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Interference and Spillover Effects · Tutorial 849 of 972

Analyzing a Two-Stage Network Experiment

Learn how to connect a network exposure mapping to weighted estimates of direct and spillover effects, then assess uncertainty using the experiment’s randomization.

Advanced 12 min read

What You'll Learn

  • Define exposure categories using own treatment and neighbors’ treatment.
  • State the assumptions needed to interpret exposure-specific effects.
  • Calculate exposure assignment probabilities in a two-stage design.
  • Construct Horvitz–Thompson estimates for exposure means and contrasts.
  • Use the actual assignment mechanism for a randomization test.
  • Recognize why network assignments cannot be treated as independent observations.

From a Two-Stage Design to an Exposure Effect

In a network experiment, a person’s outcome may depend both on their own treatment and on treatment received by people connected to them. A two-stage randomized saturation design makes these treatment patterns part of the experiment: first, groups are assigned different treatment saturation levels; then, individuals are assigned treatment within groups. An analysis must carry that design through to the exposure definition, the estimand, the weights, and the uncertainty calculation.

The analysis below uses a simple exposure mapping: a person’s exposure is their own treatment status together with whether any of their in-group neighbors are treated. This is a deliberate simplification, not a universal description of network effects. The mapping must represent the treatment features relevant to the scientific question; otherwise, the resulting estimates concern the mapped categories, not necessarily the actual mechanisms.

Definition: Let \(Z_i\) indicate whether individual \(i\) is treated, and let \(G_i\) summarize treatment among that individual’s neighbors. The exposure mapping assigns an exposure \(E_i=f(Z_i,G_i)\). Write \(Y_i(e)\) for the potential outcome of individual \(i\) under exposure \(e\). An exposure-specific finite-population mean is \(\mu(e)=N^{-1}\sum_{i=1}^{N}Y_i(e)\), where \(N\) is the number of individuals in the target population.

For the mapping used here, the categories are \(D=(1,0)\), meaning treated with no treated neighbor; \(S=(0,1)\), meaning untreated with at least one treated neighbor; and \(C=(0,0)\), meaning untreated with no treated neighbor. The direct contrast at \(G_i=0\) is \(\mu(D)-\mu(C)\). The spillover contrast among untreated individuals is \(\mu(S)-\mu(C)\). These are effects of the specified exposure contrasts; they do not summarize every possible way treatment could affect a network.

Assumptions and Assignment Probabilities

Before estimating these contrasts, specify what makes the observed outcomes interpretable. The design must be known, including both stages of randomization. For the exposure-specific interpretation, assume consistency: when individual \(i\) receives exposure \(e\), their observed outcome is \(Y_i(e)\). Also assume the mapping is sufficient for the outcome: assignments producing the same mapped exposure do not cause additional outcome differences through omitted features of the assignment. Finally, assume no interference across the groups used in the design, unless cross-group connections are explicitly incorporated into the mapping. These assumptions are substantive; randomization alone does not make a poor mapping adequate.

For an exposure \(e\), define \(\pi_i(e)=P(E_i=e)\), using the full two-stage assignment procedure. Positivity for this analysis requires \(\pi_i(e)>0\) for every individual included in the target mean. Exposure probabilities can vary across individuals, so they should be calculated from the design rather than assumed equal.

Worked Example: Defining the Target Contrasts

Suppose a hypothetical clinic network has groups of patients connected by referrals. The study’s question is whether offering a program to a patient changes that patient’s outcome when no referred contact is offered the program, and whether an untreated patient benefits when at least one referred contact is offered it.

Define \(Z_i=1\) when patient \(i\) is offered the program, and \(G_i=1\) when at least one of their in-group referral contacts is offered it. The mapping has three relevant categories: \(D=(1,0)\), \(S=(0,1)\), and \(C=(0,0)\). The direct estimand is \(\mu(1,0)-\mu(0,0)\), while the spillover estimand is \(\mu(0,1)-\mu(0,0)\).

