Why Estimate Spillovers With Assignment Probabilities?
In a randomized network study, a person's outcome may depend not only on their own assignment but also on the assignments of people connected to them. An exposure mapping summarizes the assignment information considered relevant for each unit. To estimate the mean outcome under a particular mapped exposure, we need to account for the fact that units can have different probabilities of receiving that exposure. A unit with a low probability of receiving it represents more units, on average, than a unit with a high probability.
The Horvitz–Thompson approach uses the known randomization mechanism to calculate each unit's exposure probability, then weights the observed outcome by the inverse of that probability. This gives an unbiased estimate of a finite-population exposure mean under stated conditions. Contrasting two such means estimates a spillover effect when the exposure conditions hold own treatment fixed and vary peer exposure.
Exposure-Specific Outcomes and Estimands
Index the \(N\) units by \(i=1,\ldots,N\). Let \(\mathbf{Z}\) denote the vector of treatment assignments generated by the design, and let \(G_i(\mathbf{Z})\) be the exposure mapping for unit \(i\). The mapping could record own treatment and, for example, whether any of the unit's neighbors are treated. It reduces a potentially large assignment vector to a specified exposure category.
Using \(Y_i(e)\) assumes that the exposure mapping is sufficient for the outcome: whenever two assignment vectors give unit \(i\) the same mapped exposure \(e\), they give that unit the same potential outcome. This is an exposure-mapping assumption, not a claim that there is no interference. Interference is allowed within the features captured by the mapping.
An exposure contrast is a difference between two exposure means, \(\tau(e,e')=\mu(e)-\mu(e')\). For a spillover contrast at fixed own treatment, \(e\) and \(e'\) have the same own-treatment value but differ in the peer-exposure feature. For example, one might compare untreated units with at least one treated neighbor to untreated units with no treated neighbors.
The Horvitz–Thompson Exposure Mean
Let \(I_i(e)=1\{G_i(\mathbf{Z})=e\}\) indicate whether unit \(i\) actually receives exposure \(e\). The observed outcome for unit \(i\) is \(Y_i=Y_i(G_i(\mathbf{Z}))\). If \(I_i(e)=1\), consistency gives \(Y_i=Y_i(e)\). A unit not observed under \(e\) contributes zero to the exposure-specific sum in that assignment realization, but its chance of contributing is corrected by dividing by \(\pi_i(e)\).
This estimator requires \(\pi_i(e)>0\) for every unit in the target population and exposure mean. It also requires the assignment mechanism—and therefore the probabilities \(\pi_i(e)\)—to be correctly specified. Randomization makes these probabilities known in a designed experiment. A contrast is estimated by subtracting the two exposure-mean estimates:
The estimate is not generally the average outcome among units observed in exposure \(e\). That raw average gives each observed unit equal weight, even if their probabilities of receiving \(e\) differ. The Horvitz–Thompson estimator instead targets the mean across all \(N\) units, including those not observed under that exposure in the realized assignment.
Why the Weighting Works
The key justification is that each unit's exposure indicator has expectation \(\pi_i(e)\). For a fixed unit, the potential outcome \(Y_i(e)\) is fixed; randomness comes from the assignment. Under consistency, the weighted contribution can be written as \(I_i(e)Y_i(e)/\pi_i(e)\). Therefore:
Adding these expectations over units and dividing by \(N\) gives \(E\{\widehat{\mu}_{HT}(e)\}=\mu(e)\). Subtracting the corresponding expectations for \(e\) and \(e'\) gives \(E\{\widehat{\tau}_{HT}(e,e')\}=\tau(e,e')\). This is design-based unbiasedness: the claim averages over the random assignment while treating the units' potential outcomes as fixed. It does not say that one realized estimate must be close to its target.
The argument does not require assignments or exposure indicators to be independent across units. Their dependence can affect the estimator's variance, but does not change the expectation calculation for each unit. For the estimation result, the essential ingredients are the known probabilities, positive exposure probability for every target unit, consistency, and a valid exposure mapping.
Worked Example: Unequal Exposure Probabilities on a Path
Consider four units on a line, with links \(1\text{--}2\text{--}3\text{--}4\). Each unit is independently assigned treatment with probability \(0.5\). Define high peer exposure \(e_H\) as own treatment 0 with at least one treated neighbor, and low peer exposure \(e_L\) as own treatment 0 with no treated neighbors. Suppose the realized assignment is \((0,1,0,0)\), and the observed outcomes for units 1, 3, and 4 are respectively 10, 14, and 8 outcome units.
For a unit with \(d_i\) neighbors, own treatment is 0 with probability \(0.5\), independently of neighbor assignments. Thus
Units 1 and 4 have one neighbor; units 2 and 3 have two. In the realized assignment, units 1 and 3 have high peer exposure, while unit 4 has low peer exposure. Their relevant probabilities are \(\pi_1(e_H)=0.25\), \(\pi_3(e_H)=0.375\), and \(\pi_4(e_L)=0.25\). Each exposure has positive probability for every unit in this design.
