Inverse-Probability Weighting Under Adaptive Allocation

Why Weight by the Allocation Probability?

In an adaptively randomized trial, the probability of receiving treatment can differ across participants because assignments depend on information accumulated during the trial. An ordinary comparison of the observed treatment and control averages does not account for those unequal probabilities. Inverse-probability weighting corrects for this by giving more weight to outcomes from participants who had a lower probability of receiving the treatment they received.

The correction has a precise target and assumptions. It does not automatically remove every bias caused by adaptation, and it can increase variance substantially. The key is to use the allocation probability that was known immediately before each participant’s assignment, not a probability reconstructed from the final data.

Definition: Let \(A_i=1\) denote assignment of participant \(i\) to treatment and \(A_i=0\) assignment to control. Let \(Y_i(1)\) and \(Y_i(0)\) be that participant’s potential outcomes under treatment and control, respectively. The observed outcome is \(Y_i=A_iY_i(1)+(1-A_i)Y_i(0)\). For a fixed cohort of \(N\) participants, the finite-cohort treatment effect is \(\tau_N=N^{-1}\sum_{i=1}^N\{Y_i(1)-Y_i(0)\}\).

Write \(\mathcal{F}_{i-1}\) for the information available just before participant \(i\) is randomized, including the trial history and any relevant baseline information. The known adaptive probability of treatment is \(\pi_i=P(A_i=1\mid\mathcal{F}_{i-1})\); the probability of control is \(1-\pi_i\). The probabilities may vary with the history, but must be recorded before assignment.

Assumptions: The treatment and control outcomes are well-defined for every participant, with no interference between participants, and the observed outcome equals the potential outcome under the assigned treatment (consistency). Conditional on \(\mathcal{F}_{i-1}\), the randomization uses the stated probability \(\pi_i\), independently of participant \(i\)’s potential outcomes. Both treatments have positive probability: \(0<\pi_i<1\). For variance calculations, the potential outcomes must have finite second moments.

Positivity is not merely a technical detail. If \(\pi_i=0\), a treatment outcome for participant \(i\) can never be observed, and its inverse weight is undefined. If \(\pi_i\) is positive but very close to zero, a treated participant who does occur receives a very large weight. The estimator remains unbiased under the stated randomization assumptions, but may be too unstable to be useful.

Constructing the Weighted Estimator

For each participant, the weighted contribution to the treatment mean is \(A_iY_i/\pi_i\). If the participant was treated, the observed outcome is divided by the probability of treatment; if the participant was not treated, the contribution is zero. The control contribution is \((1-A_i)Y_i/(1-\pi_i)\). Averaging each contribution over the full cohort gives the Horvitz–Thompson estimators:

\hat{\mu}_1^{HT}=\frac{1}{N}\sum_{i=1}^{N}\frac{A_iY_i}{\pi_i}, \qquad \hat{\mu}_0^{HT}=\frac{1}{N}\sum_{i=1}^{N}\frac{(1-A_i)Y_i}{1-\pi_i}, \qquad \hat{\tau}^{HT}=\hat{\mu}_1^{HT}-\hat{\mu}_0^{HT}.

The superscript \(HT\) indicates the Horvitz–Thompson form: each weighted total is divided by the known cohort size \(N\), rather than by the sum of the observed weights. The treatment-effect estimate is the difference between the two weighted means.

Worked Example: Estimating a Treatment Effect With Unequal Probabilities

Suppose a hypothetical trial enrolls four participants. Their allocation probabilities are set using the trial history available before each assignment. The table gives each participant’s probability of treatment, assignment, and observed outcome.

Participant	\(\pi_i\)	Assignment	Observed outcome
1	0.50	Treatment	8
2	0.75	Treatment	10
3	0.25	Control	4
4	0.50	Control	6

The allocation probabilities are known and lie strictly between zero and one. For the treatment mean, only participants 1 and 2 contribute:

\hat{\mu}_1^{HT} =\frac{1}{4}\left(\frac{8}{0.50}+\frac{10}{0.75}\right) =\frac{16+13.333}{4} =7.333.

For the control mean, the relevant probabilities are \(1-\pi_i\), so:

\hat{\mu}_0^{HT} =\frac{1}{4}\left(\frac{4}{1-0.25}+\frac{6}{1-0.50}\right) =\frac{5.333+12}{4} =4.333.

