Day 45: Phase 3 Capstone: Doubly Robust (AIPW) Estimation

Theory

Two ways to use unconfoundedness

With $\mu_t(x) = E[Y \mid T = t, X = x]$ and $e(x) = P(T = 1 \mid X = x)$ :

Outcome regression (imputation): $\hat\tau_{REG} = \frac1n\sum_i\big(\hat\mu_1(X_i) - \hat\mu_0(X_i)\big)$ . Right if the outcome models are right.
IPW (Day 44): right if the propensity model is right.

Augmented IPW

Start from the regression estimate and add an IPW-weighted average of the outcome-model errors:

\hat\tau_{AIPW} = \frac1n\sum_i\Big[\hat\mu_1(X_i) - \hat\mu_0(X_i) + \frac{T_i(Y_i - \hat\mu_1(X_i))}{\hat e(X_i)} - \frac{(1 - T_i)(Y_i - \hat\mu_0(X_i))}{1 - \hat e(X_i)}\Big].

The correction terms repair whatever bias the outcome model leaves.

Double robustness

$\hat\tau_{AIPW}$ is consistent if either

the outcome models $\mu_0, \mu_1$ are correct (whatever $\hat e$ is), or
the propensity model $e$ is correct (whatever $\hat\mu$ is).

When both are correct, it is also efficient: it attains the smallest possible asymptotic variance for the ATE under unconfoundedness (the semiparametric bound, Day 88).

Inference

Write $\hat\psi_i$ for the bracketed term of unit $i$ , so $\hat\tau_{AIPW} = \frac1n\sum_i\hat\psi_i$ . Its standard error is approximately

\operatorname{SE} \approx \frac{\operatorname{sd}(\hat\psi_i)}{\sqrt n}.

The $\hat\psi_i$ are (estimated) influence-function values (Day 87). With flexible machine-learning models for $\mu$ and $e$ , fit them on different folds from those used to evaluate $\hat\psi_i$ (cross-fitting): this is double/debiased machine learning (Day 90).

Phase 3 in one table

Situation	Tool (Day)
You control assignment	complete or blocked randomisation (35, 38), Neyman variance (36)
Tiny sample or sharp hypothesis	Fisher randomisation test (37)
Baseline covariates in an RCT	Lin adjustment (39)
Treatment delivered to groups or spillovers	cluster randomisation (40), saturation designs (32)
People ignore their assignment	ITT and complier effects (41)
Observational data, rich covariates	unconfoundedness + overlap (42), propensity score (43), IPW (44), AIPW (45)

Part 1 · Warm-up

Exercise 1 · One treated unit's contribution

A treated household has $Y = 20$ , $\hat\mu_1 = 19$ , $\hat\mu_0 = 15$ , $\hat e = 0.8$ . Compute its AIPW term $\hat\psi_i$ .

Show solution

Step 1. Regression part: $\hat\mu_1 - \hat\mu_0 = 4$ .

Step 2. Correction (treated): $(Y - \hat\mu_1)/\hat e = (20 - 19)/0.8 = 1.25$ .

Step 3. $\hat\psi_i = 4 + 1.25 = 5.25$ .

Exercise 2 · One control unit's contribution

A control household has $Y = 14$ , $\hat\mu_1 = 19$ , $\hat\mu_0 = 15$ , $\hat e = 0.8$ . Compute $\hat\psi_i$ .

Show solution

Step 1. Regression part: $19 - 15 = 4$ .

Step 2. Correction (control): $-(Y - \hat\mu_0)/(1 - \hat e) = -(14 - 15)/0.2 = +5$ .

Step 3. $\hat\psi_i = 4 + 5 = 9$ .

Answer: 9. A rare control (only 20% of similar households are untreated) with an outcome below its prediction pulls the estimate up strongly.

Exercise 3 · Reading the correction

What does the correction term do if the outcome model predicts the treated outcomes perfectly?

Show solution

Step 1. Then $Y_i - \hat\mu_1(X_i) = 0$ for every treated unit (and similarly for controls).

Answer: the corrections vanish and AIPW equals the regression estimator; weights matter only to the extent the outcome model errs.

Exercise 4 · A standard error

An AIPW analysis of 900 workers gives $\hat\psi_i$ values with sd 12 ($1,000). Compute the SE.

Show solution

Step 1. SE $\approx 12/\sqrt{900} = 12/30 = 0.4$ .

Answer: $400.

Exercise 5 · Which model matters?

You are confident about how a training programme selected participants (eligibility rules recorded in the files) but not about how earnings depend on covariates. Which estimator is safest?

Show solution

Step 1. Knowledge of selection means the propensity model is likely right; the outcome model may be wrong.

Step 2. IPW and AIPW both rely on a correct propensity model; AIPW also uses the outcome model to reduce variance.

