Why Normalize Inverse-Probability Weights?
Inverse-probability weighting corrects for unequal chances of receiving a treatment: an observed outcome receives more weight when its treatment assignment was less likely. Under adaptive assignment, these probabilities can vary across participants as the trial learns from accumulating outcomes. The Horvitz–Thompson (HT) estimator uses the weights directly, while the Hájek estimator divides the weighted total by the sum of the weights. That normalization can make an estimate less sensitive to an unusually large total weight, but it also makes the denominator random.
The resulting tradeoff is important: under the assumptions below, the HT estimator is exactly unbiased for a finite-population mean, whereas the normalized Hájek estimator is generally not exactly unbiased in a finite sample. Neither estimator is uniformly preferable. Their behavior depends on the assignment probabilities, the outcomes, and whether the normalization denominator is stable.
Setup and Estimators
Consider a trial with \(N\) participants and a treatment arm \(a\). Let \(Y_i(a)\) be participant \(i\)'s potential outcome under arm \(a\), and let \(A_i\) be the arm actually assigned. The finite-population treatment mean is
This target treats the \(N\) potential outcomes as fixed. Let \(H_{i-1}\) denote the observed trial history just before participant \(i\) is assigned, and let \(p_i(a)=P(A_i=a\mid H_{i-1})\) be the known probability of assigning that participant to arm \(a\). The probability may depend on earlier assignments and outcomes; it is therefore allowed to change during the trial.
The Hájek estimator is sometimes described as a stabilized weighted estimator. Multiplying every weight in a given arm by the same nonzero constant leaves this ratio unchanged, because the constant cancels between numerator and denominator. This scaling invariance does not make the estimator unbiased: the denominator still varies with the assignments and their probabilities.
Assumptions and the HT Unbiasedness Argument
The key design condition is sequential randomization: conditional on the observed history before assignment, the assignment of participant \(i\) is made using the stated probabilities, not using that participant's unobserved potential outcomes. For the HT estimator of arm \(a\), positivity is also required: \(p_i(a)>0\) at every history that can occur for each participant in the target population. The assignment probabilities must be correctly recorded, and the observed outcome for an assigned participant must equal that participant's potential outcome under the assigned arm (consistency).
Under these conditions, condition on \(H_{i-1}\). The potential outcome \(Y_i(a)\) is fixed, and the chance of assignment to arm \(a\) is \(p_i(a)\). Thus
Taking expectations and summing the \(N\) contributions gives the finite-sample result:
The Hájek estimator does not follow from the same argument. It is a ratio of random quantities, and in general the expectation of a ratio is not the ratio of expectations. Its denominator is not fixed at \(N\), and the weighted outcomes in the numerator and the weights in the denominator are linked through the same assignments.
Worked Example: A Rare Assignment and a Random Denominator
Suppose \(N=2\), with potential outcomes \(Y_1(a)=0\) and \(Y_2(a)=10\). The target is \(\mu_a=(0+10)/2=5\). For illustration, assume participant 1 has assignment probability \(p_1(a)=0.90\), participant 2 has \(p_2(a)=0.10\), and their assignments are independent. These fixed probabilities are a special case of sequential assignment. Both probabilities are positive, and the hypothetical outcomes are fixed before randomization.
Participant 1 contributes zero to the HT estimate whatever their assignment. Participant 2 contributes \(10/0.10\) before division by \(N\) if assigned to \(a\). Therefore
Its expectation is \(50(0.10)=5\), exactly the target. Its possible values, 0 and 50, also show how a rare assignment can produce a highly variable realized estimate.
For the Hájek estimate, the assignment pattern matters. If only participant 1 is assigned to \(a\), the estimate is 0. If only participant 2 is assigned, it is 10. If both are assigned, it is
The probabilities of these three nonempty patterns are 0.81, 0.01, and 0.09, respectively: for example, the probability that only participant 1 is assigned is \(0.90(1-0.10)=0.81\). Conditional on at least one participant being assigned to \(a\), the Hájek expectation is
where \(D_a\) is the number assigned to arm \(a\). This conditional expectation is not the target 5. When neither participant is assigned to \(a\), the Hájek estimate is undefined; conditioning on a nonempty arm does not restore unbiasedness. If, purely to define an unconditional numerical illustration, one sets the estimate to zero when the denominator is zero, its expectation is \(0.91\), not 5. That value depends on the arbitrary zero convention.
What Normalization Changes in a Realized Sample
The next hypothetical example shows how the estimators can differ in one realized trial. Suppose four fixed potential outcomes under arm \(a\) are 2, 4, 6, and 8 units, so the finite-population mean is 5 units. The assignment probabilities for arm \(a\) are 0.80, 0.60, 0.20, and 0.10, respectively. Imagine that participants 1 and 3 are assigned to \(a\), and their observed outcomes are 2 and 6 units. All probabilities are positive; the probabilities and outcomes are assumed known or generated consistently with the stated design.
