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Matching and Weighting Methods · Tutorial 642 of 952

Hájek Normalization for Inverse-Probability Weights

Learn how normalizing inverse-probability weights changes weighted means, and when the resulting Hájek estimates differ from Horvitz–Thompson estimates.

Advanced 12 min read

What You'll Learn

  • Distinguish Horvitz–Thompson weighted totals from Hájek normalized weighted means.
  • Derive normalized treatment and control means for the average treatment effect.
  • Construct the corresponding inverse-odds weights for the treated-group effect.
  • Calculate both estimators from hypothetical samples with unequal treatment probabilities.
  • Explain why normalization changes finite-sample estimates but does not guarantee lower error.

Why Normalize Inverse-Probability Weights?

Inverse-probability weighting uses treatment probabilities to reweight observed outcomes. For the average treatment effect, treated units receive weights proportional to the inverse of their treatment probabilities, while control units receive weights proportional to the inverse of their control probabilities. The Horvitz–Thompson estimator uses these weights to form weighted totals. The Hájek estimator divides each weighted total by the corresponding sum of weights, producing a normalized weighted mean.

That division matters in a finite sample. The sum of the weights in an arm need not equal the sample size, or the population quantity it is intended to represent. Consequently, the Horvitz–Thompson and Hájek estimates can differ even when calculated from the same observations and propensity scores. Normalization can make a weighted mean more stable in some samples, but it does not guarantee lower bias, lower variance, or better causal identification.

Setup and Identification

For unit \(i\), let \(A_i\in\{0,1\}\) indicate treatment, let \(Y_i\) be the observed outcome, and let \(X_i\) denote observed pre-treatment covariates. Write \(Y_i(1)\) and \(Y_i(0)\) for the potential outcomes under treatment and control, respectively. The observed outcome satisfies \(Y_i=A_iY_i(1)+(1-A_i)Y_i(0)\). The propensity score is \(e(X_i)=P(A_i=1\mid X_i)\).

The weighting formulas require a causal identification framework. Assume consistency, so the observed outcome under the treatment actually received equals the corresponding potential outcome. Assume conditional exchangeability, \((Y(1),Y(0))\mathbin{\perp\!\!\!\perp} A\mid X\), and positivity, \(0<e(X)<1\) for covariate values in the target population. These assumptions let observed outcomes, after weighting, represent the potential-outcome means of interest. In practice, the propensity score is usually estimated; the formulas below use \(e(X_i)\) to describe the weighting logic.

Definition: A Horvitz–Thompson weighted mean component uses inverse-probability weights to estimate a target-population total or mean without dividing by the realized sum of those weights. A Hájek weighted mean divides the weighted outcome total by the realized sum of weights. For positive weights \(w_i\), the normalized weighted mean is \(\sum_i w_iY_i/\sum_i w_i\).

For the average treatment effect (ATE), the Horvitz–Thompson estimator is

$$ \widehat{\tau}_{HT} = \frac{1}{n}\sum_{i=1}^{n} \left[ \frac{A_iY_i}{e(X_i)} - \frac{(1-A_i)Y_i}{1-e(X_i)} \right]. $$

Its first weighted total estimates the population mean of \(Y(1)\); its second estimates the population mean of \(Y(0)\). The Hájek estimator instead calculates a normalized mean within each weighted arm:

$$ \widehat{\mu}_{1,H} = \frac{\sum_{i=1}^{n} A_iY_i/e(X_i)} {\sum_{i=1}^{n} A_i/e(X_i)}, \qquad \widehat{\mu}_{0,H} = \frac{\sum_{i=1}^{n} (1-A_i)Y_i/\{1-e(X_i)\}} {\sum_{i=1}^{n} (1-A_i)/\{1-e(X_i)\}}, \qquad \widehat{\tau}_{H} = \widehat{\mu}_{1,H}-\widehat{\mu}_{0,H}. $$

Each Hájek arm estimate is a weighted average of observed outcomes in that arm, so it lies between the smallest and largest outcome with positive weight in that arm. By contrast, each Horvitz–Thompson component is a weighted sum divided by the full sample size \(n\); it is not generally a weighted average of outcomes in the observed arm.

Worked Example: ATE With Unequal Propensity Scores

Suppose four hypothetical units have the following observed treatment statuses, outcomes, and known propensity scores. Assume the identification conditions above hold for the target population. All outcomes are measured in points.

