Hájek Estimation and Variance for Exposure Effects

Why Normalize an Exposure-Weighted Estimate?

Under interference, one person’s outcome may depend both on their own treatment and on the treatments assigned to others. An exposure mapping summarizes the assignment features relevant to each person; an exposure condition might, for example, specify a person’s own treatment and the proportion of their peers treated. Once exposure conditions are defined, a central target is the average outcome that the finite population would have under a particular exposure condition.

The Horvitz–Thompson (HT) estimator weights each observed outcome by the inverse probability of that unit receiving the exposure, then divides by the population size. This makes it design-unbiased when exposure probabilities are positive, but a realized estimate can be unstable if a few observed units receive very large weights. The Hájek estimator instead divides the weighted outcome total by the weighted count of exposed units. This ratio normalization can be more stable, but it is not generally exactly unbiased, and its variance needs to account for the random denominator.

Definition: Let $Z$ be the randomized assignment, and let $E_i(Z)$ be unit $i$'s exposure under the specified exposure mapping. Write $Y_i(g)$ for unit $i$'s fixed potential outcome under exposure $g$, and define $\pi_i(g)=P(E_i(Z)=g)$. For $N$ units, the finite-population mean under exposure $g$ is $\mu(g)=N^{-1}\sum_{i=1}^N Y_i(g)$. The exposure indicator is $I_i(g)=1$ if $E_i(Z)=g$, and $0$ otherwise.

The design is the source of randomness in these definitions: the potential outcomes are fixed, while assignment and exposure indicators vary. The exposure mapping and the potential-outcome model must be aligned: for each unit, the outcome under an assignment is represented by the potential outcome for the exposure that assignment induces. Identification also requires $\pi_i(g)>0$ for every unit whose outcome contributes to $\mu(g)$. This is exposure positivity.

HT and Hájek Estimators

The HT estimator estimates the population total under exposure $g$, then divides by $N$. The Hájek estimator uses the same inverse-probability weights but estimates the total exposed population with their sum of weights, rather than fixing that denominator at $N$.

Definition: The HT estimator of $\mu(g)$ is $\hat{\mu}_{HT}(g)=N^{-1}\sum_i I_i(g)Y_i(g)/\pi_i(g)$. The Hájek estimator is $\hat{\mu}_{H}(g)=\{\sum_i I_i(g)Y_i(g)/\pi_i(g)\}/\{\sum_i I_i(g)/\pi_i(g)\}$, provided its denominator is positive. Only outcomes under assigned exposures are observed, so each displayed outcome term is calculated from observed outcomes when $I_i(g)=1$.

The expected HT weighted count $\sum_i I_i(g)/\pi_i(g)$ is $N$. The Hájek denominator is its realized value, so the Hájek estimate is a weighted average of observed outcomes and typically stays within their observed range. That can make it less sensitive to an unusually large weighted total. The tradeoff is that a ratio of random quantities is not generally unbiased, even though the weighted numerator and denominator each have the stated expectations. If no unit receives exposure $g$, the Hájek estimate is undefined; positive exposure probabilities alone do not prevent this in a particular assignment.

Worked Example: Calculating Both Exposure-Mean Estimates

Suppose a hypothetical randomized study has $N=3$ units. In the realized assignment, all three receive exposure $g$. Their observed outcomes under $g$ are 10, 8, and 6 units, and their known exposure probabilities are 0.50, 0.25, and 0.75, respectively.

For the HT estimate, the weighted outcome total is $10/0.50+8/0.25+6/0.75=20+32+8=60$. Thus $\hat{\mu}_{HT}(g)=60/3=20$ outcome units.

For the Hájek estimate, the weighted count is $1/0.50+1/0.25+1/0.75=2+4+4/3=22/3$. Therefore $\hat{\mu}_{H}(g)=60/(22/3)=90/11\approx8.182$ outcome units. The denominator is $22/3$, not 7: the third unit contributes $1/0.75=4/3$.

The HT estimate is unbiased over the randomization distribution, not required to lie within the range of the three observed outcomes. In this realization it is 20, while the Hájek estimate is a weighted average of those outcomes. Neither estimate by itself establishes a causal effect; an effect estimate requires a contrast between exposure-specific means.

Joint Exposure Probabilities and HT Variance

Assignments can create dependence between units’ exposure indicators. For example, a fixed treatment total can make it impossible for two units to receive a particular exposure simultaneously. Define the joint exposure probability $\pi_{ij}(g,h)=P(E_i(Z)=g,E_j(Z)=h)$. This quantity, rather than the product $\pi_i(g)\pi_j(h)$, determines whether the two exposure indicators are dependent.

For the HT estimator, the design variance of an exposure-specific mean is exactly expressible using these joint probabilities. Here the sum is over ordered pairs, including $i=j$; for a diagonal term, $\pi_{ii}(g,g)=\pi_i(g)$.

