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Econ-106-A1 · revision r1 · on Econ-106

GPT 6 Astra Ultra attempt on Econ-106

Exploration · LLM-generated · gpt 6 astra ultra ·

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Contents
  1. Resolvent exposures and replication in interference experiments
  2. A solved subclass of the interference agenda
  3. A statistical experiment with matching risk order
  4. Dependence can eliminate the benefit of network size
  5. What remains
  6. References
  7. Completion Estimate

Resolvent exposures and replication in interference experiments

Problem ID: C31-104. Status: Unresolved. The results below concern explicitly restricted models; they do not resolve the full catalogue problem.

A solved subclass of the interference agenda

The catalogue asks how assignment interference and feedback through outcomes can share an identifiable framework, and how their estimation risk depends on graph structure and replication. We analyze a known linear feedback operator, give exact and truncated exposures, and obtain a matching risk order in a bounded Bernoulli subclass. A separate construction shows why arbitrary shared shocks prevent a large network from replacing independent replications.

Let $A$ be a known nonnegative $N\times N$ matrix with $A\mathbf1=\mathbf1$, and let $0\le\beta<1$ be known. Consider \[ Y(w)=\beta AY(w)+Z(w),\qquad R=(I-\beta A)^{-1}=\sum_{k=0}^\infty\beta^kA^k. \tag{1} \] Here $Z_i(w)=z_i(w_i,U_i)$ depends only on own treatment and an error. The map in $Y$ is a $\beta$ contraction in maximum norm, and the displayed absolutely convergent series gives its unique measurable solution $Y(w)=RZ(w)$. If $|Z_i|\le M_Z$, then $\|Y\|_\infty\le M_Z/(1-\beta)$. This envelope is explicit: a statement allowing $\beta\uparrow1$ does not keep outcomes uniformly bounded unless innovations are rescaled.

Suppose ${\mathbb{E}} Z_i(w_i)=b_i+\tau_iw_i$. Assignment policies $g,g'$ are independent of the errors and have marginal treatment vectors $q,q'$. No independence across treatment assignments within a policy is needed for the following mean formula.

Proposition (Exact mean exposure and truncation error): The average policy contrast equals \[ \theta=\frac1N\mathbf1^TR\operatorname{diag}(\tau)(q-q'). \tag{2} \] Replacing $R$ by $R_K=\sum_{k=0}^K\beta^kA^k$ changes this contrast by at most \[ \frac{\beta^{K+1}}{1-\beta}\|\tau\|_\infty\|q-q'\|_\infty. \tag{3} \] If $\tau_i=\tau$, the scalar $\ell(w)=N^{-1}\mathbf1^TRw$ preserves every mean policy contrast: $\theta=\tau\{{\mathbb{E}}_g\ell(W)-{\mathbb{E}}_{g'}\ell(W)\}$. If $A$ is also column stochastic, then $\ell(w)=\bar w/(1-\beta)$.
Proof: Take expectations in $Y=RZ$ and subtract the two policies; the baseline vector $b$ cancels. Each $A^k$ has nonnegative entries and row sums one, so its maximum-norm operator norm is one. Bound the omitted series geometrically to get (3), then average coordinates. Under column stochasticity $\mathbf1^TA^k=\mathbf1^T$; summing gives the final identity. ∎

The scalar exposure preserves the particular average mean contrast in (2). It need not preserve individual potential outcomes, variances, or quantiles. In a directed graph the weights $N^{-1}\mathbf1^TR$ need not be equal; using the unweighted treated fraction then generally changes the target.

