diff --git a/lectures/_static/quant-econ.bib b/lectures/_static/quant-econ.bib index fb7c29ed1..7a3609776 100644 --- a/lectures/_static/quant-econ.bib +++ b/lectures/_static/quant-econ.bib @@ -5605,6 +5605,8 @@ @techreport{AlvarezArgente2026 title = {Strategic Complementarity and Persistence in Linear-Quadratic Mean Field Games}, institution = {University of Chicago and Yale University}, + type = {Working Paper}, + month = oct, year = {2026} } @@ -5698,6 +5700,62 @@ @article{AlvarezLippi2014 year = {2014} } +@article{Hayashi1982, + author = {Hayashi, Fumio}, + title = {Tobin's Marginal $q$ and Average $q$: A Neoclassical + Interpretation}, + journal = {Econometrica}, + volume = {50}, + number = {1}, + pages = {213--224}, + year = {1982}, + doi = {10.2307/1912538} +} + +@article{CooperHaltiwanger2006, + author = {Cooper, Russell W. and Haltiwanger, John C.}, + title = {On the Nature of Capital Adjustment Costs}, + journal = {Review of Economic Studies}, + volume = {73}, + number = {3}, + pages = {611--633}, + year = {2006}, + doi = {10.1111/j.1467-937X.2006.00389.x} +} + +@article{BenhabibFarmer1994, + author = {Benhabib, Jess and Farmer, Roger E. A.}, + title = {Indeterminacy and Increasing Returns}, + journal = {Journal of Economic Theory}, + volume = {63}, + number = {1}, + pages = {19--41}, + year = {1994}, + doi = {10.1006/jeth.1994.1031} +} + +@article{Wolf2023, + author = {Wolf, Christian K.}, + title = {The Missing Intercept: A Demand Equivalence Approach}, + journal = {American Economic Review}, + volume = {113}, + number = {8}, + pages = {2232--2269}, + year = {2023}, + doi = {10.1257/aer.20211751} +} + +@article{BallRomer1990, + author = {Ball, Laurence and Romer, David}, + title = {Real Rigidities and the Non-Neutrality of Money}, + journal = {Review of Economic Studies}, + volume = {57}, + number = {2}, + pages = {183--203}, + year = {1990}, + doi = {10.2307/2297377} +} + @article{GallantTauchen1996, author = {Gallant, A. Ronald and Tauchen, George}, title = {Which Moments to Match?}, diff --git a/lectures/_toc.yml b/lectures/_toc.yml index 4b01b54ea..1f2829bd6 100644 --- a/lectures/_toc.yml +++ b/lectures/_toc.yml @@ -172,6 +172,7 @@ parts: - file: two_computation - file: social_security_depletion - file: lq_mean_field_games + - file: lq_mean_field_games_extensions - caption: Asset Pricing and Finance numbered: true chapters: diff --git a/lectures/lq_mean_field_games.md b/lectures/lq_mean_field_games.md index e20a7c41f..3d21e0c9d 100644 --- a/lectures/lq_mean_field_games.md +++ b/lectures/lq_mean_field_games.md @@ -38,23 +38,18 @@ It pairs two partial differential equations: An equilibrium requires that the aggregates that agents take as given are the ones that their own decisions generate. -The same pair of equations is the workhorse of continuous-time heterogeneous-agent macroeconomics; see {cite:t}`AchdouEtAl2022`. - -Readers of {doc}`rational_expectations` will recognize that requirement. +Readers of {doc}`rational_expectations` will recognize that requirement: it is the "Big $Y$, little $y$" idea, now applied to an entire distribution rather than to a single number. -It is the "Big $Y$, little $y$" idea, now applied to an entire distribution rather than to a single number. +The same pair of equations is the workhorse of continuous-time heterogeneous-agent macroeconomics; see {cite:t}`AchdouEtAl2022`. This lecture studies a tractable special case in which the payoff is quadratic and the state evolves linearly, following {cite:t}`AlvarezArgente2026`. -Two kinds of interaction appear: - -* agents care about the cross-sectional average *state* $X$, through a matrix $\Theta_X$ -* agents care about the cross-sectional average *action* $\mathcal A$, through a matrix $\Theta_{\mathcal A}$ +Agents care about the cross-sectional average *state* $X$, through a matrix $\Theta_X$, and about the cross-sectional average *action* $\mathcal A$, through a matrix $\Theta_{\mathcal A}$. The main result is a striking simplification: ```{note} -The equilibrium of the mean field game solves the