Horvitz-Thompson Estimation of Exposure Effects

Why Weight by Exposure Probability?

With interference, one person’s outcome may depend on both that person’s treatment and other people’s treatments. An exposure mapping summarizes the parts of the assignment vector considered relevant for a unit. A randomized design may make some exposure conditions more likely for some units than for others. A simple average of the outcomes observed under one condition can then give disproportionate influence to units with a high chance of receiving it.

The Horvitz–Thompson (HT) estimator corrects this unequal representation by weighting an observed outcome inversely to the probability that its unit received the exposure of interest. The target is a finite-population mean: the average potential outcome for all units under a specified exposure condition, not merely the average among units observed in that condition.

Definition: Let $N$ be the number of units, $Z$ the randomized assignment vector, and $E_i(Z)$ the exposure assigned to unit $i$ by a specified exposure mapping. Let $Y_i(e)$ be unit $i$’s potential outcome under exposure $e$, and define its unit-level exposure probability as $\pi_i(e)=P(E_i(Z)=e)$, where the probability is determined by the randomization design. The finite-population exposure-specific mean is $$ \bar{Y}(e)=\frac{1}{N}\sum_{i=1}^{N}Y_i(e). $$

The exposure $e$ can be a combination, such as own treatment $z$ and a peer-exposure level $g$. In that notation, $Y_i(z,g)$ is the potential outcome at own treatment $z$ and peer exposure $g$, and $\bar{Y}(z,g)$ averages those potential outcomes over the same $N$ units.

The Horvitz–Thompson Estimator

Write $I\{E_i(Z)=e\}$ for an indicator that equals 1 if unit $i$ receives exposure $e$, and 0 otherwise. Under consistency, the observed outcome $Y_i^{obs}$ equals $Y_i(e)$ when $E_i(Z)=e$. The HT estimator of the finite-population mean is

\widehat{\bar{Y}}_{\mathrm{HT}}(e) = \frac{1}{N}\sum_{i=1}^{N} \frac{I\{E_i(Z)=e\}Y_i^{obs}}{\pi_i(e)}.

A unit not observed under $e$ contributes zero to that assignment’s estimate. A unit observed under $e$ contributes its outcome divided by its own probability of receiving $e$. The denominator is the population size $N$, not the number of units observed under the exposure and not the sum of the weights.

Conditions for design-unbiased estimation: The exposure mapping must define the relevant exposure conditions, and the consistency assumption must hold: a unit’s observed outcome is $Y_i(e)$ when its mapped exposure is $e$. Assignment is randomized according to a known design, so each $\pi_i(e)$ can be calculated. Finally, $\pi_i(e)>0$ for every unit whose potential outcome contributes to the target mean. This positivity condition ensures that every target unit has a chance to receive the exposure.

Randomization can make exposure assignments across units dependent. That does not prevent the estimator from being unbiased: the formula uses each unit’s marginal probability $\pi_i(e)$, and does not require unit exposures to be independent. Correct probabilities must come from the assignment design, however, not from the proportion of units who happened to receive an exposure in one realized assignment.

Why the Estimator Is Unbiased

The key calculation is unit-specific. For a fixed unit $i$, the exposure indicator is 1 with probability $\pi_i(e)$. When it is 1, consistency gives $Y_i^{obs}=Y_i(e)$. Therefore,

\begin{aligned} \mathbb{E}_Z\left[ \frac{I\{E_i(Z)=e\}Y_i^{obs}}{\pi_i(e)} \right] &= \frac{1}{\pi_i(e)} \mathbb{E}_Z\left[I\{E_i(Z)=e\}Y_i(e)\right]\\ &= \frac{1}{\pi_i(e)}\pi_i(e)Y_i(e)\\ &=Y_i(e). \end{aligned}

Here the expectation is over the randomization distribution, holding the units’ potential outcomes fixed. Taking the average over units and using linearity of expectation gives

\mathbb{E}_Z\left[\widehat{\bar{Y}}_{\mathrm{HT}}(e)\right] = \frac{1}{N}\sum_{i=1}^{N}Y_i(e) = \bar{Y}(e).

Thus the estimator is design-unbiased for the finite-population mean under the stated conditions. Unbiasedness is a property over repeated assignments from the design. It does not guarantee that one realized estimate will be close to its target, particularly when exposure probabilities are small.

