--- title: "Finite convex certificates for quadratic informational substitutes" date: 2026-10-15 version: "0.1.0-candidate" doi: 10.5281/zenodo.23401650 pdf: https://github.com/ipitchford/finite-signal-quadratic-substitutes/releases/download/v0.1.0-candidate/paper.pdf repository: https://github.com/ipitchford/finite-signal-quadratic-substitutes archive: https://zenodo.org/records/23401650 license: CC0-1.0 status: unrefereed (internally replayed; not peer reviewed, not independently reproduced, not formally verified) --- # Finite convex certificates for quadratic informational substitutes ## Summary Two reports can overlap without being interchangeable. Suppose a forecaster can reveal all of one report, none of it, or a randomly chosen fragment. Is the second report at least as valuable after every such partial disclosure as it is after the first report is revealed completely? This research candidate gives a finite mathematical test for that question when both reports have finitely many possible outcomes and prediction quality is measured by squared error for one numerical target. The test allows impossible combinations of report outcomes. It replaces an unlimited collection of disclosure choices with finitely many convex feasibility problems. The result also explains why checking two outcomes at a time can be misleading. In one disclosure orientation, every two-state posterior restriction of the first report passes, but a three-state mixture fails. A different example shows that this failure can occur even without impossible signal combinations when the target has several components. This is an analytic result with supplementary exact arithmetic checks, not a study of real forecasting performance. It remains an unrefereed, AI-assisted candidate. ## Summary for specialists Let $A$ and $B$ be finite signals, with positive source prior and arbitrary joint support. Write $r_{ib}=\Pr(B=b\mid A=i)$, $v_{ib}=\mathbb E[Y\mid A=i,B=b]$ on supported cells, and $x_i=\sum_b r_{ib}v_{ib}$ for a scalar square-integrable target $Y$. A disclosure $T$ is generated from $A$ using fresh randomness. For quadratic value $V(S)=\mathbb E[\mathbb E(Y\mid S)^2]$, the oriented substitution condition is $$\begin{aligned}G(T)&=V(T,B)-V(T)\\&\quad-V(A,B)+V(A)\geq0.\end{aligned}$$ for every admissible $T$. The two-sided condition applies the criterion in both orientations. These are the two-base-signal quantifiers in Chen–Waggoner Definition 2.2.3, not submodularity between arbitrary pairs of partially disclosed signals. Define $L_i(z)=\sum_{b:r_{ib}>0}r_{ib}(v_{ib}-z_b)^2$. Fix a finite $C\geq0$. For each distinct source mean $t$, find one auxiliary vector $z^t$ satisfying $$L_i(z^t)\leq C(x_i-t)^2$$ for every source state $i$. The paper proves that these finitely many completion systems are feasible exactly when $N(q)\leq C D(q)$ for every posterior $q$, where $D$ is the variance of the source means and $N$ is the residual conditional variance after observing $B$. Setting $C=1$ characterises substitution. Minimising $C$ gives the sharp coefficient through an attained convex quadratically constrained program whenever its value is finite. ## Technical account The proof has two different jobs. Compact minimax duality shows that a universal posterior inequality forces a common completion vector at each source mean. Interpolation between consecutive scalar means then turns those finitely many certificates into a bound for every posterior. The auxiliary completion coordinates are proof variables, not assigned conditional expectations on impossible cells. Under full rectangular support, the completion at $x_j$ is forced to be row $v_j$. The coefficient therefore reduces to $$R=\max_{i,j:x_i\ne x_j}\frac{\sum_b r_{ib}(v_{ib}-v_{jb})^2}{(x_i-x_j)^2}.$$ Equal-mean rows with incompatible conditional means give an infinite coefficient. If all source means agree and the rows are compatible, the coefficient is zero (the maximum over distinct means is empty). With sparse support, the compatibility requirement concerns only cells shared by the equal-mean rows. The sharp comparison concerns a **normalised gap**: under the stated nondegeneracy and full-support hypotheses, the infimum of $G(T)/(V(A)-V(T))$ is $1-R$. It is not a minimum absolute loss or a general optimal-disclosure solution. In limiting disclosures, both numerator and denominator can approach zero. ### One sparse family, two different answers Take channel rows $(1/2,1/2,0)$, $(0,1/2,1/2)$ and $(1/2,0,1/2)$, indexing rows and columns by $0,1,2$. The supported target means are $v_{00}=v_{01}=0$, $v_{11}=v_{20}=a$ and $v_{12}=v_{22}=b$. For $0 2. Anonymous, Four tests for binary informational substitutes: binary-source predecessor, including sparse support and vector targets. 3. Ordentlich and Polyanskiy, Strong Data Processing Constant is Achieved by Binary Inputs: adjacent binary extremal reduction, not the same fixed-target coefficient.