Problem ID: C31-129 Model: GPT 6 Astra Ultra Version: 1
Causal inference under interference and network spillovers
Status. Unresolved for unrestricted learned interference. We give simultaneous honest inference over a finite local exposure class, an explicit dependence/complexity rate, and a nonvanishing-width example when dependence is unrestricted.
Short target and sampling assumptions
Fix $n$ units, a pretreatment network and covariates. All probabilities below condition on these fixed objects. Assignments $Z_1,\ldots,Z_n$ are independent Bernoulli with known probabilities. Potential outcomes are fixed functions of assignments. For each unit specify a set $B_i\subseteq[n]$ such that $Y_i(z)$ depends only on $z_{B_i}$ and $|Y_i(z)|\leq B$ for every assignment. The locality restriction is imposed on the actual outcome, not only on its exposure approximation.
Let $\mathcal H$ be a prespecified finite set of $H$ exposure maps, each $h_i(z)$ depending only on $z_{B_i}$. For each $h$ fix two distinct exposure labels $e_h,e'_h$, and assume their probabilities $p_i^h(e_h),p_i^h(e'_h)$ are at least $p>0$. Set \[
\tau_h=\frac1n\sum_i\left\{{\mathbb{E}}[Y_i(Z)\mid h_i(Z)=e_h]
-{\mathbb{E}}[Y_i(Z)\mid h_i(Z)=e'_h]\right\}.
\] This is a design-specific contrast even for insufficient maps. A learner may select $\widehat h$ using the entire realized experiment, provided it selects from $\mathcal H$.
Join distinct units $i,j$ in a dependency graph when $B_i\cap B_j\ne\varnothing$. Let $\chi$ be any number of colors in a proper coloring of this graph; $\chi\leq\Delta+1$ for maximum graph degree $\Delta$. Define \[
\widehat\tau_h=\frac1n\sum_iY_i(Z)
\left\{\frac{{\mathbf 1}\{h_i(Z)=e_h\}}{p_i^h(e_h)}
-\frac{{\mathbf 1}\{h_i(Z)=e'_h\}}{p_i^h(e'_h)}\right\}.
\]
Lemma (Colored concentration): Suppose random variables $X_1,\ldots,X_n$ are independent within each class of a partition into $\chi$ classes and each lies in an interval of length at most $L$. Then, for every $t>0$, \[
{\mathbb{P}}\left(\left|n^{-1}\sum_i(X_i-{\mathbb{E}} X_i)\right|>t\right)
\leq2\exp\left(-\frac{2nt^2}{\chi L^2}\right).
\]
Proof: Write $S_c=\sum_{i\in c}(X_i-{\mathbb{E}} X_i)$. Hölder's inequality with exponents all equal to $\chi$ and the independent-variable Hoeffding moment bound give \[
{\mathbb{E}} e^{\lambda\sum_cS_c}
\leq\prod_c({\mathbb{E}} e^{\chi\lambda S_c})^{1/\chi}
\leq\exp(\chi\lambda^2 nL^2/8).
\] Markov's inequality with $\lambda=4t/(\chi L^2)$ gives the upper tail $e^{-2nt^2/(\chi L^2)}$. Apply the same argument to negatives and add the two tail probabilities. Independence between different color classes is not required. ∎
Theorem (A valid interval after exposure-map selection): Under the assumptions above, with probability at least $1-\alpha$, every $h\in\mathcal H$ satisfies \[
|\widehat\tau_h-\tau_h|\leq
r_n:=\frac{B}{p}\sqrt{\frac{2\chi\log(2H/\alpha)}n}.
\] Consequently $[\widehat\tau_{\widehat h}-r_n,\widehat\tau_{\widehat h}+r_n]$ covers the selected contrast $\tau_{\widehat h}$ with the same probability, even when selection uses outcomes.