algebraic Riccati equation of a *single-agent* linear quadratic regulator problem, in which the curvature matrices $Q$ and $\Gamma$ are replaced by +The equilibrium of the mean field game solves the algebraic Riccati equation of a *single-agent* linear quadratic regulator problem, in which the curvature matrices $Q$ and $\Gamma$ of the return in an agent's own state and own action are replaced by the **effective curvatures** $$ Q + \Theta_X \qquad\text{and}\qquad \Gamma + \Theta_{\mathcal A} . @@ -63,7 +58,16 @@ $$ Everything we know about the linear regulator can therefore be brought to bear on the equilibrium: existence conditions, uniqueness, comparative statics, and numerical methods. -We then put the framework to work on two economic examples from {cite:t}`AlvarezArgente2026`: an industry equilibrium with capital accumulation, and a multiproduct pricing problem with Kimball demand. +Four lessons organize the lecture: + +* **Effective curvature** ({ref}`mfg_effective`). Aggregate dynamics depend on strategic interaction only through the effective curvatures, which measure how fast an agent's marginal returns fall when everybody moves together. +* **Individual versus aggregate** ({ref}`mfg_decomp`). An agent's deviation from the cross-sectional mean depends only on $Q$ and $\Gamma$, which measure how fast her marginal returns fall when she moves alone, so individual (micro) and aggregate (macro) adjustment speeds differ: three years against five in our capital example. +* **The missing intercept** ({ref}`mfg_identification`). Regressions with time effects see only deviations from the mean, while strategic interaction lives entirely in the time effects. +* **Marginal versus total** ({ref}`mfg_planner`). A planner responds to how the *total* return changes when everybody moves together, not to how an agent's marginal return changes, and that can make the planner's allocation either slower or faster than the equilibrium. + +We illustrate them with two economic examples from {cite:t}`AlvarezArgente2026`: an industry equilibrium with capital accumulation, and a multiproduct pricing problem with Kimball demand. + +A sequel, {doc}`lq_mean_field_games_extensions`, adds cross terms between states and actions, forcing processes, and interactions through higher moments of the cross-sectional distribution. Riccati equations appear in several other QuantEcon lectures: @@ -77,7 +81,10 @@ Let's start with some imports: ```{code-cell} ipython3 import numpy as np import matplotlib.pyplot as plt +from matplotlib.patches import Patch +from matplotlib.lines import Line2D from scipy.linalg import solve_continuous_are, expm +from scipy.optimize import brentq ``` ## The environment @@ -90,12 +97,12 @@ The cross-sectional average state and action are $$ X = \int x \, m(x) dx , \qquad -\mathcal A = \int \alpha_*(x) \, m(x) dx , +\mathcal A = \int \alpha^*(x) \, m(x) dx , $$ -where $m$ is the density of the state across agents. +where $m$ is the density of the state across agents and $\alpha^*$ is the optimal policy. -The period return is quadratic in all four objects: +The period return is ```{math} :label: mfg_return @@ -104,6 +111,14 @@ F(x,X) + R(\alpha, \mathcal A) -\tfrac12 \alpha^\top \Gamma \alpha - \mathcal A^\top \Theta_{\mathcal A}\alpha . ``` +This is not the most general quadratic return. + +A second-order approximation to a general return also produces cross terms between states and actions, which {doc}`lq_mean_field_games_extensions` takes up. + +It also produces terms that involve only the aggregates, which do not affect an individual's choices but do matter for a planner, as we will see in {ref}`mfg_planner`. + +We omit constants, which do not affect choices, and linear terms, which affect steady-state levels but not the dynamics of deviations from them. + The individual state follows ```{math} @@ -117,6 +132,12 @@ Agents discount at rate $\rho > 0$. We assume that $Q$ and $\Gamma$ are positive definite, that $\Sigma$ is positive semi-definite, that $B$ has full row rank, and that $\Theta_X$ and $\Theta_{\mathcal A}$ are symmetric. +Symmetry of the interaction matrices is a genuine restriction. + +The own-curvature matrix $Q$ can always be taken to be symmetric, because $x^\top Q x$ depends only on the symmetric part of $Q$, but $X^\top\Theta_X x$ pairs two different vectors, and the antisymmetric part of $\Theta_X$ does affect choices. + +It is, however, a restriction on payoffs rather than on behavior: the pricing example below has asymmetric best responses but a symmetric $\Theta_X$. + ### Complements and substitutes From {eq}`mfg_return`, the cross derivative between an agent's own state and the average state is $-\Theta_X$, and between her own action and the average action it is $-\Theta_{\mathcal A}$. @@ -128,12 +149,16 @@ So and similarly for actions. -In the Loewner ordering, *smaller* $\Theta$ means *more* complementarity. +We compare symmetric matrices with the Loewner order, in which $\Theta_1 \preceq \Theta_2$ means that $\Theta_2 - \Theta_1$ is positive semi-definite. + +In that order, *smaller* $\Theta$ means *more* complementarity. ```{note} -The monotonicity condition that {cite:t}`LasryLions2007` impose to obtain uniqueness corresponds here to $-\Theta_X$ being negative semi-definite, that is, to strategic *substitutability* in states. +The monotonicity condition that {cite:t}`LasryLions2007` impose to obtain uniqueness corresponds here to $\Theta_X$ being positive semi-definite, that is, to strategic *substitutability* in states. + +We will not need it: under the symmetry assumptions above, and provided $\Gamma + \Theta_{\mathcal A}$ is positive definite, there is at most one equilibrium whether interactions are complements or substitutes. -We will not need it: in this linear quadratic setting there is at most one equilibrium whether interactions are complements or substitutes. +{doc}`lq_mean_field_games_extensions` shows how uniqueness can fail once interactions between states and actions are allowed to be asymmetric. ``` ### Equilibrium @@ -155,7 +180,7 @@ H(p, x, \mathcal A) = \max_{\alpha} + p^\top(B\alpha - Ax) \right\} , $$ -with maximizer $\alpha_*(p,\mathcal A) = \Gamma^{-1}(B^\top p - \Theta_{\mathcal A}\mathcal A)$. +with maximizer $\alpha^*(p,\mathcal A) = \Gamma^{-1}(B^\top p - \Theta_{\mathcal A}\mathcal A)$. The density evolves according to the Kolmogorov forward equation @@ -200,7 +225,7 @@ This is the familiar algebraic Riccati equation of {doc}`lqcontrol`, written in The optimal action is $$ -\alpha_*(x,t) = \Gamma^{-1}\left[B^\top(\beta_1(t) + \bar\beta_2 x) - \Theta_{\mathcal A}\mathcal A(t)\right] . +\alpha^*(x,t) = \Gamma^{-1}\left[B^\top(\beta_1(t) + \bar\beta_2 x) - \Theta_{\mathcal A}\mathcal A(t)\right] . $$ Averaging across agents and solving the resulting fixed point in $\mathcal A$ gives @@ -216,7 +241,15 @@ Equation {eq}`mfg_aggregate_action` is where the action interaction first bites: Two objects now remain: the linear coefficient $\beta_1(t)$ of the value function, and the aggregate state $X(t)$. -Differentiating and aggregating gives a pair of linear differential equations, +Matching the terms in the HJB equation that are linear in $x$ gives + +$$ +\dot\beta_1 = (\rho I + A^\top)\beta_1 + \Theta_X X - \bar\beta_2 B\Gamma^{-1}\left(B^\top\beta_1 - \Theta_{\mathcal A}\mathcal A\right) , +$$ + +and averaging the state equation across agents gives $\dot X = B\mathcal A - AX$, because the idiosyncratic shocks average out. + +Substituting {eq}`mfg_aggregate_action` into both equations gives a pair of linear differential equations, ```{math} :label: mfg_hamiltonian_system @@ -225,15 +258,15 @@ Differentiating and aggregating gives a pair of linear differential equations, \qquad \mathcal H = \begin{bmatrix} -\rho I + A^\top - \bar\beta_2 \Lambda & \Theta_X + \bar\beta_2(B\Gamma^{-1}B^\top - \Lambda)\bar\beta_2 \\ -\Lambda & -A + \Lambda\bar\beta_2 +\rho I + A^\top - \bar\beta_2 D & \Theta_X + \bar\beta_2(B\Gamma^{-1}B^\top - D)\bar\beta_2 \\ +D & -A + D\bar\beta_2 \end{bmatrix}, ``` where we abbreviate $$ -\Lambda \equiv B(\Gamma + \Theta_{\mathcal