Answer: AIPW (or IPW): both stay consistent through the propensity model, and AIPW is more precise whenever the outcome model explains a good share of earnings variation.

Part 2 · Proofs

Exercise 6 · Robust to a wrong propensity model

Assume unconfoundedness and that $\tilde\mu_1(x) = \mu_1(x)$ is correct, while $\tilde e(x)$ is any function in $(0, 1)$ . Prove $E\big[\tilde\mu_1(X) + T(Y - \tilde\mu_1(X))/\tilde e(X)\big] = E[Y(1)]$ .

Show solution

Step 1. $E[\mu_1(X)] = E[E[Y \mid T = 1, X]] = E[Y(1)]$ (Day 42).

Step 2. Correction term, by the LIE: $E\Big[\frac{T(Y - \mu_1(X))}{\tilde e(X)}\Big] = E\Big[\frac{E[T(Y - \mu_1(X)) \mid X]}{\tilde e(X)}\Big]$ .

Step 3. $E[T(Y - \mu_1(X)) \mid X] = e(X)\,\big(E[Y \mid T = 1, X] - \mu_1(X)\big) = 0$ .

Step 4. So the correction has mean 0 whatever $\tilde e$ is, and the total is $E[Y(1)]$ . $\blacksquare$

Exercise 7 · Robust to a wrong outcome model

Now assume $\tilde e(x) = e(x)$ is correct and $\tilde\mu_1$ is any function. Prove the same identity.

Show solution

Step 1. Rearrange: $\tilde\mu_1(X) + \frac{T(Y - \tilde\mu_1(X))}{e(X)} = \frac{TY}{e(X)} + \tilde\mu_1(X)\Big(1 - \frac{T}{e(X)}\Big)$ .

Step 2. $E[TY/e(X)] = E[Y(1)]$ (Day 44, Exercise 6).

Step 3. $E\big[\tilde\mu_1(X)(1 - T/e(X)) \mid X\big] = \tilde\mu_1(X)\big(1 - e(X)/e(X)\big) = 0$ .

Step 4. So the total has mean $E[Y(1)]$ . $\blacksquare$ The same arguments for $Y(0)$ give double robustness for the ATE.

Exercise 8 · AIPW nests IPW

Show that AIPW with $\hat\mu_0 = \hat\mu_1 = 0$ is the Horvitz–Thompson IPW estimator.

Show solution

Step 1. With $\hat\mu \equiv 0$ , the regression part is 0.

Step 2. The corrections become $T_iY_i/\hat e_i - (1 - T_i)Y_i/(1 - \hat e_i)$ .

Step 3. Their average is $\hat\tau_{HT}$ (Day 44). $\blacksquare$

A good outcome model shrinks the residuals that get weighted, which is why AIPW is typically less variable than IPW when the outcome model fits well.

Exercise 9 · AIPW in a randomised experiment

In an RCT with known $e = p$ and linear outcome models fitted separately by arm, explain why AIPW is consistent whatever the outcome model, and how it relates to Lin's estimator (Day 39).

Show solution

Step 1. The propensity is known exactly ( $e = p$ ), so Exercise 7 applies: consistent for any $\hat\mu$ .

Step 2. With linear $\hat\mu_t$ fitted by OLS in each arm, residuals average to zero within each arm, so the correction terms average to (nearly) zero.

Step 3. AIPW then reduces to $\frac1n\sum(\hat\mu_1(X_i) - \hat\mu_0(X_i))$ : each arm's regression line evaluated at every unit's covariates and averaged, which equals Lin's estimator.

Answer: in an RCT, covariate adjustment is a special case of AIPW: it can only improve precision, never introduce bias asymptotically.

Exercise 10 · The product of errors

Show that, for the $Y(1)$ part, the bias of AIPW with wrong models $\tilde\mu_1$ and $\tilde e$ is $E\Big[\frac{(e(X) - \tilde e(X))(\mu_1(X) - \tilde\mu_1(X))}{\tilde e(X)}\Big]$ .

Show solution

Step 1. Condition on $X$ : $E[T(Y - \tilde\mu_1) \mid X] = e(X)(\mu_1(X) - \tilde\mu_1(X))$ .

Step 2. So $E\big[\tilde\mu_1 + T(Y - \tilde\mu_1)/\tilde e\big] = E\Big[\tilde\mu_1 + \frac{e}{\tilde e}(\mu_1 - \tilde\mu_1)\Big]$ .

Step 3. Subtract $E[\mu_1(X)] = E[Y(1)]$ : bias $= E\Big[(\mu_1 - \tilde\mu_1)\Big(\frac{e}{\tilde e} - 1\Big)\Big] = E\Big[\frac{(e - \tilde e)(\mu_1 - \tilde\mu_1)}{\tilde e}\Big]$ . $\blacksquare$

The bias is a product of the two errors: if each model is moderately good, the bias is very small. This is why AIPW works with machine-learning nuisance models (Day 90).

