Exposure Mappings and Horvitz–Thompson
A treatment vector has one coordinate for every unit, but an outcome may depend on only a few features of that vector. An exposure mapping records those features. If the mapping is sufficient for the outcome and each unit has a known, positive probability of the exposure of interest, inverse exposure probabilities yield an unbiased mean estimator. The analysis of (Aronow and Samii 2017) also shows why estimating its variance is a harder problem.
1. The main identity
Consider a fixed population of \(N\) units. Let \(D_i=f(\mathbf Z,\theta_i)\) be unit \(i\)’s exposure, where \(\theta_i\) contains fixed information such as its network position. Assume assignments producing the same exposure also produce the same fixed potential outcome \(y_i(d)\), and that the observed outcome satisfies \(Y_i=y_i(D_i)\). All probabilities, expectations, and variances below refer to the random assignment. For each exposure being compared, require known probabilities \(\pi_i(d)=\Pr(D_i=d)>0\) for every unit in the target population.
Proposition 1. The estimator
is unbiased for \(\mu(d)=N^{-1}\sum_iy_i(d)\). Differences of these estimators are unbiased for the corresponding average exposure effects, provided both exposures have positive probability for every unit in the same target population.
On the event \(D_i=d\), consistency allows us to replace \(Y_i\) by the fixed value \(y_i(d)\). Taking expectation then cancels the indicator probability with the denominator. The algebra is short because the exposure assumption has already specified which potential outcome is being averaged.
If some units have zero probability of a target exposure, this estimator cannot recover the full-population mean without further assumptions. One option is to restrict the target to units with positive probability of every exposure being compared, adjusting the denominator to that population’s size.
2. Pair probabilities determine the variance
Put \(\pi_{ij}(d)=\Pr(D_i=d,D_j=d)\), including \(\pi_{ii}(d)=\pi_i(d)\). Then
Assignments and exposures have different dependence structures. Even if the \(Z_i\) are independent, two \(D_i\) can share the same neighbors’ assignments. Their joint probabilities must therefore be calculated from the mapping.
For an exposure contrast, include the cross-exposure covariance as well. In particular, a single unit cannot simultaneously have two different exposures, so its joint probability for that pair is zero.
3. Zero pair probabilities are an identification issue
The inverse-probability construction works for a product only if that pair can be observed together. A zero joint probability leaves a variance term that this construction cannot estimate. Without additional restrictions on the potential outcomes, an unbiased variance estimator is generally unavailable in this case.
A conservative correction can bound such products using
The square terms on the right can each be estimated from a marginal exposure event with positive probability. Replacing missing variance contributions by suitable upper bounds gives an estimator whose expectation is at least the true variance. This does not guarantee that the estimate exceeds the true variance in every realized assignment.
For a known design, exposure probabilities can be found by enumerating assignments or approximated by independent simulations. Rare events need enough simulated draws to control the error in their inverse probabilities. Structural zeros should be checked from the design, since simulation alone cannot distinguish zero probability from an event too rare to appear.
4. If the mapping is misspecified
Without the sufficiency condition, the weighted estimator still averages outcomes over assignments conditional on the reported exposure. That average can depend on the assignment design, so it is no longer the fixed \(y_i(d)\) estimand originally claimed.
A finer mapping can capture additional features of interference, but it need not be sufficient and can create rare exposure cells and unstable weights. The two-stage allocation approach defines effects averaged over a specified distribution of peer assignments.
5. Computing exposure probabilities on a small graph
Take a three-vertex path \(1-2-3\) with independent Bernoulli\((p)\) treatment assignment, where \(0<p<1\) and \(N=3\). For illustration, suppose each outcome depends on treatment assignments only through whether the unit has at least one treated neighbor. Then
The exposure probabilities are
The two leaves always have the same exposure, so \(\pi_{13}(1)=p\), not \(p^2\). Their contribution to the two ordered off-diagonal terms in the variance of \(\widehat\mu(1)\) is
Both leaves use the same middle vertex’s treatment, which explains their perfect exposure dependence. If their outcomes also depend on their own treatment, the exposure label must include it. Otherwise the proposed outcome model is not sufficient, regardless of whether the probabilities have been computed correctly.