A})^{-1}B^\top . +D \equiv B(\Gamma + \Theta_{\mathcal A})^{-1}B^\top . $$ This is a **state-costate** system of exactly the kind studied in {doc}`lagrangian_lqdp`. @@ -242,9 +275,9 @@ The aggregate state $X$ has an initial condition, namely the mean of the initial The costate $\beta_1$ does not: it must be chosen so that the solution does not violate the agent's transversality condition, which here requires the eigenvalues governing the path to have real parts below $\rho/2$. -Because $\Theta_X$, $B\Gamma^{-1}B^\top$ and $\Lambda$ are symmetric, $\mathcal H - \tfrac\rho2 I$ is a Hamiltonian matrix, so its eigenvalues are symmetric about the origin. +Because $\Theta_X$, $B\Gamma^{-1}B^\top$ and $D$ are symmetric, $\mathcal H - \tfrac\rho2 I$ is a Hamiltonian matrix, so its eigenvalues are symmetric about the origin. -Equivalently: +It follows that: ```{prf:proposition} :label: mfg_prop_roots @@ -252,7 +285,9 @@ Equivalently: If $\lambda$ is an eigenvalue of $\mathcal H$, then so are $\rho - \lambda$, $\bar\lambda$, and $\rho - \bar\lambda$. ``` -Exactly $n$ eigenvalues can therefore have real parts below $\rho/2$, which pins down a unique stable invariant subspace and hence at most one equilibrium. +So at most $n$ eigenvalues can have real parts below $\rho/2$, and exactly $n$ do when none lies on the line $\operatorname{Re}\lambda = \rho/2$. + +That pins down a unique stable invariant subspace, and when $D$ is positive definite that subspace can be written as $\beta_1 = SX$ for a single matrix $S$, so there is at most one equilibrium. This is the standard connection between algebraic Riccati equations and invariant subspaces of Hamiltonian matrices, treated at length by {cite:t}`LancasterRodman1995`. @@ -260,9 +295,7 @@ This is the standard connection between algebraic Riccati equations and invarian We look for a saddle path along which the costate is a linear function of the state, $\beta_1(t) = S X(t)$. -Substituting into {eq}`mfg_hamiltonian_system` gives a quadratic matrix equation for $S$ whose coefficients involve $\bar\beta_2$. - -That equation looks forbidding, but a change of variable transforms it. +Substituting into {eq}`mfg_hamiltonian_system` gives a quadratic matrix equation for $S$ whose coefficients involve $\bar\beta_2$, and a change of variable simplifies it. Define @@ -270,10 +303,14 @@ $$ P \equiv S + \bar\beta_2 . $$ +The new matrix has a natural meaning. + +An agent at the mean has marginal value of the state $u_x = \beta_1 + \bar\beta_2 X = (S + \bar\beta_2)X = PX$, so $P$ maps the aggregate state into the costate of the average agent. + ```{prf:proposition} :label: mfg_prop_riccati -An equilibrium is characterized by a matrix $P$ solving +An equilibrium is characterized by a symmetric matrix $P$ solving $$ P \, B(\Gamma+\Theta_{\mathcal A})^{-1}B^\top \, P @@ -298,14 +335,35 @@ The only difference is that $Q$ has become $Q + \Theta_X$ and $\Gamma$ has becom ```{prf:proposition} :label: mfg_prop_equivalence -The equilibrium of a linear quadratic mean field game with interaction matrices $\Theta_X$ and $\Theta_{\mathcal A}$, and its aggregate law of motion, coincide with the solution of a single-agent linear quadratic control problem whose state curvature is $Q+\Theta_X$ and whose action curvature is $\Gamma+\Theta_{\mathcal A}$. +Suppose $Q + \Theta_X$ and $\Gamma + \Theta_{\mathcal A}$ are positive definite. + +Then the equilibrium of a linear quadratic mean field game with interaction matrices $\Theta_X$ and $\Theta_{\mathcal A}$, and its aggregate law of motion, coincide with the solution of a single-agent linear quadratic control problem whose state curvature is $Q+\Theta_X$ and whose action curvature is $\Gamma+\Theta_{\mathcal A}$. ``` This is the organizing result of the lecture. -It says that strategic interaction does not change the *form* of the problem that determines aggregate dynamics; it changes the *curvatures* that enter it. +(mfg_effective)= +### Effective curvature + +We call $Q + \Theta_X$ and $\Gamma + \Theta_{\mathcal A}$ the **effective curvatures**. + +They