Worked Example: Unequal Probabilities in Two Pairs

Suppose four units form two pairs, and each unit’s exposure is $(Z_i,Z_j)$, where $j$ is that unit’s partner. Treatments are assigned independently within and across pairs. In Pair A, the treatment probabilities are $P(Z_1=1)=0.8$ and $P(Z_2=1)=0.5$. In Pair B, they are $P(Z_3=1)=0.6$ and $P(Z_4=1)=0.25$. In a hypothetical assignment, the observed pair assignments are $(Z_1,Z_2)=(1,0)$ and $(Z_3,Z_4)=(0,1)$. The observed outcomes for units 1 through 4 are 20, 9, 11, and 12 points, respectively. Estimate the mean outcome under exposure $e=(1,0)$.

For unit 1, receiving $(1,0)$ means that unit 1 is treated and unit 2 is untreated. Independence within Pair A gives $\pi_1(1,0)=0.8(1-0.5)=0.4$. For unit 2, the corresponding probability is $\pi_2(1,0)=0.5(1-0.8)=0.1$.

For unit 3, the probability of own treatment and an untreated partner is $\pi_3(1,0)=0.6(1-0.25)=0.45$. For unit 4, it is $\pi_4(1,0)=0.25(1-0.6)=0.1$. All four units therefore have positive probability of receiving the target exposure.

The observed exposure is $(1,0)$ for units 1 and 4. Units 2 and 3 have exposure $(0,1)$, so they contribute zero to this estimate. With $N=4$,

\widehat{\bar{Y}}_{\mathrm{HT}}(1,0) = \frac{1}{4}\left( \frac{20}{0.4}+\frac{0}{0.1}+\frac{0}{0.45}+\frac{12}{0.1} \right) = \frac{50+120}{4} = 42.5\text{ points}.

The large estimate reflects the low probability that unit 4 receives this exposure: its observed outcome receives a weight of $1/0.1=10$. The result is not a claim that the average outcome is exactly 42.5 points. It is one randomized-design estimate of a finite-population mean; a single realization can be far from that target.

Direct and Spillover Contrasts

Once exposure-specific means have been estimated, contrasts can be formed by subtracting the relevant means. A direct effect compares own-treatment conditions while holding peer exposure fixed. A spillover effect compares peer-exposure conditions while holding own treatment fixed. These are contrasts between exposure-specific means in the same finite population.

Definition: For binary own treatment $z$ and peer exposure $g$, the finite-population direct contrast at fixed peer exposure $g$ is $$ \tau_D(g)=\bar{Y}(1,g)-\bar{Y}(0,g). $$ The spillover contrast at fixed own treatment $z$, comparing peer exposures $g_1$ and $g_0$, is $$ \tau_S(z;g_1,g_0)=\bar{Y}(z,g_1)-\bar{Y}(z,g_0). $$ Their HT estimates are obtained by replacing each exposure-specific mean with its HT estimate.

Worked Example: Direct and Spillover Contrasts in Four Pairs

Consider eight hypothetical units in four pairs. Treatment is independently assigned to every unit with probability $0.5$. A unit’s exposure is $(Z_i,Z_j)$, with its own treatment first and its partner’s treatment second. In the realized assignment, the four pairs have assignments $(1,1)$, $(0,0)$, $(1,0)$, and $(0,1)$. The observed outcomes, in unit order within those pairs, are respectively (18, 16), (10, 12), (15, 9), and (11, 13), measured in points. Estimate the direct contrast at peer treatment 0, both direct contrasts, and the spillover contrasts at each own-treatment level.

For every unit and every possible pair exposure, independence gives probability $\pi_i(z,g)=0.5(0.5)=0.25$. Thus, for each exposure condition, the HT estimate is the sum of the outcomes observed in that condition divided by $N(0.25)=8(0.25)=2$.

For exposure $(1,0)$, units 5 and 8 are observed, with outcomes 15 and 13. For $(0,0)$, units 3 and 4 are observed, with outcomes 10 and 12. The two means needed for the direct contrast at peer treatment 0 are

\widehat{\bar{Y}}_{\mathrm{HT}}(1,0) = \frac{1}{8}\left(\frac{15}{0.25}+\frac{13}{0.25}\right) = 14\text{ points}, \qquad \widehat{\bar{Y}}_{\mathrm{HT}}(0,0) = \frac{1}{8}\left(\frac{10}{0.25}+\frac{12}{0.25}\right) = 11\text{ points}.

The estimated direct contrast at peer treatment 0 is $14-11=3$ points. At peer treatment 1, units 1 and 2 are observed under $(1,1)$, and units 6 and 7 under $(0,1)$:

\widehat{\bar{Y}}_{\mathrm{HT}}(1,1)=\frac{18+16}{2}=17, \qquad \widehat{\bar{Y}}_{\mathrm{HT}}(0,1)=\frac{9+11}{2}=10.

