2$ .

Step 2. Never-takers: untreated even with $Z = 1$ : $\pi_n = 1 - 0.7 = 0.3$ .

Step 3. Compliers: $\pi_c = 1 - 0.2 - 0.3 = 0.5$ (= first stage $0.7 - 0.2$ ).

Exercise 3 · LATE from means

With Exercise 2's take-up and mean outcomes 57 ( $Z = 1$ ) and 50.8 ( $Z = 0$ ), compute the LATE.

Show solution

Step 1. Reduced form: $57 - 50.8 = 6.2$ .

Step 2. LATE $= 6.2/0.5 = 12.4$ .

Exercise 4 · Why defiers matter

Explain in words why defiers would bias the Wald ratio.

Show solution

Step 1. Winning the instrument switches compliers into treatment and defiers out of it.

Step 2. The reduced form becomes (compliers' effects) minus (defiers' effects), and the first stage becomes $\pi_c - \pi_d$ .

Answer: the ratio is no longer an average of anyone's effects and can even fall outside the range of all individual effects (Exercise 14).

Exercise 5 · Who complies?

For the draft lottery (Day 61), describe the compliers, always-takers and never-takers.

Show solution

Step 1. Compliers: men who served because they were drafted but would not have volunteered.

Step 2. Always-takers: volunteers.

Step 3. Never-takers: men exempted (health, deferments) or who avoided service even if drafted.

Answer: the IV estimate is the effect of service for draftees who would not have volunteered.

Part 2 · Proofs

Exercise 6 · The LATE theorem

Prove the LATE theorem.

Show solution

Step 1. By exclusion, $Y = Y(0) + D(Z)\,(Y(1) - Y(0))$ .

Step 2. By independence, $E[Y \mid Z = 1] - E[Y \mid Z = 0] = E\big[(D(1) - D(0))(Y(1) - Y(0))\big]$ .

Step 3. By monotonicity, $D(1) - D(0) \in \{0, 1\}$ , equal to 1 exactly for compliers. So the reduced form is $\pi_c\,E[Y(1) - Y(0) \mid c]$ .

Step 4. First stage: $E[D(1) - D(0)] = \pi_c$ .

Step 5. Divide. $\blacksquare$

Exercise 7 · Identifying the shares

Prove $\pi_a = P(D = 1 \mid Z = 0)$ and $\pi_n = P(D = 0 \mid Z = 1)$ under independence and monotonicity.

Show solution

Step 1. With $Z = 0$ , the treated are those with $D(0) = 1$ : always-takers and defiers. With no defiers, only always-takers.

Step 2. Independence: the type distribution is the same for $Z = 0$ and $Z = 1$ , so $P(D = 1 \mid Z = 0) = \pi_a$ .

Step 3. With $Z = 1$ , the untreated are never-takers (and defiers, excluded): $P(D = 0 \mid Z = 1) = \pi_n$ . $\blacksquare$

Exercise 8 · Complier mean of Y(1)

Prove $E[Y(1) \mid c] = (E[YD \mid Z = 1] - E[YD \mid Z = 0])/\pi_c$ .

Show solution

Step 1. With $Z = 1$ , treated units are compliers and always-takers: $E[YD \mid Z = 1] = \pi_c E[Y(1) \mid c] + \pi_a E[Y(1) \mid a]$ .

Step 2. With $Z = 0$ , treated units are always-takers only: $E[YD \mid Z = 0] = \pi_a E[Y(1) \mid a]$ .

Step 3. Subtract and divide by $\pi_c$ . $\blacksquare$

Exercise 9 · Wald with defiers

Without monotonicity, show the Wald estimand is $\dfrac{\pi_c\,LATE_c - \pi_d\,LATE_d}{\pi_c - \pi_d}$ .

Show solution

Step 1. From Exercise 6, Step 2: the reduced form is $E[(D(1) - D(0))(Y(1) - Y(0))]$ .

Step 2. $D(1) - D(0) = 1$ for compliers, $-1$ for defiers, 0 otherwise: reduced form $= \pi_c LATE_c - \pi_d LATE_d$ .

Step 3. First stage $= \pi_c - \pi_d$ . $\blacksquare$

Exercise 10 · Complier characteristics

For a pre-treatment covariate $X$ , prove $E[X \mid c] = (E[XD \mid Z = 1] - E[XD \mid Z = 0])/\pi_c$ .

Show solution

Step 1. $X$ is pre-treatment, so it is not affected by $Z$ and, by independence, its distribution within each type does not depend on $Z$ .

Step 2. $E[XD \mid Z = 1] = \pi_c E[X \mid c] + \pi_a E[X \mid a]$ and $E[XD \mid Z = 0] = \pi_a E[X \mid a]$ .

Step 3. Subtract and divide by $\pi_c$ . $\blacksquare$

Part 3 · Applied

Exercise 11 · A school-voucher lottery

Among lottery winners ( $Z = 1$ ), 70% use a voucher; their mean test score is 60, and non-users score 50. Among losers, 20% attend a private school anyway, scoring 62; the rest score 48. Compute the type shares, the LATE, the compliers' mean scores with and without the voucher, and compare with always-takers and never-takers.

Show solution

Step 1. Shares: $\pi_a = 0.2$ , $\pi_n = 0.3$ , $\pi_c = 0.5$ .

Step 2. $E[Y(1) \mid c] = (0.7 \times 60 - 0.2 \times 62)/0.5 = (42 - 12.4)/0.5 = 59.2$ .

Step 3. $E[Y(0) \mid c] = (0.8 \times 48 - 0.3 \times 50)/0.5 = (38.4 - 15)/0.5 = 46.8$ .

Step 4. LATE $= 59.2 - 46.8 = 12.4$ (matches Exercise 3).