Proof: Each summand is between $-B/p$ and $B/p$, because the two exposure indicators are disjoint. Its expectation is the corresponding difference of conditional means by the definition of conditional expectation. Within a color class the $B_i$ are disjoint; the summands are therefore independent functions of disjoint collections of independently assigned treatments. Apply the lemma with $L=2B/p$, then take a union bound over $H$ maps. The simultaneous event implies the selected-map statement for every possible selection rule. ∎
Corollary (Stable responses with bounded exposure error): For every $h$ for which functions $y_i^h$ satisfy $|Y_i(z)-y_i^h(h_i(z))|\leq\varepsilon_h$ for all assignments, put $\tau_h^*=n^{-1}\sum_i[y_i^h(e_h)-y_i^h(e'_h)]$. On the simultaneous event of the theorem, \[
|\widehat\tau_h-\tau_h^*|\leq r_n+2\varepsilon_h
\] for all such maps. In particular this covers an outcome-selected map if its error bound holds uniformly and is known.
Proof: Conditional averaging preserves the error bound at each exposure, so $|\tau_h-\tau_h^*|\leq2\varepsilon_h$. Add the simultaneous estimation bound. The error assumption is substantive and cannot be verified merely by selecting a map with a small observed residual. ∎
For an undirected network of maximum degree $d$, if each $B_i$ is its closed neighborhood, each has at most $d+1$ coordinates and any coordinate occurs in at most $d+1$ such sets. Hence each dependency-graph vertex has degree at most $(d+1)^2-1$. One obtains informative diameter whenever \[
\frac{(d+1)^2\log(2H/\alpha)}{np^2}\longrightarrow0,
\qquad\sup_h\varepsilon_h\longrightarrow0
\] for the stable-response target. This sufficient condition need not be necessary, and better variance-based bounds can improve the $p$ dependence.
Proposition (Overlap alone cannot make confidence sets shrink): For every $n$ there is a local-outcome experiment with two exposures of probability $1/2$, bounded outcomes in $[0,1]$, and parameter $\theta\in\{0,1\}$, such that every confidence set $C$ satisfying ${\mathbb{P}}_\theta(\theta\in C)\geq1-\alpha$ for both parameters obeys \[
{\mathbb{E}}_0{\operatorname{diam}} C\geq\frac12-2\alpha.
\]
Proof: Use $B_i=\{1\}$ for all $i$, $Z_1\sim\mathrm{Ber}(1/2)$, and $Y_i(z)=\theta z_1$. Other assignments can be independent Bernoulli and are irrelevant. The exposure map is $h_i(z)=z_1$, and the average exposure contrast is exactly $\theta$. On event $E=\{Z_1=0\}$ the full data laws coincide under $\theta=0$ and $\theta=1$; both give $E$ probability $1/2$. Thus, under $P_0$, the probability of $E$ and failure to include 1 is at most $\alpha$, by transferring that identical subprobability law from $P_1$. The probability of failure to include 0 is also at most $\alpha$. With probability at least $1/2-2\alpha$, both points belong to $C$, forcing diameter at least one. The dependency graph is complete despite exposure overlap being fixed. ∎
Remaining gap and reference
The result pays for outcome-based map selection through $\log H$ and for local dependence through $\chi$. It does not handle an unrestricted map learner, unknown global interference, treatment-induced network formation, or unidentified observational treatment assignment. The target after map selection is explicitly $\tau_{\widehat h}$; selecting a convenient map does not establish that its contrast equals a policy effect.
Peter M. Aronow and Cyrus Samii (2017), Estimating Average Causal Effects Under General Interference, with Application to a Social Network Experiment, Annals of Applied Statistics 11(4), 1912--1947. https://doi.org/10.1214/16-AOAS1005; author preprint record https://arxiv.org/abs/1305.6156. The primary abstract and bibliographic record were verified: they establish the role of known exposure mappings and local dependence. The elementary finite-class concentration and lower-bound calculations are proved here and not attributed as their exact theorems.
Completion Estimate
55%