For a unit with \(k\) independently assigned neighbors, the probability of no treated neighbor is \((1-p)^k\). Thus a high-degree unit can have an extremely rare unexposed cell even though its own treatment probability is well away from zero and one. Fine exposure definitions create a precision issue through their design probabilities.
6. Deriving the covariance formula
Let \(I_i(d)=\mathbf1\{D_i=d\}\). Conditional on the fixed potential outcomes, the covariance between two weighted terms is
Summing and dividing by \(N^2\) gives the result. For two distinct exposures \(d,e\), define \(\pi_{ij}(d,e)=\Pr(D_i=d,D_j=e)\). The same calculation gives
The variance of a contrast is the sum of its two variances minus twice this covariance.
When \(\pi_{ij}(d,e)>0\), an unbiased estimator of the potential-outcome product is
Substituting these estimates into the covariance expansion gives the usual inverse-joint-probability construction for observable pairs. The diagonal case with the same exposure reduces to \(I_i(d)Y_i^2/\pi_i(d)\).
7. Why some variance terms lack an unbiased estimator
For the same individual and distinct exposures, \(I_i(d)I_i(e)=0\) always, so \(\pi_{ii}(d,e)=0\). The two potential outcomes are never observed together. Some cross-individual pairs can also be impossible under constrained randomization or the graph mapping.
To see a conservative repair, consider an unobservable contribution with magnitude \(2|y_i(d)y_j(e)|\). Young’s inequality gives
For a missing term \(c\,y_i(d)y_j(e)\), use the upper bound \(\lvert c\rvert\{y_i(d)^2+y_j(e)^2\}/2\), where \(c\) includes the term’s coefficient in the variance expansion. Estimate each square with its marginal inverse-probability weight. The absolute value of the coefficient makes this an upper bound for either sign. The bound can be loose when the design rules out many exposure pairs.
For the contrast \(\widehat\mu(d)-\widehat\mu(e)\), the same-unit missing product has an especially useful interpretation. The covariance of the two weighted contributions from unit \(i\) is \(-y_i(d)y_i(e)\), because the exposure indicators cannot both equal one. Subtracting twice that covariance contributes
to the contrast variance. Replacing it by \(N^{-2}(y_i(d)^2+y_i(e)^2)\) overestimates this particular term by
The upper bound is valid for positive or negative outcomes because its excess is a square. Feasible marginal-square contributions are \(I_i(d)Y_i^2/\pi_i(d)\) and \(I_i(e)Y_i^2/\pi_i(e)\). For impossible pairs involving different units, apply the bound to each term with its own coefficient and exposure labels. In a same-exposure sum over ordered pairs, both \((i,j)\) and \((j,i)\) appear; a sum over unordered pairs instead needs a factor of two. Cross-exposure pairs need separate care: \(\pi_{ij}(d,e)\) need not equal \(\pi_{ji}(d,e)\).
The difference between first and second moments is now explicit. A mean uses one potential outcome at a time; a variance uses products, some of which the design never reveals. The same obstacle arises in an ordinary randomized experiment without interference: a unit’s treated and untreated potential outcomes are never observed together. Interference can also make joint exposures impossible across different people.
A confidence interval based on a normal approximation additionally requires a central limit theorem and a variance estimate that converges to the true variance or a conservative bound at the relevant asymptotic scale. Conservativeness in expectation alone is insufficient. Unbiasedness of \(\widehat\mu\) controls its expectation, but says nothing by itself about the shape of its distribution.
8. What the estimator targets under a coarser mapping
Without exposure sufficiency, let the full potential outcome be \(Y_i(\mathbf z)\). The expectation of the same weighted estimator is
The last expression is a design-specific conditional average. It may change when the assignment law changes, even if the label \(d\) does not. If exposure sufficiency holds over the combined support of the designs being compared, and the relevant exposure probabilities remain positive, these averages reduce to the same fixed exposure potential outcomes.
Simulating assignments from a known design can approximate the required probabilities. For a positive probability \(\pi\) estimated from \(M\) independent draws, the relative standard deviation of its frequency is \(\sqrt{(1-\pi)/(M\pi)}\). Rare exposures require many draws for accurate inverse weights. Replacing exact probabilities with simulated estimates generally loses the exact unbiasedness in Proposition 1. Zero observed frequency is compatible with a small positive probability, while structural impossibility should be established from the design itself.