Step 5. Always-takers' treated mean: 62; never-takers' untreated mean: 50.

Answer: vouchers raise compliers' scores by 12.4 points. Compliers would score less than never-takers without the voucher (46.8 versus 50): they are relatively disadvantaged families for whom the voucher made private school affordable.

Exercise 12 · Who are the compliers?

In the same lottery, 50% of children are girls. Among winners, girls who use the voucher make up 45% of all winners; among losers, girls in private school make up 10% of all losers. What share of compliers are girls?

Show solution

Step 1. $E[\text{girl} \cdot D \mid Z = 1] = 0.45$ and $E[\text{girl} \cdot D \mid Z = 0] = 0.10$ .

Step 2. $E[\text{girl} \mid c] = (0.45 - 0.10)/0.5 = 0.70$ .

Answer: 70% of compliers are girls, versus 50% overall. Generalising the LATE to boys requires care.

Exercise 13 · Two instruments, two answers

Using distance to college, the return to schooling is 0.13; using a tuition subsidy, 0.07. Both are valid. Can both be right?

Show solution

Step 1. Each instrument has its own compliers: people for whom distance matters (often from poorer, rural families), and people for whom tuition matters.

Step 2. With heterogeneous returns, their LATEs can differ.

Answer: yes. The disagreement says returns vary across people, not that one instrument is invalid. An over-identification test (Day 66) would reject even if both are valid.

Exercise 14 · Defiers in a fine policy

A city sends warning letters ( $Z$ ) about a recycling fine. The letter makes 40% of households start recycling (compliers, effect on waste of 10 kg) but makes 10% stop in protest (defiers, effect 4 kg). Compute the Wald estimate.

Show solution

Step 1. Reduced form: $0.4 \times 10 - 0.1 \times 4 = 3.6$ .

Step 2. First stage: $0.4 - 0.1 = 0.3$ .

Step 3. Wald: $3.6/0.3 = 12$ .

Answer: 12, larger than every individual effect in the population (10 and 4): without monotonicity the ratio is not an average treatment effect for anyone.

Exercise 15 · A mandate versus a lottery

A health insurance lottery gives a LATE of 0.4 extra doctor visits per year. A ministry asks what a universal mandate would do. Explain what extra information is needed.

Show solution

Step 1. A mandate changes coverage for never-takers and (if it forces nothing on them) leaves always-takers unchanged.

Step 2. The LATE describes compliers only; never-takers may be healthier or less inclined to use care, so their effect may be smaller.

Step 3. Evidence needed: complier versus never-taker characteristics (Exercise 10), effect heterogeneity by observables among compliers, or a second instrument that moves never-takers.

Answer: the LATE is informative but not the answer; extrapolation needs assumptions about how effects vary.

Verify in Python

import numpy as np

rng = np.random.default_rng(63); n = 2_000_000
Z = rng.random(n) < .5
typ = rng.choice(["c", "a", "n"], size=n, p=[.5, .2, .3])
girl = rng.random(n) < np.select([typ == "c", typ == "a", typ == "n"], [.7, .3, .33])
Y0 = np.select([typ == "c", typ == "a", typ == "n"], [46.8, 50, 50]) + rng.normal(0, 8, n)
gain = np.select([typ == "c", typ == "a", typ == "n"], [12.4, 12.0, 5.0]) + rng.normal(0, 3, n)
D = (typ == "a") | ((typ == "c") & Z)
Y = Y0 + D * gain

fs = D[Z].mean() - D[~Z].mean(); rf = Y[Z].mean() - Y[~Z].mean()
print(f"shares: always {D[~Z].mean():.3f}, never {1 - D[Z].mean():.3f}, compliers {fs:.3f}")
print(f"Wald {rf / fs:.2f};  true LATE {gain[typ == 'c'].mean():.2f};  ATE {gain.mean():.2f}")
EY1c = ((Y * D)[Z].mean() - (Y * D)[~Z].mean()) / fs
EY0c = ((Y * ~D)[~Z].mean() - (Y * ~D)[Z].mean()) / fs
print(f"complier means: Y(1) {EY1c:.2f}, Y(0) {EY0c:.2f};  share girls among compliers {((girl * D)[Z].mean() - (girl * D)[~Z].mean()) / fs:.3f}")

# Exercise 11 from the summary statistics, and Exercise 14
print("voucher:", round((.7 * 60 - .2 * 62) / .5, 2), round((.8 * 48 - .3 * 50) / .5, 2), " defier Wald:", round((.4 * 10 - .1 * 4) / .3, 3))

Key takeaways

Under independence, exclusion and monotonicity, IV estimates the average effect for compliers (LATE), not the ATE.
Type shares, complier mean potential outcomes and complier characteristics are identified from simple moments.
Different instruments define different compliers; defiers break the interpretation entirely.

Day 63. LATE & Compliers (Imbens–Angrist)

Theory

Potential treatments

Assumptions

The LATE theorem (Imbens and Angrist, 1994)

What is identified

Interpretation

Part 1 · Warm-up

Exercise 1 · Classify the types

Exercise 2 · Type shares

Exercise 3 · LATE from means

Exercise 4 · Why defiers matter

Exercise 5 · Who complies?

Part 2 · Proofs

Exercise 6 · The LATE theorem

Exercise 7 · Identifying the shares

Exercise 8 · Complier mean of Y(1)

Exercise 9 · Wald with defiers

Exercise 10 · Complier characteristics

Part 3 · Applied

Exercise 11 · A school-voucher lottery

Exercise 12 · Who are the compliers?

Exercise 13 · Two instruments, two answers

Exercise 14 · Defiers in a fine policy

Exercise 15 · A mandate versus a lottery

Verify in Python

Key takeaways

Simulation Lab: Compliers & the LATE