Worked Example: Direct HT and Hájek Calculations
The HT estimate divides the weighted total by the fixed population size:
The Hájek estimate divides that same weighted total by the sum of the observed weights:
Here the normalized estimate is closer to the target of 5 units, but this single realized comparison does not establish that it is better in repeated trials. The HT estimate remains unbiased over the assignment mechanism under the stated assumptions; the Hájek estimate has no general finite-sample unbiasedness guarantee. A different realization could reverse which estimate is closer.
Finite-Sample Bias and Variability
Normalization can reduce the influence of an unusually large total weight. When only a few participants receive a treatment, one or two small assignment probabilities can make the HT total large. The Hájek denominator rescales that total according to the observed weight sum, which may yield a more stable-looking estimate. But this rescaling can shift the center of the estimator: the denominator and numerator co-vary, and the ratio can be biased even when each weighted contribution has the right expectation.
A small calculation makes the tradeoff concrete. Suppose two participants have potential outcomes 0 and 10 under arm \(a\), each with assignment probability 0.50, independently. The target mean is 5. The HT estimate is \(10\mathbb{1}(A_2=a)\), so it is 0 or 10 with equal probability, has expectation 5, and has variance \(25\) squared units. The Hájek estimate is 0 if only participant 1 is assigned, 10 if only participant 2 is assigned, and 5 if both are assigned. It is undefined if neither is assigned. Among nonempty-arm trials, its mean is 5 and its variance is \((25+25+0)/3=50/3\) squared units. This conditional comparison suggests lower variability in this example, but it also changes the conditioning event: the HT result above is unconditional, while the Hájek calculation excludes trials with an empty arm. If the Hájek estimate is set to zero for the empty-arm case, it has expectation \((0+0+10+5)/4=3.75\), introducing bias.
Thus, “stabilized” should not be read as “always less variable,” “unbiased,” or “safer for inference.” Normalization may reduce variability in some designs and samples, but its bias can offset that gain. A small or highly variable denominator is a warning sign: the ratio may be unstable, and the probability of an empty arm may be substantial.
Diagnostics, Alternatives, and Limitations
Before using either estimator, inspect the assignment probabilities and the realized weights \(1/p_i(a)\) among participants assigned to arm \(a\). Check for probabilities near zero, a few weights that dominate the weighted total, and histories in which an arm has little chance of receiving new participants. For Hájek estimation, also inspect the denominator \(\sum_i\mathbb{1}(A_i=a)/p_i(a)\) and how often it is zero or unusually small. These checks diagnose potential instability; they do not by themselves correct it.
The main alternatives answer related but not identical needs:
| Estimator | Normalization | Finite-sample feature | Main concern |
|---|---|---|---|
| HT | Divide weighted total by fixed \(N\) | Unbiased for the finite-population mean under the stated design assumptions | Can be highly variable when some assignment probabilities are small |
| Hájek | Divide weighted total by observed weight sum | Scale-invariant and often less affected by an extreme total weight | Generally biased in finite samples; undefined if the denominator is zero |
| Ordinary arm mean | Divide observed outcomes by the number assigned to the arm | Simple description of outcomes among those assigned | Does not generally correct for unequal adaptive assignment probabilities |
The ordinary arm mean can be useful as a descriptive summary of the participants assigned to that arm, but under unequal probabilities it does not generally target the finite-population mean. Neither HT unbiasedness nor Hájek normalization compensates for incorrect probabilities, violations of the assignment mechanism, or a failure to observe outcomes. Positivity is especially consequential: if a participant has zero probability of assignment to arm \(a\), inverse weighting cannot recover that participant's contribution from the trial.
This tutorial compares point estimators, not the procedures for estimating their variances or constructing tests and intervals. In particular, lower variability in an illustrative calculation is not sufficient to establish good uncertainty quantification. The choice of estimator should follow the stated estimand and design, with its bias and variability assessed under the actual adaptive policy.
Common Mistake
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- For three participants with fixed potential outcomes 3, 6, and 9 units under arm \(a\), suppose their assignment probabilities are 0.50, 0.25, and 0.75. What is the target mean? State the assumptions needed for HT unbiasedness.
- In a hypothetical trial, one participant with outcome 4 units and assignment probability 0.50 is assigned to arm \(a\), and no other participant is assigned to that arm. The arm’s total population size is \(N=2\). Calculate the HT contribution to the mean estimate and the Hájek estimate.
- Why does a zero Hájek denominator require special treatment? Does assigning zero to that case preserve the estimator’s general unbiasedness?
- Give one circumstance in which normalization might help and one reason it might still produce a poor estimate.
Solutions. (1) The target is \((3+6+9)/3=6\) units. HT unbiasedness requires sequential randomization using the stated probabilities, positive assignment probability for each participant, consistency, and correctly known probabilities. (2) The HT estimate is \(\{4/0.50\}/2=4\) units. The Hájek estimate is \(4/0.50\) divided by \(1/0.50\), or 4 units. (3) With no assigned participants, both its weighted outcome sum and weight sum are zero, so the ratio is undefined. Setting it to zero is a convention, not a general unbiasedness result; it can introduce bias, as the examples show. (4) Normalization may limit the effect of an unusually large realized total weight. It can still perform poorly if the denominator is zero or unstable, or if the ratio’s finite-sample bias is substantial.