The high-exposure estimate for the mean across all four units is
For low exposure, only unit 4 contributes in this realization:
The estimated spillover contrast at own treatment 0 is \(19.33-8.000=11.33\) outcome units. The high-exposure estimate is not the unweighted average \((10+14)/2=12\): the two observed units had different probabilities of receiving that exposure. The estimate may be variable in a small network, but its design expectation equals the target finite-population contrast.
Worked Example: Exposure Probability Under Complete Randomization
Four units form a cycle, so each unit has two neighbors. Exactly two of the four units are assigned treatment, with each of the six possible assignments equally likely. Let \(e\) mean own treatment 0 and exactly one treated neighbor. For a specified unit, there are six possible pairs of treated units overall. To receive \(e\), the specified unit must be untreated, one of its two neighbors must be treated, and its other treated unit must be the one unit opposite it on the cycle. There are two such assignments, so
Suppose the realized assignment is \((1,1,0,0)\), and the observed outcomes for units 3 and 4 are 9 and 15 outcome units. Each of these untreated units has exactly one treated neighbor, so both receive \(e\). Since all four units have the same probability, the estimate is
The observed outcomes average 12 outcome units, but that average describes only the two units observed under \(e\). The Horvitz–Thompson estimate uses the known probability and the target denominator \(N=4\) to estimate the exposure mean for all four units. The result is not required to equal the observed-subset average.
Worked Example: A Fixed-Own-Treatment Spillover Contrast
Suppose a network consists of two disconnected pairs, \((1,2)\) and \((3,4)\), and each unit's only neighbor is its partner. Assignments are independent Bernoulli with probability \(0.5\). Compare \(e_H=(0,1)\), meaning untreated with a treated partner, against \(e_L=(0,0)\), untreated with an untreated partner. For every unit, each exposure requires a particular assignment of the unit and partner, so \(\pi_i(e_H)=\pi_i(e_L)=0.5(0.5)=0.25\).
Suppose the realized assignment is \((0,1,0,0)\), with observed outcomes \(Y_1=10\), \(Y_3=8\), and \(Y_4=12\) outcome units. Unit 1 receives \(e_H\); units 3 and 4 receive \(e_L\). The two estimates, each for all four units, are
The estimated spillover contrast is \(10.00-20.00=-10.00\) outcome units. In this realization, the estimate is based on one observed unit under \(e_H\) and two under \(e_L\), so it can be unstable. Its design-based interpretation is still clear: it estimates the finite-population mean outcome under being untreated with a treated partner minus the mean under being untreated with an untreated partner.
Common Mistakes and Practical Interpretation
- Using an unweighted observed-exposure average. This is generally not the all-unit exposure mean when units have unequal probabilities of receiving the exposure. Divide each observed outcome by its own exposure probability and use the target population size \(N\).
- Using the treatment probability as the exposure probability. The event \(G_i(\mathbf{Z})=e\) can involve own assignment and several neighbors' assignments. Calculate the probability of that entire event under the design.
- Assuming an exposure absent from the realized data has probability zero. A condition may fail to occur in one assignment even though its design probability is positive. The estimator is then based on the units observed under the exposure; a zero design probability is a different issue and prevents estimating that unit's contribution this way.
- Interpreting the contrast as a universal effect. A contrast between mapped exposures applies to the defined mapping and target population. It does not establish that every possible feature of the network is irrelevant to outcomes.
- Reading one estimate as precise because it is unbiased. Unbiasedness is an average-over-randomizations property. Sparse exposure observations and highly unequal weights can still produce substantial uncertainty.
The practical sequence is to specify the exposure categories, compute each unit's probability under the actual assignment mechanism, check that the required probabilities are positive, and then calculate the weighted exposure means and their contrast. Exposure probabilities are specific to both the network and the randomization design; they cannot be substituted from a different design without justification.
Check Your Understanding
Use the definitions and calculations in this tutorial to answer the following questions.
- Why does the Horvitz–Thompson exposure mean divide each observed outcome by its unit-specific exposure probability?
- On a path network under independent assignment with probability \(0.5\), what are the high- and low-peer-exposure probabilities for an untreated unit with two neighbors?
- Under complete randomization of exactly two treated units among four, why is the probability of own treatment 0 and exactly one treated neighbor equal to \(1/3\) on a four-cycle?
- What exposure-mapping assumption lets \(Y_i(e)\) be used even though many full assignment vectors can produce exposure \(e\)?
- Does design-based unbiasedness guarantee that a single estimated spillover contrast is close to its target? Explain.