These definitions answer exactly which comparisons the study targets. They do not cover the outcome for a treated patient with a treated contact, exposure \((1,1)\). Nor do they establish that “any treated contact” is an adequate summary if the number or identity of treated contacts matters. Those are questions about the mapping and target, not details to be repaired by the estimator.

Horvitz–Thompson Estimation

Once the mapping and probabilities are specified, an exposure mean can be estimated by weighting observed outcomes inversely by their exposure probabilities. This Horvitz–Thompson estimator targets the finite-population mean, rather than simply averaging the people observed in that exposure category.

Definition: For exposure \(e\), the Horvitz–Thompson estimate of the exposure-specific mean is \(\widehat{\mu}_{HT}(e)=N^{-1}\sum_{i=1}^{N} I(E_i=e)Y_i/\pi_i(e)\), where \(I(E_i=e)\) is 1 when individual \(i\) receives exposure \(e\) and 0 otherwise. An estimated contrast is the difference between the corresponding estimated means.

The reason for the weighting is visible from a short expectation calculation. Under the randomization distribution and consistency, if \(E_i=e\), then the observed outcome is \(Y_i(e)\). Thus the weighted contribution for individual \(i\) is \(I(E_i=e)Y_i(e)/\pi_i(e)\). Its expectation is

$$ E\left[\frac{I(E_i=e)Y_i(e)}{\pi_i(e)}\right] = \frac{P(E_i=e)Y_i(e)}{\pi_i(e)} = Y_i(e). $$

Averaging over the \(N\) individuals gives \(E[\widehat{\mu}_{HT}(e)]=\mu(e)\). Subtracting two such estimates gives an unbiased estimator of the corresponding mean contrast. This result relies on correct exposure probabilities, positivity, and the consistency and mapping assumptions stated above; it does not claim that a particular estimate is free of randomization error.

Worked Example: Probabilities and Weighted Exposure Means

Consider two hypothetical groups, \(A,B,C\) and \(D,E,F\), each with three individuals. Within each group, everyone is a neighbor of everyone else. In stage one, exactly one group is assigned high saturation and the other low saturation, with each group equally likely to be high. In stage two, exactly one of the three individuals in the high-saturation group is selected for treatment, each with probability \(1/3\); nobody in the low-saturation group is treated. There are six equally likely complete assignments.

Use the mapping \(D=(1,0)\), \(S=(0,1)\), and \(C=(0,0)\), where \(G_i=1\) means at least one in-group neighbor is treated. Any individual has probability \(1/2\) of belonging to the high-saturation group. Conditional on that, their probability of being the selected treated person is \(1/3\), so \(\pi_i(D)=(1/2)(1/3)=1/6\). Their probability of being untreated while another group member is treated is \(\pi_i(S)=(1/2)(2/3)=1/3\). If their group is low saturation, they are in category \(C\), giving \(\pi_i(C)=1/2\). All three probabilities are positive.

Suppose the realized assignment makes \(A,B,C\) the high-saturation group and selects \(A\) for treatment. The observed outcomes are 8 for \(A\), 6 for \(B\), 7 for \(C\), 5 for \(D\), 4 for \(E\), and 6 for \(F\), in hypothetical outcome units. The exposures are \(D\) for \(A\), \(S\) for \(B,C\), and \(C\) for \(D,E,F\). With \(N=6\), the weighted means are

$$ \widehat{\mu}_{HT}(D) =\frac{1}{6}\frac{8}{1/6} =\frac{8}{6/6}=8, \qquad \widehat{\mu}_{HT}(S) =\frac{1}{6}\left(\frac{6}{1/3}+\frac{7}{1/3}\right) =\frac{39}{6}=6.5, $$

and

$$ \widehat{\mu}_{HT}(C) =\frac{1}{6}\left(\frac{5}{1/2}+\frac{4}{1/2}+\frac{6}{1/2}\right) =\frac{30}{6}=5. $$

The estimated direct contrast is \(8-5=3\) outcome units, and the estimated spillover contrast is \(6.5-5=1.5\) outcome units. These are estimates under the stated mapping and randomization, not proof that treatment caused these differences in this particular realized assignment.

