Missing Network Links and Spillover Bias

econometricsnetworks

Suppose treatment is randomized but the observed network omits some links. A person recorded as having no treated neighbors may in fact have several. Randomization identifies the assignment probabilities; it does not reveal these missing exposures. (Zhang 2024) expresses the observed response as a mixture over the true network and derives the implications for spillover estimates.

1. The latent and observed regressions

Let \(N_i^*\) be true degree and \(S_i^*\) the number of treated true neighbors. Write the structural conditional mean as \(m^*(d,s^*,n^*,z)\). We instead observe degree \(N_i\) and treated-neighbor count \(S_i\).

Iterated expectations gives a mixture in any case. The substantive step is using the assumptions to show that its component mean is still \(m^*\) after conditioning on the incomplete network.

2. Why spillover bias has no automatic sign

Let \(w_s(s^*,n^*)\) denote the mixture weights at observed exposure \(s\). Then

\[ m(d,s,n,z)-m(d,s',n,z) =\sum_{s^*,n^*}m^*(d,s^*,n^*,z) (w_s(s^*,n^*)-w_{s'}(s^*,n^*)). \]

Each set of mixture weights sums to one. Their difference is signed and shifts weight across latent exposures and degrees, so the contrast need not equal the structural spillover at \((s,s')\). In particular, missing links need not attenuate every spillover estimate.

3. What randomization identifies

Suppose links are missing but not spuriously added, and treatment is independently assigned with common probability \(p\). Conditional on \(N_i^*=n^*\) and observed degree \(n\), there are \(n^*-n\) missing neighbors. Nondifferential missingness leaves their treatment assignments independent Bernoulli draws, so

\[ S_i^*-S_i\mid N_i^*-N_i=n^*-n \sim\operatorname{Binomial}(n^*-n,p). \]

Randomization supplies the binomial law of treatment among a given number of missing neighbors. The number itself remains unknown. This separates the known part of the mixture kernel from the unknown true-degree distribution.

Validation data, repeated measurement, or assumptions on link observation can identify that distribution. Recovering the response then requires enough distinct mixture equations, expressed as a rank or completeness condition. A written mixture equation alone does not establish that its components can be recovered.

4. Implications for exposure choices

Using the treated-neighbor share \(S_i/N_i\) introduces error in both numerator and denominator. Dropping individuals with observed degree zero may additionally select on true degree and the missingness process.

The exposure-mapping estimator presumes the stated exposure is sufficient for outcomes. When the network is incomplete, that presumption needs an argument about the latent network or an explicit sensitivity analysis. Random treatment assignment by itself cannot certify that the measured exposure is the causal one.

5. Factor the mixture under independent assignment

Fix observed degree \(n\) and true degree \(n^*\geq n\). Under independent treatment with common probability \(p\) and link observation independent of treatment, the observed and missing neighbors’ treatment counts are independent binomials conditional on the neighbor sets:

\[ S\sim\operatorname{Binomial}(n,p),\qquad B\triangleq S^*-S\sim\operatorname{Binomial}(n^*-n,p). \]

In this stronger simple design, \(S\) gives no additional information about \(N^*\) once \(N=n\) and the relevant baseline covariates are fixed. Write \(q_n(n^*)=\Pr(N^*=n^*\mid N=n,Z=z)\). Then

\[ \begin{aligned} m(d,s,n,z) =\sum_{n^*\geq n}q_n(n^*) \sum_{b=0}^{n^*-n}&m^*(d,s+b,n^*,z)\\ &\times\binom{n^*-n}{b}p^b(1-p)^{n^*-n-b}. \end{aligned} \]

The treatment design fixes the binomial kernel, while the observation model determines how much weight to put on each missing-neighbor count. If assignments or missingness follow a different model, this factorization must be derived again.

6. Linear counts need not attenuate at all

Suppose the latent mean is \(m^*(d,s^*,n^*)=a+\tau d+\beta s^*+\gamma n^*\). Applying the mixture formula gives

\[ m(d,s,n)=a+\tau d+\beta s +\beta p(\mathbb{E}[N^*\mid N=n]-n) +\gamma\mathbb{E}[N^*\mid N=n]. \]

At fixed observed degree, the contrast between \(s+1\) and \(s\) is exactly \(\beta\). Missing neighbors change the intercept of this conditional regression but not its slope in the observed count. This is one reason that a blanket attenuation claim is wrong.