The estimated treatment effect is \(7.333-4.333=3.000\) outcome units. This is an estimate for the four-participant cohort, not the unweighted difference between observed arm averages, which is \(9-5=4\) units. A single realized estimate need not equal the true cohort effect; the unbiasedness result concerns repeated randomization under the specified design.

Why the Estimator Is Unbiased

The derivation uses the known probability for the current assignment. Conditional on the pre-assignment history, \(Y_i(1)\) and \(Y_i(0)\) are fixed potential outcomes, while the random assignment has probability \(\pi_i\) of treatment. By consistency, \(A_iY_i=A_iY_i(1)\). Therefore:

E\!\left[\frac{A_iY_i}{\pi_i}\,\middle|\,\mathcal{F}_{i-1}\right] =\frac{Y_i(1)}{\pi_i}E[A_i\mid\mathcal{F}_{i-1}] =Y_i(1).

Likewise, \(E[1-A_i\mid\mathcal{F}_{i-1}]=1-\pi_i\), so the weighted control contribution has conditional expectation \(Y_i(0)\). Subtracting the two expectations and averaging over participants yields the result:

Result: Under the stated sequential randomization, consistency, and positivity assumptions, \(E(\hat{\mu}_1^{HT})=\mu_1\), \(E(\hat{\mu}_0^{HT})=\mu_0\), and \(E(\hat{\tau}^{HT})=\tau_N\), where \(\mu_a=N^{-1}\sum_iY_i(a)\). These expectations average over the trial’s random assignments; the potential outcomes and cohort are treated as fixed.

Adaptation does not invalidate this argument simply because later probabilities depend on earlier assignments or outcomes. The conditional expectation is taken at each assignment, given the history that determined that assignment’s probability. The argument does require that the recorded probability is the actual probability used by the randomization procedure.

Variance: The Cost of Unequal Probabilities

For a single participant, define the weighted contrast contribution \(D_i=A_iY_i/\pi_i-(1-A_i)Y_i/(1-\pi_i)\). Its conditional mean is \(Y_i(1)-Y_i(0)\). Its conditional second moment follows from the fact that treatment and control are mutually exclusive:

E(D_i^2\mid\mathcal{F}_{i-1}) =\frac{Y_i(1)^2}{\pi_i}+\frac{Y_i(0)^2}{1-\pi_i}.

Subtracting the square of the conditional mean gives the conditional variance:

\operatorname{Var}(D_i\mid\mathcal{F}_{i-1}) =\frac{Y_i(1)^2}{\pi_i}+\frac{Y_i(0)^2}{1-\pi_i} -\{Y_i(1)-Y_i(0)\}^2.

This expression shows why extreme probabilities can be costly: the treatment-outcome term is divided by \(\pi_i\), and the control-outcome term by \(1-\pi_i\). Under sequential randomization, each contribution minus its conditional mean has conditional expectation zero. These successive mean-zero increments are uncorrelated, so for the fixed-cohort estimator the design variance is the sum of their expected conditional variances, divided by \(N^2\):

\operatorname{Var}(\hat{\tau}^{HT}) =\frac{1}{N^2}\sum_{i=1}^{N} E\!\left[ \frac{Y_i(1)^2}{\pi_i}+\frac{Y_i(0)^2}{1-\pi_i} -\{Y_i(1)-Y_i(0)\}^2 \right].

Here the expectation also averages over the histories that can arise in the adaptive trial, and thus over the corresponding history-dependent probabilities. The formula is a design-variance expression, not automatically an estimable standard error: for each participant, one of the two potential outcomes is unobserved. Estimating variance in practice requires an appropriate method and may require additional modeling or conservative procedures.

Worked Example: Comparing Variability at Two Allocation Probabilities

For illustration, suppose one hypothetical participant has fixed potential outcomes \(Y_i(1)=10\) and \(Y_i(0)=6\). Compare the conditional variance of the weighted contrast contribution when \(\pi_i=0.50\) and when \(\pi_i=0.90\). The treatment effect for this participant is \(10-6=4\) outcome units.