The estimated direct contrast at peer treatment 1 is $17-10=7$ points. The two direct contrasts differ in this realization; they compare different exposure conditions and need not be equal.

For spillovers at own treatment 1, compare peer treatment 1 with peer treatment 0: $\widehat{\tau}_S(1;1,0)=17-14=3$ points. At own treatment 0, the same peer-exposure comparison gives $\widehat{\tau}_S(0;1,0)=10-11=-1$ point. The negative estimate means the estimated mean outcome under peer treatment 1 is 1 point lower than under peer treatment 0 among units held at own treatment 0. These are randomized-design estimates of contrasts, not proof that the outcomes of particular units would change by exactly those amounts.

Calculating Probabilities From the Design

The required probability is the probability that a particular unit receives the target mapped exposure, summing over assignment vectors that produce it. It is not always the probability of a single assignment vector. A mapping that groups multiple assignments into the same exposure condition requires adding their probabilities.

Worked Example: “Any Treated Peer” as an Exposure

Suppose a group has three units. Treatments are assigned independently with probability $0.5$, and a unit’s exposure is $(Z_i,G_i)$, where $G_i=1$ if at least one of its two peers is treated and $G_i=0$ otherwise. In a hypothetical assignment, the treatment vector is $(1,0,0)$, and the observed outcomes are 6, 4, and 5 points. Estimate the mean under exposure $(1,0)$.

For any unit, exposure $(1,0)$ occurs if that unit is treated and both peers are untreated. By independence,

\pi_i(1,0)=0.5(0.5)(0.5)=0.125.

For comparison, exposure $(1,1)$ requires own treatment and at least one of the two peers treated, so $\pi_i(1,1)=0.5[1-(0.5)^2]=0.375$. The probability calculation follows the exposure definition, not just the unit’s own treatment probability.

Under the realized assignment, unit 1 has exposure $(1,0)$; units 2 and 3 have exposure $(0,1)$. Therefore,

\widehat{\bar{Y}}_{\mathrm{HT}}(1,0) = \frac{1}{3}\left(\frac{6}{0.125}+0+0\right) = 16\text{ points}.

Only one unit contributes in this assignment, and the estimate is much larger than its observed outcome. That is compatible with the estimator’s definition: it uses the design probability to represent the units that could have received $(1,0)$, rather than treating the one observed unit as the entire target population.

Common Mistakes and Interpretation

Using the observed fraction as the probability. The exposure probability comes from the randomization design. A realized assignment’s exposure count is random and is not a substitute for $\pi_i(e)$.
Using one common probability when probabilities differ by unit. The HT formula uses $\pi_i(e)$ for each unit. Replacing these with a pooled probability generally changes the estimator.
Dividing by the number observed under the exposure. The HT mean has the fixed denominator $N$. Dividing by the observed count instead defines a different estimator.
Interpreting a zero contribution as a zero potential outcome. A unit not observed under $e$ contributes zero to that assignment’s calculation, but its unobserved $Y_i(e)$ need not be zero. Even if every unit has positive exposure probability, an exposure may be absent in one realized assignment.
Treating a large weight as extra information. A small $\pi_i(e)$ creates a large weight and can make an estimate highly variable. Unbiasedness does not guarantee precision for a particular assignment.
Changing the exposure mapping without changing the estimand. “Any treated peer” and “number of treated peers,” for example, define different exposure conditions and can produce different means and contrasts. The mapping must be stated and scientifically appropriate for the question.

The HT approach provides a direct connection between a randomized exposure design and finite-population causal contrasts under interference. Its central requirements are a well-defined exposure mapping, consistent exposure-specific potential outcomes, and correct positive unit-level probabilities. Direct and spillover estimates then follow by first estimating each relevant exposure-specific mean and subtracting the appropriate pair.

Check Your Understanding

Use the definitions and calculations in this tutorial to answer the following questions.

In the HT estimator, why is the contribution from a unit observed under exposure $e$ divided by $\pi_i(e)$?
A design gives two units different probabilities of receiving the same exposure. Which probability belongs in each unit’s contribution?
For an exposure $e=(z,g)$, write the direct contrast at fixed $g$ and the spillover contrast at fixed $z$ when comparing $g_1$ with $g_0$.
If an exposure has positive probability for every unit but is not observed in one realized assignment, what does its HT estimate equal in that assignment, and what does that value not imply?
Why can an HT estimate be much larger than all the outcomes observed under the target exposure in one randomized assignment?