Unit\(A_i\)\(Y_i\)\(e(X_i)\)
11100.20
2040.40
3180.60
4020.80

For the treated Horvitz–Thompson component, the inverse-probability contributions are \(10/0.20=50\) and \(8/0.60=13.333\). For the control component, they are \(4/(1-0.40)=6.667\) and \(2/(1-0.80)=10\). Thus

$$ \widehat{\tau}_{HT} = \frac{50+13.333-(6.667+10)}{4} = \frac{46.666}{4} = 11.67\text{ points}. $$

For the Hájek treated mean, the weights are \(1/0.20=5\) and \(1/0.60=1.667\), so \(\widehat{\mu}_{1,H}=(50+13.333)/(5+1.667)=9.500\) points. For the control mean, the weights are \(1/0.60=1.667\) and \(1/0.20=5\), giving \(\widehat{\mu}_{0,H}=(6.667+10)/(1.667+5)=2.500\) points. Therefore

$$ \widehat{\tau}_{H}=9.500-2.500=7.000\text{ points}. $$

The estimates differ substantially in this small sample because the realized sums of inverse-probability weights are not the denominators used by the Horvitz–Thompson estimator. The calculation alone does not establish which estimate is closer to the causal effect.

Why the Normalization Is a Ratio

The normalized form follows from expressing each target mean as a ratio of expectations. For example, conditional exchangeability and positivity imply

$$ E\left[\frac{AY}{e(X)}\right]=E[Y(1)], \qquad E\left[\frac{A}{e(X)}\right]=1. $$

To see the first equality, condition on \(X\). Since \(Y=Y(1)\) when \(A=1\), exchangeability gives

$$ E\left[\left.\frac{AY}{e(X)}\right|X\right] = \frac{E[AY(1)\mid X]}{e(X)} = \frac{e(X)E[Y(1)\mid X]}{e(X)} = E[Y(1)\mid X]. $$

Averaging over \(X\) yields \(E[Y(1)]\). The denominator identity follows in the same way: \(E[A/e(X)\mid X]=e(X)/e(X)=1\). Therefore the treated Hájek mean estimates the ratio \(E[AY/e(X)]/E[A/e(X)]=E[Y(1)]\). Replacing these expectations by their sample averages produces the observed normalized weighted mean. The common factor \(1/n\) cancels from its numerator and denominator.

The same argument applies to controls: \(E[(1-A)Y/\{1-e(X)\}]=E[Y(0)]\) and \(E[(1-A)/\{1-e(X)\}]=1\). These population identities explain why both approaches target the same arm means under the assumptions. They do not imply that their finite-sample estimates are equal.

Normalization for the Treated-Group Effect

The average treatment effect for the treated (ATT) is \(E[Y(1)-Y(0)\mid A=1]\). The observed treated mean estimates \(E[Y(1)\mid A=1]\). To represent the treated group’s covariate distribution among controls, a control unit receives the inverse-odds weight \(e(X_i)/\{1-e(X_i)\}\). Under conditional exchangeability, weighting control outcomes this way targets \(E[Y(0)\mid A=1]\).

Let \(n_1=\sum_i A_i\), the number of treated units in the sample. A Horvitz–Thompson-style ATT estimator uses the treated sample mean and scales the weighted control total by \(n_1\). The Hájek form normalizes the control weights by their observed sum:

$$ \widehat{\tau}_{ATT,HT} = \frac{\sum_i A_iY_i}{n_1} - \frac{\sum_i (1-A_i)\{e(X_i)/[1-e(X_i)]\}Y_i}{n_1}, \qquad \widehat{\tau}_{ATT,H} = \frac{\sum_i A_iY_i}{n_1} - \frac{\sum_i (1-A_i)\{e(X_i)/[1-e(X_i)]\}Y_i} {\sum_i (1-A_i)e(X_i)/[1-e(X_i)]}. $$

Here the treated mean is already normalized by the number of treated units; the key difference is the denominator for the reweighted control mean. The Horvitz–Thompson scaling uses \(n_1\), while the Hájek scaling uses the realized sum of inverse-odds weights.

Worked Example: Normalizing Control Weights for the ATT

Consider a hypothetical sample with two treated units and two controls. Their outcomes are measured in points. Assume the propensity scores and identification conditions are appropriate for targeting the treated group.

Unit\(A_i\)\(Y_i\)\(e(X_i)\)
11100.20
2180.60
3040.40
4020.80

The treated mean is \((10+8)/2=9\) points. The control inverse-odds weights are \(0.40/0.60=0.667\) and \(0.80/0.20=4\). Their weighted outcome total is \((0.667)(4)+(4)(2)=2.668+8=10.668\), and their weight sum is \(0.667+4=4.667\). Using the displayed rounded values, the Hájek reweighted control mean is approximately \(10.668/4.667=2.286\) points. Thus

$$ \widehat{\tau}_{ATT,H}=9-2.286=6.714\text{ points}. $$

For the Horvitz–Thompson-style estimate, divide that weighted control total by \(n_1=2\), not by the control weight sum. It gives \(10.668/2=5.334\) points for the counterfactual control mean and hence \(\widehat{\tau}_{ATT,HT}=9-5.334=3.666\) points. The difference reflects normalization in this small sample; neither calculation by itself verifies covariate balance or causal identification.