Assessing Uncertainty With the Randomization

Individuals’ exposures are dependent: assigning one person to treatment changes the exposure possibilities for their neighbors, and the first-stage group assignment affects several people at once. Therefore, treating all observed individuals as independent observations does not reproduce the experiment’s uncertainty. A design-based assessment instead uses the actual assignment mechanism. For a small design, an exact randomization test can enumerate every permitted assignment.

For the example above, consider the sharp null hypothesis that no individual’s outcome changes under any assignment. Under this null, each individual’s observed outcome would be the same under all six possible assignments. A two-sided randomization test compares the observed absolute value of a statistic with its values under those assignments. The test’s p-value is the proportion of permitted assignments whose absolute statistic is at least as large as the observed one. This test concerns the sharp null; it is not automatically a test of only the average direct contrast.

Worked Example: Exact Randomization Test for a Direct Contrast

Use the six outcomes from the previous example and the statistic \(\widehat{\mu}_{HT}(D)-\widehat{\mu}_{HT}(C)\). The observed statistic is \(8-5=3\) outcome units. Under the sharp null, keep the six outcomes fixed and recompute the statistic for each allowed assignment.

If \(A,B,C\) is the high-saturation group, the mean outcome in the low-saturation group \(D,E,F\) is \((5+4+6)/3=5\). Selecting \(A\), \(B\), or \(C\) in turn gives direct-contrast statistics \(8-5=3\), \(6-5=1\), and \(7-5=2\).

If \(D,E,F\) is the high-saturation group, the low-saturation group mean is \((8+6+7)/3=7\). Selecting \(D\), \(E\), or \(F\) gives statistics \(5-7=-2\), \(4-7=-3\), and \(6-7=-1\). Thus, among the six equally likely assignments, two statistics have absolute value at least 3. The exact two-sided p-value is \(2/6=0.3333\).

At a significance level of \(\alpha=0.05\), this test does not reject the sharp null. The p-value means that, if the sharp null were true and the stated randomization were repeated, one third of assignments would produce a statistic at least as extreme in absolute value as the observed statistic. It is not the probability that the null is true. With only six assignments, the test also has limited p-value resolution.

Common Pitfalls and Interpretation

A common mistake is to calculate an unweighted average among people observed in each exposure category and interpret differences as finite-population exposure effects. In this example, the category counts happen to be one, two, and three, and the relevant probabilities are equal across individuals. In other designs, individuals can have different chances of entering a category. An unweighted category mean can then overrepresent individuals with higher exposure probabilities; the Horvitz–Thompson weights correct for that unequal chance under the stated assumptions.

A second mistake is to treat the network as if each person were independently randomized. Here the six complete assignments are the randomization units for the exact test; the six people do not have six independent exposure assignments. In larger studies, uncertainty methods must likewise reflect the two-stage design and the dependence induced by the network. A test or interval also needs a clearly stated target: inference about a sharp null, an average exposure contrast, or another quantity are not interchangeable.

Finally, a precise calculation cannot compensate for an unsuitable exposure mapping. If outcomes depend on the number of treated neighbors, for example, “at least one” may combine consequentially different situations. If connections cross groups, assuming no cross-group interference may also fail. Report the mapping and design assumptions alongside estimates so readers know which causal contrasts the analysis supports.

Key takeaway: A two-stage network analysis proceeds from a defensible exposure mapping to design-based exposure probabilities, weighted estimates, and uncertainty calculations that respect the actual randomization. Direct and spillover contrasts are interpretable only for the specified exposures and under the mapping and interference assumptions.

Check Your Understanding

Answer each question using the design and definitions in this tutorial.

  1. In the six-person example, why is the probability of exposure \(S\) equal to \(1/3\), rather than \(2/3\)?
  2. State the direct and spillover contrasts for categories \(D=(1,0)\), \(S=(0,1)\), and \(C=(0,0)\).
  3. What positivity condition is required to estimate an exposure-specific mean for every individual?
  4. In the worked randomization test, how many assignments produce an absolute statistic at least as large as 3, and what is the two-sided p-value?
  5. Why does the exact sharp-null test not automatically test only the average direct contrast?
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