GPT 6 Astra Ultra attempt on Econ-36
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Scalable heterogeneous-agent macroeconomics A coupled residual certificate and its conditioning limits
Problem ID:C63-35. Status: Unresolved; rigorous restricted results.Short recap and restricted result
The target asks for globally certified heterogeneous-agent computation, including binding constraints, and propagation of solver error into estimation. We give a complete certificate for a coupled update whose two blocks satisfy explicit Lipschitz inequalities. The missing economic step is verification of those inequalities uniformly over an invariant domain; household discounting alone does not supply them.The sequence-space method of Auclert et al. [ssj] is an established efficient approach to solution and estimation. Krusell--Smith [ks] provides a foundational aggregate-risk computation. Neither reference is being invoked for the global certificate proved here. The certificate is an elementary contraction argument, specialized to expose the feedback constants that matter for a coupled economy.
Coupled residuals
Let $X,Y$ be nonempty complete metric spaces with distances $d_X,d_Y$. They may consist of value functions and, in the second block, prices, policies and distribution-transition functions. A domain is fixed in advance, and all distances below are uniform norms on that domain when the objects are functions. Let $T=(T_1,T_2):X\times Y\to X\times Y$ map the domain into itself. Assume its fixed points are precisely equilibria of the declared restricted model, with feasibility and complementarity included in this equivalence. Suppose for all $x,x',y,y'$, \begin{align} d_X(T_1(x,y),T_1(x',y'))&\leq\beta d_X(x,x')+b d_Y(y,y'),\\ d_Y(T_2(x,y),T_2(x',y'))&\leq c d_X(x,x')+d d_Y(y,y'), \end{align} where $0\leq\beta,d<1$ and $b,c\geq0$. Put $D=(1-\beta)(1-d)-bc$.Theorem (Global two-block error certificate): If $D>0$, $T$ has exactly one fixed point $(x^*,y^*)$. For any candidate $(x,y)$ define the exact residual magnitudes $r_1=d_X(x,T_1(x,y))$ and $r_2=d_Y(y,T_2(x,y))$. Then \[
\begin{pmatrix}d_X(x,x^*)\\d_Y(y,y^*)\end{pmatrix}
\leq\frac1D
\begin{pmatrix}1-d&b\\c&1-\beta\end{pmatrix}
\begin{pmatrix}r_1\\r_2\end{pmatrix}
\] coordinatewise. There is $t>0$ such that $q=\max\{\beta+bt,d+c/t\}<1$. In the distance $d_t((x,y),(x',y'))=\max\{d_X(x,x'),d_Y(y,y')/t\}$, iteration satisfies \[
d_t(T^k(x,y),(x^*,y^*))\leq
\frac{q^k}{1-q}d_t((x,y),T(x,y)).
\]
Proof: When $b,c>0$, choose $c/(1-d)<t<(1-\beta)/b$; the interval is nonempty exactly because $D>0$. If $b=0$, choose $t$ large enough that $d+c/t<1$. If $c=0,b>0$, choose $t<(1-\beta)/b$. These choices also cover $b=c=0$. The two Lipschitz inequalities prove $T$ is a $q$-contraction in $d_t$. Consecutive iterates have distances at most $q^k d_t(u,Tu)$; summing this geometric series makes them Cauchy in the complete product space. Their limit is fixed by continuity of $T$. Two fixed points have distance at most $q$ times their distance, so coincide. The tail sum gives the iteration estimate. For the coordinate estimate write $a=d_X(x,x^*)$, $e=d_Y(y,y^*)$. The triangle inequality gives $(1-\beta)a-be\leq r_1$ and $(1-d)e-ca\leq r_2$. Multiply the first by $1-d$ and add $b$ times the second to get $Da\leq(1-d)r_1+br_2$. Multiply the second by $1-\beta$ and add $c$ times the first to obtain the other bound. ∎
Proposition (Approximate evaluation and global validation): Suppose only $\widetilde T(u)$ is evaluated, with verified block errors $d_X(T_1(u),\widetilde T_1(u))\leq\delta_1$ and $d_Y(T_2(u),\widetilde T_2(u))\leq\delta_2$. The preceding coordinate bound remains valid after replacing $r_i$ by $d_i(u_i,\widetilde T_i(u))+\delta_i$. If a scalar residual function $r(s)$ is $L$-Lipschitz on a metric domain $\mathcal D$ and $G$ is an $h$-net, then \[
\sup_{s\in\mathcal D}|r(s)|\leq\max_{g\in G}|r(g)|+Lh.