Part 3 · Applied

Return to Day 44's six households, now with outcome-model predictions:

Household	$T$	$\hat e$	$Y$	$\hat\mu_1$	$\hat\mu_0$
1	1	0.8	20	19	15
2	1	0.5	16	16	12
3	1	0.2	10	12	9
4	0	0.8	14	19	15
5	0	0.5	12	16	12
6	0	0.2	8	12	9

Exercise 11 · Regression and AIPW estimates

Compute the regression estimate and the AIPW estimate of the ATE.

Show solution

Step 1. Regression: $\hat\mu_1 - \hat\mu_0$ = 4, 4, 3, 4, 4, 3; mean $22/6 \approx 3.67$ .

Step 2. AIPW terms, treated ( $+\,(Y - \hat\mu_1)/\hat e$ ): household 1: $4 + 1/0.8 = 5.25$ ; household 2: $4 + 0 = 4$ ; household 3: $3 + (-2)/0.2 = -7$ .

Step 3. Controls ( $-\,(Y - \hat\mu_0)/(1 - \hat e)$ ): household 4: $4 - (-1)/0.2 = 9$ ; household 5: $4 - 0 = 4$ ; household 6: $3 - (-1)/0.8 = 4.25$ .

Step 4. Mean: $(5.25 + 4 - 7 + 9 + 4 + 4.25)/6 = 19.5/6 = 3.25$ .

Answer: regression 3.67, AIPW 3.25 (Day 44: naive 4.0, Hájek 0.36). With six households the estimators disagree a lot; their disagreement is itself a warning about model dependence and overlap.

Exercise 12 · A wrong outcome model, a right propensity model

In simulation (code below), the true earnings model is non-linear in age, the analyst fits a linear outcome model, and the propensity model is correct. Which estimators are biased?

Show solution

Step 1. Regression relies on the (wrong) linear outcome model: biased.

Step 2. IPW relies on the (correct) propensity model: consistent.

Step 3. AIPW is consistent through the propensity model (Exercise 7).

Step 4. The simulation confirms it: regression averages about 1.44 against a truth of 2, while IPW and AIPW average about 1.98.

Step 5. Precision: here AIPW's SD (≈ 0.43) is slightly larger than IPW's (≈ 0.38). The linear model misses the U-shape in age, so its residuals are large, and weighting large residuals adds noise.

Answer: double robustness delivered consistency; the precision gain of AIPW needs a reasonably good outcome model. Flexible outcome models (splines, machine learning) fix this, which is the motivation for Day 90.

Exercise 13 · Both models wrong

What happens if both models are misspecified, and how can analysts reduce that risk?

Show solution

Step 1. No guarantee: the bias is the product of the two errors (Exercise 10), which can still be substantial.

Step 2. Remedies: flexible models (splines, interactions, machine learning) for both nuisances, with cross-fitting; balance diagnostics for the propensity model; and checking that regression, IPW and AIPW roughly agree.

Answer: double robustness gives two chances, not immunity.

Exercise 14 · Writing the policy report

A ministry asks whether a youth employment subsidy raised employment. You have administrative data on all eligible youths, rich pre-programme records and the eligibility rules. Outline the Phase 3 analysis plan.

Show solution

Step 1. Estimand: ATT (is the subsidy worth it for those who used it?) and, for an expansion, the ATE or CATEs (Day 33).

Step 2. Identification: argue unconfoundedness from the eligibility rules and include all variables officials used, plus lagged employment (Day 42).

Step 3. Overlap: estimate propensities, inspect histograms and normalised differences, and trim if needed (Days 42–43).

Step 4. Estimation: AIPW with flexible models and cross-fitting; report regression and IPW as checks (Days 44–45).

Step 5. Credibility: placebo test on pre-programme employment, sensitivity analysis for unobserved confounding.

Answer: a transparent chain from estimand to assumptions to diagnostics to estimator, with robustness checks.

Exercise 15 · Match the tool to the problem

For each case, name the Phase 3 tool: (a) 12 clinics, 6 randomised to a new protocol, outcome is patient satisfaction; (b) a lottery for vouchers where 30% of winners do not use them; (c) a trial randomising households within villages where neighbours share information; (d) an observational study with 50 baseline covariates and good overlap; (e) an RCT with a strong baseline measure of the outcome.

Show solution

Step 1. (a) Cluster randomisation with clinic-level randomisation inference (Days 37, 40).

Step 2. (b) ITT for the voucher offer and CACE = ITT_Y / ITT_D for users (Day 41).