Design-based variance: For an exposure condition $g$, $$ \operatorname{Var}\{\hat{\mu}_{HT}(g)\} = \frac{1}{N^2} \sum_{i=1}^N\sum_{j=1}^N \frac{\pi_{ij}(g,g)-\pi_i(g)\pi_j(g)} {\pi_i(g)\pi_j(g)} Y_i(g)Y_j(g). $$

This formula follows by writing the estimator’s error as a sum of centered exposure indicators: $\hat{\mu}_{HT}(g)-\mu(g)=N^{-1}\sum_i\{I_i(g)/\pi_i(g)-1\}Y_i(g)$. Taking its variance expands into covariances between every pair of these terms. For each pair, the covariance of the indicators is $\pi_{ij}(g,g)-\pi_i(g)\pi_j(g)$; dividing by the two marginal probabilities gives the factor in the displayed formula. The terms for $i\ne j$ are precisely where dependence between units enters.

The formula is an exact variance, but it involves potential outcomes that may not all be observed. A pairwise HT variance estimator can estimate it when the required joint probabilities are positive: it weights observed pairs by their inverse joint exposure probability. When some $\pi_{ij}(g,g)=0$, the corresponding pair is never jointly observed, so that direct estimator cannot recover its contribution without additional assumptions. Thus, knowing all the individual exposure probabilities is not by itself enough to guarantee an unbiased, estimable variance.

Worked Example: Dependence Under Mutually Exclusive Exposures

Consider a hypothetical population of two units and an assignment design that gives exactly one of them exposure $g$. Each unit has probability $0.50$ of receiving $g$, so $\pi_1(g)=\pi_2(g)=0.50$, but $\pi_{12}(g,g)=0$. Suppose the fixed potential outcomes under $g$ are 4 and 10 units.

If unit 1 receives $g$, the HT estimate of the population mean is $(4/0.50)/2=4$. If unit 2 receives $g$, it is $(10/0.50)/2=10$. Each assignment has probability $0.50$, and the true mean is $(4+10)/2=7$. The design variance is therefore $\{0.50(4-7)^2+0.50(10-7)^2\}=9$ square outcome units.

The joint-probability formula gives the same result. The two diagonal contributions to the variance of the estimated total are $4^2$ and $10^2$, since $(0.50-0.25)/0.25=1$. Each off-diagonal contribution is $(-0.25)(4)(10)/(0.50\cdot0.50)=-40$. The total variance for the estimated total is $16+100-40-40=36$; dividing by $N^2=4$ gives $9$.

The negative off-diagonal terms reflect the design: if one unit receives $g$, the other cannot. Although this hypothetical calculation uses both fixed potential outcomes to show the exact variance, in an actual experiment only one is observed. Because the joint probability is zero, the standard pairwise variance estimator cannot observe their product; in general, design-unbiased variance estimation may then require additional structure.

Hájek Variance by Linearization

For a Hájek estimate, the random weighted denominator prevents the exact HT variance expression from applying directly. A first-order Taylor expansion gives a useful design-based approximation. Write $A_g=\sum_i I_i(g)Y_i(g)/\pi_i(g)$ and $B_g=\sum_i I_i(g)/\pi_i(g)$, so that $\hat{\mu}_H(g)=A_g/B_g$. Both $A_g$ and $B_g$ have expectations $N\mu(g)$ and $N$, respectively.

Linearization: When $B_g$ is near $N$, a first-order expansion of the ratio around $(N\mu(g),N)$ gives $$ \hat{\mu}_H(g)-\mu(g) \ \approx\ \frac{1}{N}\sum_i \left\{\frac{I_i(g)}{\pi_i(g)}-1\right\} \{Y_i(g)-\mu(g)\}. $$

To see why the outcomes are centered, the first-order ratio expansion is $\hat{\mu}_H(g)-\mu(g)\approx\{(A_g-N\mu(g))-\mu(g)(B_g-N)\}/N$. Combining the two terms inside the numerator gives $\sum_i\{I_i(g)/\pi_i(g)\}\{Y_i(g)-\mu(g)\}$. Since $\sum_i\{Y_i(g)-\mu(g)\}=0$, this is also the centered-indicator expression in the display. Its variance has the same joint-probability structure as the HT variance, but uses residuals $Y_i(g)-\mu(g)$ instead of the uncentered outcomes. Replacing the unknown mean with the Hájek estimate and $N$ with $B_g$ gives a common plug-in linearization variance estimate:

\widehat{\operatorname{Var}}_{\mathrm{lin}}\{\hat{\mu}_{H}(g)\} = \frac{1}{B_g^2} \sum_{i=1}^N\sum_{j=1}^N I_i(g)I_j(g) \frac{\pi_{ij}(g,g)-\pi_i(g)\pi_j(g)} {\pi_{ij}(g,g)} \frac{\{Y_i(g)-\hat{\mu}_H(g)\}\{Y_j(g)-\hat{\mu}_H(g)\}} {\pi_i(g)\pi_j(g)}.

