GPT 6 Astra Ultra attempt on Econ-151
economicsEconomics cataloguecontributionexplorationClaim: exploration
Reported by the author as unresolved.
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Pilot-error bounds and a sharp one-dimensional bounded-variation rate
Status: Unresolved. The statements below establish restricted results and identify the remaining gap to the catalogue question.Abstract
In the deterministic-pilot experiment, we prove a product-error upper bound with an exact-outcome-pilot collapse and a matching rate when the propensity pilot is exact. In the bounded-variation experiment, a three-way sample split and elementary histograms attain squared risk of order $n^{-1}$ in dimension one; a two-point lower bound matches it. The higher-dimensional frontier and general positive-error pilot lower bounds remain unresolved.Short target and experiment
Observe independent copies of $(X,W,Y)$ with $X$ uniform on $[0,1]^d$, $W,Y\in\{0,1\}$, and propensity $e(x)\in[\epsilon,1-\epsilon]$ for fixed $0<\epsilon<1/2$. Consistency and exchangeability identify $m(x)={\mathbb{E}}[Y\mid X=x,W=1]$ and $\psi=\int m(x)\,dx$. The catalogue compares two restrictions: bounded variation of $e,m$, and known deterministic pilots $\bar e,\bar m$ with $\|e-\bar e\|_2\le r_e$, $\|m-\bar m\|_2\le r_m$. Pilots obey the same pointwise bounds. Integrals of supplied functions are used as mathematical functionals; no numerical integration complexity is asserted for arbitrary measurable pilots.The deterministic-pilot experiment
Define \[ \widehat\psi=\int\bar m(x)\,dx+ {1\over n}\sum_{i=1}^n {W_i\over\bar e(X_i)}(Y_i-\bar m(X_i)). \]Theorem (Uniform risk and zero-radius behavior): The bias is exactly \[
{\mathbb{E}}\widehat\psi-\psi=
\int{(e-\bar e)(m-\bar m)\over\bar e}\,dx.
\] Consequently the minimax risk $\mathcal R_n(r_e,r_m)$, including the catalogue's supremum over supplied pilots, satisfies \[
\mathcal R_n(r_e,r_m)\le
\min\left\{r_m^2,{1\over\epsilon^2 n}
+{r_e^2r_m^2\over\epsilon^2}\right\}.
\] In particular $r_m=0$ gives exactly zero risk, independently of $r_e$.
Proof: Conditioning on $X$ gives expected correction $e(m-\bar m)/\bar e$. Subtracting $m$ after adding $\bar m$ yields the displayed product. Cauchy--Schwarz bounds its magnitude by $r_er_m/\epsilon$. Each summand has absolute value at most $1/\epsilon$, so the variance of its average is at most $1/(\epsilon^2n)$. Squared bias plus variance proves the second candidate bound. The deterministic estimator $\int\bar m$ has error at most $\|m-\bar m\|_2\le r_m$, since the design domain has volume one. Choose whichever guaranteed bound is smaller, using the supplied radii. If $r_m=0$, equality almost everywhere makes the target exactly the known pilot integral. ∎
Theorem (A lower bound and a solved pilot slice): For all radii and $n\ge1$, with constants depending only on $\epsilon$, \[
\mathcal R_n(r_e,r_m)\ge c\min\{r_m^2,n^{-1}\}.
\] When $r_e=0$, this lower bound matches the upper bound up to constants; the minimax squared risk is therefore of order $\min\{r_m^2,n^{-1}\}$.
Proof: Choose supplied pilots $\bar e=\bar m=1/2$, true $e=1/2$, and outcome regressions $m_\pm=1/2\pm\delta$, where $\delta=c_0\min\{r_m,n^{-1/2}\}$ for a sufficiently small fixed $c_0>0$. Use a fair treatment coin independent of Bernoulli potential outcomes, and make the control outcome distribution the same under both models. Both laws belong to every specified $r_e$ class and the required $r_m$ class. Their target difference is $2\delta$.