If the latent response is instead \((s^*)^2\), then

\[ \mathbb{E}[(s+B)^2\mid n] =s^2+2s\mathbb{E}[B\mid n]+\mathbb{E}[B^2\mid n], \]

and the observed one-neighbor contrast is \(2s+1+2\mathbb{E}[B\mid n]\). It exceeds the latent contrast \(2s+1\) evaluated at the observed count whenever missing treated neighbors have positive expected number. Curvature turns the omitted exposure into an upward shift in the contrast.

These examples keep latent degree composition fixed across the observed exposure cells. Allowing that composition to change creates another contribution to the contrast. Its sign depends on the response and observation models, which is why the linear-count example cannot justify a general attenuation rule.

7. Shares give a different calculation

Suppose true degree is positive and \(m^*=a+\beta S^*/N^*\). For fixed observed degree \(n>0\), independent assignment gives

\[ \mathbb{E}\left[\frac{S^*}{N^*}\,\middle|\,S=s,N=n\right] =s\mathbb{E}[1/N^*\mid n]+p(1-n\mathbb{E}[1/N^*\mid n]). \]

For the observed treated share \(s/n\), the attenuation factor is \(n\mathbb{E}[1/N^*\mid n]\in(0,1]\), multiplying the structural slope. Thus attenuation does hold in this linear-share model. It follows from the ratio exposure and the stated independence assumptions; it does not extend automatically to the count and nonlinear examples above.

Observed degree zero needs separate handling: \(S/N\) is undefined there, yet the individual may have positive true degree and substantial exposure. Dropping these individuals changes the population under study and can distort a policy target.

8. What an observation model can identify

For illustration, suppose each true link is observed independently with known probability \(q>0\). Conditional on true degree, \(N\mid N^*=k\) is binomial\((k,q)\). The probability-generating functions satisfy

\[ G_N(t)=\mathbb{E}[(1-q+qt)^{N^*}] =G_{N^*}(1-q+qt). \]

For a known finite maximum degree, this is a triangular linear system in the true-degree probabilities, with diagonal entries \(q^k>0\). It is therefore algebraically invertible. If \(q\) is unknown, one observed degree distribution generally cannot separate degree heterogeneity from observation probability without additional information.

For a maximum true degree of two, write \(a_j=\Pr(N=j)\) and \(u_j=\Pr(N^*=j)\). Binomial thinning gives

\[ a_2=q^2u_2,\qquad a_1=qu_1+2q(1-q)u_2,\qquad a_0=u_0+(1-q)u_1+(1-q)^2u_2. \]

Back substitution gives the explicit inverse

\[ u_2=\frac{a_2}{q^2},\qquad u_1=\frac{a_1}{q}-\frac{2(1-q)a_2}{q^2},\qquad u_0=1-u_1-u_2. \]

An error in the observed frequency \(a_2\) is amplified by \(q^{-2}\) and also enters the estimate of \(u_1\) with the opposite sign. Population probabilities generated by this observation model give a nonnegative inverse. Empirical frequencies can produce negative estimates, signaling sampling error or model incompatibility rather than negative latent probabilities.

The implied posterior used in the spillover mixture is then

\[ \Pr(N^*=k\mid N=n) =\frac{\binom{k}{n}q^n(1-q)^{k-n}u_k}{a_n}, \qquad a_n>0. \]

The observation model now supplies the posterior weights needed in the spillover mixture. There are still two recovery problems: infer the missing-neighbor distribution, then infer responses at latent exposures. Solving the first does not ensure that the second system has enough independent equations.

Even with known \(q\), small diagonal entries can make inversion unstable for large degrees. Identification and statistical precision are separate issues. A validation sample or repeated network measurements can help estimate the observation law and constrain the latent distribution, but the rank and compatibility conditions must still be checked.

9. Own-treatment effects and policy evaluation

Under randomized own treatment and the stipulated nondifferential observation mechanism, the mixture weights do not change with \(d\). Therefore

\[ m(1,s,n,z)-m(0,s,n,z) =\sum_{s^*,n^*}w_s(s^*,n^*) (m^*(1,s^*,n^*,z)-m^*(0,s^*,n^*,z)). \]

The own-treatment effect is therefore averaged with nonnegative weights summing to one. Changing \(s\) changes the exposure mixture itself, so a spillover contrast also compares different latent configurations. This is the source of the difference between the two comparisons.

For a policy that changes neighbor assignments, the latent distribution induced by that policy is needed to evaluate welfare. Substituting the observed network into the rule can miss both neighbors and their induced outcomes. The network-policy notes presume the exposure information needed by their identifying model; incomplete networks require this additional observation-model step.

Zhang, Lina. 2024. “Spillovers of Program Benefits with Missing Network Links.” arXiv. https://doi.org/10.48550/arXiv.2009.09614.

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