Step 3. (c) Spillovers: estimate direct and spillover effects, or randomise villages (Days 32, 40).

Step 4. (d) Propensity score with AIPW and balance checks (Days 43–45).

Step 5. (e) Lin's covariate adjustment, or block on the baseline measure (Days 38–39).

Verify in Python

import numpy as np
from scipy import optimize

T = np.array([1, 1, 1, 0, 0, 0]); e = np.array([.8, .5, .2, .8, .5, .2]); Y = np.array([20, 16, 10, 14, 12, 8.])
m1 = np.array([19, 16, 12, 19, 16, 12.]); m0 = np.array([15, 12, 9, 15, 12, 9.])
psi = m1 - m0 + T * (Y - m1) / e - (1 - T) * (Y - m0) / (1 - e)
print("psi:", psi, " regression:", round((m1 - m0).mean(), 3), " AIPW:", round(psi.mean(), 3))

# Monte Carlo: wrong outcome model, correct propensity model
rng = np.random.default_rng(45)
out = {"regression": [], "IPW (Hajek)": [], "AIPW": []}
for _ in range(1500):
    n = 1000; age = rng.uniform(18, 60, n); X = np.column_stack([np.ones(n), age])
    e_true = 1 / (1 + np.exp(-(3 - 0.1 * age))); T = rng.random(n) < e_true
    y = 10 + 0.02 * (age - 35)**2 + 2 * T + rng.normal(0, 2, n)          # non-linear in age; effect 2
    b = optimize.minimize(lambda b: -np.sum(T * (X @ b) - np.log1p(np.exp(X @ b))), np.zeros(2)).x
    eh = 1 / (1 + np.exp(-X @ b))
    b1 = np.linalg.lstsq(X[T], y[T], rcond=None)[0]; b0 = np.linalg.lstsq(X[~T], y[~T], rcond=None)[0]
    mu1, mu0 = X @ b1, X @ b0                                              # linear (wrong) outcome models
    out["regression"].append((mu1 - mu0).mean())
    out["IPW (Hajek)"].append((T * y / eh).sum() / (T / eh).sum() - ((~T) * y / (1 - eh)).sum() / ((~T) / (1 - eh)).sum())
    out["AIPW"].append(np.mean(mu1 - mu0 + T * (y - mu1) / eh - (~T) * (y - mu0) / (1 - eh)))
for k, v in out.items():
    print(f"{k:12s} mean {np.mean(v):.3f} (truth 2)  SD {np.std(v):.3f}")

Key takeaways

AIPW adds an inverse-probability-weighted correction to an outcome-model estimate.
It is consistent if either model is right, and efficient if both are; its bias is a product of the two models' errors.
Phase 3 gives you the design-based and model-based toolkit; Phase 4 adds graphs to decide what to condition on.

Day 45. Phase 3 Capstone: Doubly Robust (AIPW) Estimation

Theory

Two ways to use unconfoundedness

Augmented IPW

Double robustness

Inference

Phase 3 in one table

Part 1 · Warm-up

Exercise 1 · One treated unit's contribution

Exercise 2 · One control unit's contribution

Exercise 3 · Reading the correction

Exercise 4 · A standard error

Exercise 5 · Which model matters?

Part 2 · Proofs

Exercise 6 · Robust to a wrong propensity model

Exercise 7 · Robust to a wrong outcome model

Exercise 8 · AIPW nests IPW

Exercise 9 · AIPW in a randomised experiment

Exercise 10 · The product of errors

Part 3 · Applied

Exercise 11 · Regression and AIPW estimates

Exercise 12 · A wrong outcome model, a right propensity model

Exercise 13 · Both models wrong

Exercise 14 · Writing the policy report

Exercise 15 · Match the tool to the problem

Verify in Python

Key takeaways

Simulation Lab: Double Robustness (AIPW)

Household	$T$	$\hat e$	$Y$	$\hat\mu_1$	$\hat\mu_0$
1	1	0.8	20	19	15
2	1	0.5	16	16	12
3	1	0.2	10	12	9
4	0	0.8	14	19	15
5	0	0.5	12	16	12
6	0	0.2	8	12	9

Household	$T$	$\hat e$	$Y$	$\hat\mu_1$	$\hat\mu_0$
1	1	0.8	20	19	15
2	1	0.5	16	16	12
3	1	0.2	10	12	9
4	0	0.8	14	19	15
5	0	0.5	12	16	12
6	0	0.2	8	12	9

Household	$T$	$\hat e$	$Y$	$\hat\mu_1$	$\hat\mu_0$
1	1	0.8	20	19	15
2	1	0.5	16	16	12
3	1	0.2	10	12	9
4	0	0.8	14	19	15
5	0	0.5	12	16	12
6	0	0.2	8	12	9