Finite-Sample Behavior: What Normalization Does and Does Not Do

For an ATE arm, the Hájek estimator can be written as a ratio of a weighted outcome total to a weight total. If a sample average in the numerator converges to its expectation and the corresponding denominator converges to a positive expectation, the ratio converges to the ratio of those expectations. This is the usual ratio argument: if \(U_n\) converges to \(u\) and \(V_n\) converges to \(v\ne0\), then \(U_n/V_n\) converges to \(u/v\). Applied to the treated arm, \(u=E[Y(1)]\) and \(v=1\); the control arm follows analogously. This explains consistency under the identification assumptions and appropriate sampling conditions, not exact finite-sample equality.

In finite samples, the denominator is random. A ratio involving random quantities is generally not equal to the ratio of their expectations, so Hájek estimates are not generally exactly unbiased just because the weighted totals have useful expectation identities. Conversely, an unusually large inverse-probability weight can make a Horvitz–Thompson total highly variable. Dividing by the realized weight sum can dampen the impact of a mismatch between the weight sum and its target, but can also introduce ratio bias. Which estimate performs better depends on the data-generating process and sample.

For the normalized weighted mean, positive weights also give a useful arithmetic check: the estimate must lie between the minimum and maximum observed outcomes among units with positive weight in that arm. If it does not, there is likely a calculation or denominator error. That check does not establish that the estimate is causally valid or precise.

Worked Example: A Normalized Mean Stays Within the Observed Range

Suppose three hypothetical control units used to estimate a treated-group counterfactual have outcomes \(3\), \(7\), and \(11\) points, and inverse-odds weights \(0.25\), \(1\), and \(4\), respectively. The normalized weighted mean is

$$ \frac{(0.25)(3)+(1)(7)+(4)(11)}{0.25+1+4} = \frac{0.75+7+44}{5.25} = \frac{51.75}{5.25} = 9.857\text{ points}. $$

This result is between the smallest outcome, \(3\) points, and the largest, \(11\) points. To verify why that must hold for positive weights, write the normalized weights as \(q_i=w_i/\sum_j w_j\). Then \(q_i>0\), \(\sum_i q_i=1\), and the estimate is \(\sum_iq_iY_i\). Since every \(Y_i\) lies between the minimum and maximum outcomes, their weighted average does too. The largest outcome receives most of the weight here, so the estimate is near \(11\), but remains a weighted mean rather than an unnormalized total.

Common Mistakes and Practical Interpretation

  • Calling the Hájek estimator unbiased by definition. Normalizing creates a ratio with a random denominator. It may have finite-sample bias; its attraction is not a universal guarantee of unbiasedness or lower error.
  • Dividing by the wrong quantity. For the ATE Hájek estimator, divide each arm’s weighted outcome total by that arm’s sum of weights. For the ATT form shown here, divide the control weighted total by the control inverse-odds weight sum. Do not substitute the total sample size or the treated count for these Hájek denominators.
  • Interpreting the weight sum as a sample size. A sum of weights is a normalization quantity, not the number of observed people. It can be fractional and can vary substantially across samples.
  • Taking a difference as proof of a causal effect. Weighting identifies a causal target only when the relevant identification assumptions hold and the weights are constructed appropriately. Normalization itself does not make confounding disappear.
Key takeaway: Horvitz–Thompson estimators use inverse-probability-weighted totals with a fixed scaling; Hájek estimators normalize each weighted total by its realized weight sum. Under the required identification assumptions they target the same population means, but their finite-sample values need not agree, and normalization is not automatically superior.

Check Your Understanding

Answer each question using the formulas and distinctions in this tutorial.

  1. For an ATE Hájek estimate, what denominator is used for the treated outcome total, and what denominator is used for the control outcome total?
  2. Why can the Horvitz–Thompson and Hájek estimates differ in a finite sample even when they use the same propensity scores?
  3. For the ATT, what inverse-odds weight is assigned to a control unit with propensity score \(e(X_i)\)?
  4. With positive weights, explain why a normalized weighted mean cannot exceed the largest observed outcome in its weighted group.
  5. Does Hájek normalization guarantee lower finite-sample bias or variance than Horvitz–Thompson weighting? Explain briefly.
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