\] The same statement holds for norm-valued residuals using their norm Lipschitz bound. Truncation, quadrature and distribution-compression errors must be included in the $\delta_i$ or in the grid bound.
Proof: The first claim is the triangle inequality for each block. For every $s$ choose $g\in G$ with distance at most $h$; then $|r(s)|\leq |r(g)|+Lh$. Take the supremum. For vectors apply the reverse triangle inequality to their norms. ∎
This validates all states only when the asserted net and Lipschitz bound are themselves established. Finite simulation paths need not form a net of a distribution space.
Explicit failures of unjustified certificates
Proposition (Conditioning cannot be omitted): For each $\gamma\in(0,1)$ the scalar fixed-point equation $x=(1-\gamma)x$ has unique equilibrium zero and residual $r(x)=\gamma|x|$. Every error bound $|x|\leq C r(x)$ valid for nonzero $x$ requires $C\geq\gamma^{-1}$. For the equation $F(x)=x^2=0$ on $[0,1]$, no finite constant $C$ satisfies $|x|\leq C|F(x)|$ throughout any neighborhood of zero. Finally fixed-price contraction of a household block does not imply contraction of the coupled update.
Proof: The first ratio is identically $\gamma^{-1}$. The second ratio is $1/x$ for $x>0$, unbounded near zero. For the last assertion take $T(v,p)=(\tfrac12v+p,v)$ on ${\mathbb{R}}^2$. For fixed $p$ the first component contracts in $v$ with coefficient $1/2$. Its update matrix has positive eigenvalue $(1+\sqrt{17})/4>1$, so iteration expands along its eigenvector in every norm. Here the feedback product $bc=1$ violates $D>0$. ∎
Binding constraints do not invalidate the theorem when the update is Lipschitz but nondifferentiable. They do invalidate applications that silently replace a complementarity residual with an equality residual on slack inequalities. No differentiability of $T$ was used.
Propagation into estimation
Let $\Theta$ be a parameter set with metric $d_\Theta$, true parameter $\theta_0$, and model moments $M(\theta)\in{\mathbb{R}}^k$. Numerical moments $\widetilde M$ and empirical moments $\widehat m$ satisfy \[ \sup_{\theta\in\Theta}\|\widetilde M(\theta)-M(\theta)\|\leq\eta, \quad \|\widehat m-M(\theta_0)\|\leq\zeta. \] Assume quantitative identification $\|M(\theta)-M(\theta_0)\|\geq\kappa d_\Theta(\theta,\theta_0)$ for all $\theta$, with $\kappa>0$.Theorem (Solver and sampling errors remain separately visible): If $\widehat\theta$ is a $\xi$-approximate minimum-distance fit, meaning \[
\|\widetilde M(\widehat\theta)-\widehat m\|
\leq\inf_\theta\|\widetilde M(\theta)-\widehat m\|+\xi,
\] then $d_\Theta(\widehat\theta,\theta_0)\leq
[2\eta+2\zeta+\xi]/\kappa$. If moments are $L_M$-Lipschitz in the equilibrium object and that object has a uniform numerical error bound $\Delta$, with direct integration error at most $\eta_0$, one may take $\eta=L_M\Delta+\eta_0$.
Proof: Use $\theta_0$ as a feasible comparison to bound the fitted discrepancy by $\eta+\zeta+\xi$. A second triangle inequality bounds $\|M(\widehat\theta)-M(\theta_0)\|$ by $\eta+(\eta+\zeta+\xi)+\zeta$. Divide by $\kappa$. The last assertion follows by adding the Lipschitz change in the moment to its direct numerical integration error. ∎
Verification and unresolved step
All displayed bounds are deterministic and require no empirical benchmark claims. In particular the exact inverse contains the feedback denominator $D$, which can be small even when $\beta<1$ is well below one. The first unresolved economic step is construction of an equilibrium- equivalent invariant update with certified $b,c,d$ satisfying $D>0$ for a nontrivial globally nonlinear aggregate-risk model across binding constraints. The theorem supplies no general complexity guarantee for covering its infinite-dimensional distribution domain. Thus the complete catalogue computation and benchmarking programme remains unresolved.References
- [ssj] A. Auclert, B. Bardóczy, M. Rognlie and L. Straub (2021), Using the Sequence-Space Jacobian to Solve and Estimate Heterogeneous-Agent Models, Econometrica 89(5), 2375--2408. https://doi.org/10.3982/ECTA17434.
- [ks] P. Krusell and A. A. Smith, Jr. (1998), Income and Wealth Heterogeneity in the Macroeconomy, Journal of Political Economy 106(5), 867--896. https://doi.org/10.1086/250034.