This is a linearization-based estimate, not an exact finite-sample variance formula for the ratio. It requires positive joint probabilities for the observed pairs used in the sum; as with HT variance estimation, zero joint probabilities can prevent direct estimation. The estimate can also be negative in finite samples, which signals a limitation of this plug-in approximation, not a negative true variance. A standard error is the square root only when the estimated variance is nonnegative.

Worked Example: A Linearized Hájek Standard Error

Suppose a hypothetical finite population has $N=3$ units, and each independently receives exposure $g$ with probability $0.50$. In one realized assignment, units 1 and 2 receive $g$, with observed outcomes 4 and 10; unit 3 does not receive $g$. Independence gives $\pi_{ij}(g,g)=0.25$ for distinct units, while the diagonal probabilities are 0.50. All these pair probabilities are positive.

The weighted outcome total is $4/0.50+10/0.50=28$, and the weighted count is $1/0.50+1/0.50=4$. Thus $\hat{\mu}_H(g)=28/4=7$. The observed residuals are $4-7=-3$ and $10-7=3$.

For distinct units, the covariance factor $\pi_{ij}-\pi_i\pi_j$ is $0.25-0.25=0$, so the off-diagonal contributions vanish. Each diagonal contribution has pair factor $(0.50-0.25)/0.50=0.50$. For either observed unit, the residual divided by its exposure probability is $-3/0.50=-6$ or $3/0.50=6$. The sum of the diagonal terms is therefore $0.50(36)+0.50(36)=36$. Dividing by $B_g^2=4^2=16$ gives an estimated linearized variance of $2.25$ square outcome units and a linearized standard error of $\sqrt{2.25}=1.50$ outcome units.

This calculation is an approximation based on the ratio linearization and the observed residuals; it is not the exact randomization variance of the Hájek ratio. The independence assumption is justified here by the stated design. Under a design with shared assignments, the off-diagonal terms would generally not vanish and must be calculated from the actual joint exposure probabilities.

Exposure Contrasts and Practical Interpretation

An exposure effect commonly compares two conditions, such as $\tau(g,h)=\mu(g)-\mu(h)$. For HT estimates, its variance is the variance under $g$, plus the variance under $h$, minus twice their covariance. That covariance depends on cross-exposure probabilities $\pi_{ij}(g,h)=P(E_i=g,E_j=h)$, including cases where the two exposures concern the same unit. Therefore, adding two exposure-specific standard errors as if the estimates were independent is generally unjustified.

For a Hájek contrast, the same principle applies to the linearized estimates: include the two variance estimates and subtract twice their estimated covariance. The covariance calculation uses residuals centered at their respective exposure-specific means, the joint probabilities for one unit exposed to $g$ and another to $h$, and the corresponding weighted denominators $B_g$ and $B_h$. If the assignment design makes exposure indicators dependent, those cross-terms carry that dependence into the uncertainty calculation.

Hájek normalization can be useful when the realized weighted exposure count differs substantially from $N$, but it does not repair a poorly supported exposure condition. Very small probabilities create large weights; rare or mutually incompatible joint exposures can also make uncertainty difficult to estimate. The point estimate, the exposure probabilities, the joint exposure probabilities, and any limitations of variance estimation should therefore be reported together.

Common Mistakes

Dropping the Hájek denominator. Dividing the weighted outcome total by $N$ gives the HT estimate, not the Hájek estimate. For the Hájek estimate, sum every observed unit’s inverse probability in the denominator.
Assuming exposure indicators are independent. Shared assignments can make joint exposure probabilities differ from products of marginal probabilities. Use $\pi_{ij}(g,h)$ from the design rather than substituting $\pi_i(g)\pi_j(h)$ without justification.
Calling the Hájek estimate unbiased. The weighted numerator and denominator have useful expectations, but their ratio is generally not unbiased. Describe its variance approximation as linearization-based.
Ignoring zero joint probabilities. A unit may have positive probability of an exposure even when a particular pair of units can never receive it together. A pairwise variance estimator then cannot directly estimate that pair’s contribution.

Key takeaway: HT estimates normalize by the known population size; Hájek estimates normalize by a random, inverse-probability-weighted exposure count. For either estimator’s uncertainty under interference, individual exposure probabilities are not enough: dependence induced by shared assignments is represented by joint exposure probabilities.

Check Your Understanding

Answer each question using the definitions and design-based reasoning developed above.

In what way does the Hájek estimator’s denominator differ from the HT estimator’s denominator?
For three observed outcomes 12, 9, and 6 with exposure probabilities 0.50, 0.50, and 1.00, calculate the weighted outcome total and weighted count used in the Hájek estimate.
Why can two units’ exposure indicators be dependent even when both units have positive marginal exposure probabilities?
What design feature makes the pairwise HT variance estimator unavailable for a pair of units?
Why is the linearized Hájek variance estimate not generally an exact finite-sample variance of the ratio?