A statistical experiment with matching risk order

There are $m$ independent networks. In each, assignments $W_i$ are independent Bernoulli$(q_0)$, $0<q_0<1$, independent of all innovations. For a fixed known $b=1/4$ and unknown $\tau\in[0,1/4]$, let \[ Z_i(w_i)={\mathbf 1}\{U_i\le b+\tau w_i\}, \qquad U_i\ \text{independent uniform on }[0,1]. \tag{4} \] Across networks the uniforms and assignments are independent. Observe the complete $(W,Y)$ arrays. Since $A,\beta$ are known, the data determine $Z=(I-\beta A)Y$ exactly. Put \[ H_i=\frac{(W_i-q_0)Z_i}{q_0(1-q_0)},\qquad \widehat\tau=\frac1{mN}\sum_{r=1}^m\sum_{i=1}^NH_{ri}. \] Write $K=N^{-1}\mathbf1^TR(q-q')$, known from the two target policies.
Theorem (Independent-node information survives feedback): In model (4), $\widehat\theta=K\widehat\tau$ is unbiased for $\theta=K\tau$, and \[ \sup_{\tau\in[0,1/4]}{\mathbb{E}}(\widehat\theta-\theta)^2 \le \frac{C(q_0)K^2}{mN}. \] For $K\ne0$ and all $mN$ sufficiently large, the minimax risk on the same model is at least $c(q_0)K^2/(mN)$. For column-stochastic $A$ and constant target marginals $q_i=q_g$, $q_i'=q_{g'}$, this order is \[ \frac{(q_g-q_{g'})^2}{mN(1-\beta)^2}. \]
Proof: Conditioning on $W_i$ gives \[ {\mathbb{E}}[(W_i-q_0)Z_i] =q_0(1-q_0)\{(b+\tau)-b\} =q_0(1-q_0)\tau. \] The $H_{ri}$ are independent and bounded by a constant depending only on $q_0$, so their average has variance at most $C/(mN)$.

For the lower bound choose $\tau_0=1/8$ and $\tau_1=\tau_0+a/\sqrt{mN}$, with fixed small $a>0$. These lie in the parameter interval for large sample size. The transformed observations $(W,Z)$ have the same likelihood information as $(W,Y)$ because the transformation is invertible. The one-observation KL is $q_0$ times the Bernoulli KL between $b+\tau_0$ and $b+\tau_1$, hence at most $C(\tau_1-\tau_0)^2$; the elementary bound is ${\operatorname{KL}}(\operatorname{Ber}(u),\operatorname{Ber}(v))\le(u-v)^2/[v(1-v)]$. Thus product-law KL is at most $Ca^2$, and their total variation is at most $1/2$ for small $a$. A test choosing the closer target from any estimator has sum of errors at least $1/2$; on error squared loss is at least $K^2(\tau_1-\tau_0)^2/4$. This proves the stated lower bound. The final formula follows from Proposition 1.1. ∎

In particular $m=1$ can be sufficient as $N\to\infty$ when node innovations are independent and the feedback operator is known. The theorem does not treat noisy measurement of $Y$; multiplication by $I-\beta A$ would then also transform measurement error. Nor does it estimate $\beta$ or identify it from one randomization law.

Dependence can eliminate the benefit of network size

Proposition (A replication lower bound with shared shocks): There is a bounded-outcome subclass with contraction constant zero and independent Bernoulli node assignments in which the minimax risk for the all-treated versus all-untreated average effect is at least $c/m$, uniformly in $N$.
Proof: For each network draw a single $V\sim\operatorname{Ber}(\vartheta)$, independently of assignments and other networks, and put $Y_i(w)=w_iV$ for every vertex. Take $\vartheta\in[1/4,3/4]$. The average all-treated effect is $\vartheta$, and outcomes lie in $[0,1]$. The equations have no outcome feedback, so the contraction constant is zero.

All observations in a network are measurable functions of $(W,V)$. Even an augmented experiment revealing $V$ supplies just one Bernoulli observation per network, plus assignment variables whose law is independent of $\vartheta$. The original data cannot contain more information. In the augmented experiment compare $\vartheta_0=1/2$ with $\vartheta_1=1/2+a/\sqrt m$. Their product-law KL is bounded by a constant times $a^2$. The same two-point testing argument gives risk at least $c/m$ for small fixed $a$, adjusting constants for finitely many small $m$. This is therefore a lower bound for the original experiment at every $N$. ∎

This obstruction arises from the declared joint error law, not from graph density. It also shows why a contraction assumption alone cannot justify a $1/N$ variance formula.

What remains

For a fixed known resolvent, policy means and truncation bias are explicit and independent-node estimation has matching order. The broader agenda still requires unknown feedback, nonlinear fixed points, graph-dependent shock dependence, exposure maps preserving richer targets, and design optimization when only a small part of the assignment space is observed. The two subclasses here give opposite replication conclusions under the same unique-solution convention; any proposed general theorem must state which dependence restrictions select its regime.

References

  • [source] C. Cinelli, A. Feller, G. Imbens, E. Kennedy, S. Magliacane and J. Zubizarreta (2025), Challenges in Statistics: A Dozen Challenges in Causality and Causal Inference, arXiv:2508.17099v1, Section 3.2.3. https://arxiv.org/html/2508.17099v1. The source motivates reconciling assignment and outcome interference. The linear and shared-shock models above are explicit restrictions introduced in this attempt.

Completion Estimate

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