With equal allocation:

\operatorname{Var}(D_i\mid\mathcal{F}_{i-1}) =\frac{10^2}{0.50}+\frac{6^2}{0.50}-4^2 =200+72-16 =256

in squared outcome units. With a 0.90 probability of treatment:

\operatorname{Var}(D_i\mid\mathcal{F}_{i-1}) =\frac{10^2}{0.90}+\frac{6^2}{0.10}-4^2 =111.111+360-16 =455.111.

The more treatment-favoring allocation gives a larger variance for this weighted contrast: control is uncommon, and its observed outcomes carry large inverse weights. For an \(N\)-participant estimator, each participant’s contribution is scaled by \(1/N\), so its variance contribution is scaled by \(1/N^2\). The calculation illustrates a general risk, not a claim that equal allocation is always optimal; the best allocation depends on the design goal and outcome distributions.

Diagnostics and Alternative Weighting

A useful review begins with the allocation probabilities themselves. Check that each recorded probability matches the randomization algorithm and the information available at assignment. Summarize the smallest and largest probabilities, inspect the resulting inverse weights, and identify whether a few participants dominate the weighted totals. Also check that the outcome is recorded consistently with the stated treatment assignment and that the intended target is the enrolled cohort, not a post hoc subset defined by trial results.

The Horvitz–Thompson estimator is not the only way to use inverse probabilities. A Hájek estimator divides the weighted outcome total by the sum of weights in that arm. This ratio often stabilizes the scale of the estimate, but because its denominator is random, it is generally not exactly unbiased in finite samples. The ordinary unweighted mean is simpler, but does not correct for unequal allocation probabilities in the same way. These alternatives trade properties; none removes the need to examine the design and the weights.

Worked Example: Horvitz–Thompson Versus Hájek Estimates

Using the four hypothetical participants above, the Horvitz–Thompson treatment mean is \(29.333/4=7.333\). The sum of treatment-arm weights is \(1/0.50+1/0.75=3.333\), so the Hájek treatment mean is:

\hat{\mu}_1^{H} =\frac{8/0.50+10/0.75}{1/0.50+1/0.75} =\frac{29.333}{3.333} =8.800.

For control, the weighted outcome total is \(17.333\), and its weight sum is \(1/0.75+1/0.50=3.333\). Thus \(\hat{\mu}_0^{H}=17.333/3.333=5.200\), giving a Hájek difference of \(8.800-5.200=3.600\) outcome units. It differs from the Horvitz–Thompson estimate of \(3.000\) units because the Hájek estimator uses random arm-specific weight sums as denominators. In this small example, neither observed difference should be mistaken for the unknown true effect.

Common warning signs include extreme weights, large changes in the estimate after removing one heavily weighted observation, and very small effective sample sizes in one arm. Truncating weights can reduce variability, but changes the estimator and can introduce bias; it is not a harmless repair. Likewise, replacing the recorded probabilities with estimates based on the final trial data can undermine the design-based justification above. Inverse-probability weighting corrects for unequal known assignment probabilities; it does not by itself resolve every source of selection or analysis bias.

Key takeaway: Use each participant’s known, pre-assignment probability to form the weighted contribution. This yields an unbiased Horvitz–Thompson treatment-effect estimate for the fixed cohort under sequential randomization and positivity, while small allocation probabilities can produce severe variance and instability.

Check Your Understanding

Answer each question, then compare with the solutions.

A participant has probability \(0.20\) of treatment and is assigned treatment with observed outcome 5. What is the participant’s contribution to the treatment weighted total before division by \(N\)?
Why does the unbiasedness argument still apply when \(\pi_i\) depends on earlier trial outcomes?
What happens to the control-outcome term in the variance expression as \(\pi_i\) approaches 1?
Why is a Hájek estimate not generally exactly unbiased in finite samples?

Solutions. (1) The contribution is \(5/0.20=25\). (2) Conditional on the pre-assignment history, the assignment still has the recorded probability \(\pi_i\); the conditional expectation of the weighted contribution is \(Y_i(1)\). (3) The term \(Y_i(0)^2/(1-\pi_i)\) can become very large when \(Y_i(0)\ne 0\), increasing variability. (4) Its denominator is a random sum of weights, so the expectation of the ratio is not generally the ratio of expectations.