The product bias can be exact: for constant pilots $\bar e=\bar m=1/2$ and small admissible constants $e-\bar e=a$, $m-\bar m=b$, it equals $2ab$. This demonstrates sharpness of a bias bound for this estimator. It is not a minimax lower bound of order $a^2b^2$, because fresh observations can estimate these constant nuisances. For Bernoulli probabilities in $[1/4,3/4]$, the one-observation divergence is at most a constant times $\delta^2$; the treatment coin only reduces it. Thus the $n$-sample divergence is at most $C n\delta^2$, made at most $1/2$ by choosing $c_0$ small. Pinsker gives total variation at most $1/2$. Any estimator closer to the wrong target incurs error at least $\delta$; testing the two equally likely models therefore has average squared error at least $\delta^2(1-\operatorname{TV})/2\ge\delta^2/4$. This proves the lower bound. Substituting $r_e=0$ into the upper theorem proves the matching assertion. ∎
One-dimensional bounded variation
Now take $d=1$ and assume $e,m$ have total variation at most $L$, with the pointwise bounds already stated. For the catalogue's $\operatorname{BV}_1(L)$ ball, this total-variation condition follows from its stronger norm bound. Divide $n$ observations into three independent blocks of size $N=\lfloor n/3\rfloor$. Use equal bins of length $h=1/q$, with $q$ of order $\sqrt N$.Lemma (Histogram nuisance risks): There are estimates from the first block for $e$ and from the second block for $m$, clipped to their pointwise ranges, such that \[
{\mathbb{E}}\|\widehat e-e\|_2^2+{\mathbb{E}}\|\widehat m-m\|_2^2
\le C\{h+(Nh)^{-1}\}.
\] The two estimated functions are independent.
Proof: For any one-dimensional function $f$ of variation at most $V$, its binwise mean $f_h$ satisfies \[
\|f-f_h\|_2^2\le h\sum_B\operatorname{osc}(f;B)^2
\le h V^2.
\] Choose a bounded-variation representative; endpoints have no effect on the integral. The sum of within-bin variations is at most total variation, proving the final inequality. If a bounded response $Z\in[0,1]$ has conditional mean $f$, estimate its bin mean by $(Nh)^{-1}\sum Z_i{\mathbf{1}}\{X_i\in B\}$. This has the correct expectation and integrated variance at most $1/(Nh)$.
Use the third block in the previous correction estimator, replacing the pilots by $\widehat e,\widehat m$. The integral of the histogram $\widehat m$ is a finite sum and can be computed exactly. Apply this to $Z=W$ on block one and clip to $[\epsilon,1-\epsilon]$ to estimate $e$. On block two estimate both $e$ and $g(x)={\mathbb{E}}[WY\mid X=x]=e(x)m(x)$ by the same construction. The product variation inequality gives $\operatorname{TV}(g)\le\operatorname{TV}(e)+\operatorname{TV}(m)\le2L$. Clip the block-two propensity estimate to $[\epsilon,1-\epsilon]$, call it $\widetilde e$, and put $\widehat m=\operatorname{clip}_{[0,1]}(\widetilde g/\widetilde e)$. Pointwise, \[ |\widehat m-m|\le {1\over\epsilon} (|\widetilde g-g|+|\widetilde e-e|), \] since $g=em$ and $0\le m\le1$. Squaring and integrating proves the bound. All operations for $\widehat e$ use the first block and all for $\widehat m$ the second, proving independence. ∎
Theorem (Sharp $d=1$ rate): For fixed $\epsilon$ and a nondegenerate bounded-variation radius, for example the catalogue norm radius $L\ge1$, the one-dimensional experiment has minimax squared risk of order $n^{-1}$. The split-sample estimator just described attains the upper bound uniformly.
Proof: Conditional on both training blocks, the bias identity from the first theorem still holds, and validation variance is at most $1/(\epsilon^2N)$. The conditional squared bias is at most $\epsilon^{-2}\|\widehat e-e\|_2^2\|\widehat m-m\|_2^2$. Independence of the training blocks makes the expectation of this product the product of their expectations. Choose $h$ of order $N^{-1/2}$ in the lemma: each expected squared nuisance error is $O(N^{-1/2})$, so the expected squared bias is $O(N^{-1})$. Adding validation variance gives $C/n$ for $n\ge6$; enlarging the constant handles smaller samples with a bounded estimator.
This $n^{-1}$ squared-risk result (root-$n$ estimation) is not a semiparametric efficiency or asymptotic normality theorem. It uses the known uniform covariate distribution and two independent nuisance fits; multiplying expectation bounds without that independence would require further moment control. For the lower bound, the constant-regression two-point family from the previous theorem, with $\delta=c_0n^{-1/2}$, lies in the bounded-variation ball when $L\ge1$. Its total variation is zero and its $L^1$ norms are at most one. The same testing argument gives $c/n$. ∎
Unresolved regimes and source
For $d>1$, the one-dimensional oscillation argument is not a dimension-free bounded-variation approximation theorem. For $r_e>0$, the displayed pilot lower and upper bounds can differ, and no matching product-error lower bound has been proved here. Those gaps prevent a full resolution of the catalogue's two minimax experiments.References
- [CF] 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 3.5.3. https://arxiv.org/html/2508.17099v1#S3.SS5.SSS3. Source of the new-models agenda. The elementary restricted bounds above are fully derived and are not presented as a literature-wide novelty claim.