have a simple reading. + +An agent's marginal return to her state is $-Qx - \Theta_X X$. + +If she moves alone, holding $X$ fixed, it falls with slope $Q$. + +If everybody moves together, so that $x = X$, it equals $-(Q + \Theta_X)X$ and falls with slope $Q + \Theta_X$. + +The first slope governs an agent's deviations from the crowd, and the second governs the crowd. -An immediate economic implication is that two models with very different strategic interactions can generate identical aggregate dynamics, provided the effective curvatures agree. +The same reading applies to actions, with $\Gamma$ and $\Gamma + \Theta_{\mathcal A}$. + +{numref}`fig-mfg-marginal-returns` below draws the two slopes for our capital accumulation example. + +So strategic interaction does not change the *form* of the problem that determines aggregate dynamics; it changes the *curvatures* that enter it. + +An immediate implication is that two models with very different strategic interactions can generate identical aggregate dynamics, provided their effective curvatures agree, and {ref}`mfg_identification` shows what that means for estimation. ### Existence and uniqueness @@ -323,18 +381,48 @@ E \equiv Q + \Theta_X + \left(A^\top + \tfrac\rho2 I\right) ```{prf:proposition} :label: mfg_prop_existence -Suppose $B(\Gamma+\Theta_{\mathcal A})^{-1}B^\top$ is invertible. +Suppose $B(\Gamma+\Theta_{\mathcal A})^{-1}B^\top$ is positive definite. -1. A necessary condition for an equilibrium is that $E$ be positive semi-definite. -1. If $Q + \Theta_X$ and $\Gamma + \Theta_{\mathcal A}$ are positive definite, an equilibrium exists and is unique. 1. There is at most one equilibrium. +1. A necessary condition for a symmetric solution of the equilibrium Riccati equation is that $E$ be positive semi-definite. +1. If $Q + \Theta_X$ is positive definite, an equilibrium exists. ``` The sufficient condition has a clean reading: *effective* curvature must remain positive in both states and actions. +For states that is $Q + \Theta_X \succ 0$; for actions it is the maintained assumption on $B(\Gamma+\Theta_{\mathcal A})^{-1}B^\top$, which holds whenever $\Gamma + \Theta_{\mathcal A} \succ 0$. + Complementarity is therefore permissible, but only up to a point. -Notice also that the two interactions enter $E$ additively, so substitutability in actions can offset complementarity in states. +The two interactions enter $E$ through separate terms, and a larger $\Theta_{\mathcal A}$ raises the second, so substitutability in actions can offset complementarity in states. + +(mfg_decomp)= +### Aggregate and individual dynamics + +The equilibrium splits each agent's state into the cross-sectional mean $X$ and her deviation from it, $z = x - X$. + +Using $\beta_1 = (P - \bar\beta_2)X$ in the optimal action shows that $\alpha^* - \mathcal A = \Gamma^{-1}B^\top\bar\beta_2 z$. + +Subtracting the aggregate law of motion from the individual one then gives + +```{math} +:label: mfg_decomposition +\begin{aligned} +\dot X &= \left(D P - A\right) X , \\ +dz &= \left(B\Gamma^{-1}B^\top\bar\beta_2 - A\right) z \, dt + \Sigma^{1/2}dW , \\ +\alpha^* &= (\Gamma+\Theta_{\mathcal A})^{-1}B^\top P X + \Gamma^{-1}B^\top\bar\beta_2 z . +\end{aligned} +``` + +The aggregate follows the equilibrium matrix $P$. + +An agent's deviation from the mean follows the single-agent matrix $\bar\beta_2$, and neither interaction matrix appears, because strategic interaction operates through $X$ and $\mathcal A$, which drop out of deviations from the mean. + +So the deviation of any agent from the crowd moves exactly as the whole economy would move if there were no interactions. + +From now on we call these **individual** dynamics, as opposed to **aggregate** dynamics, and we label them "no interactions" in tables and figures. + +This split is also the source of the missing intercept that we study in {ref}`mfg_identification`: data on deviations from the mean reveal only the individual dynamics. ## Computing equilibria @@ -344,27 +432,42 @@ $$ A^\top \tilde P + \tilde P A - \tilde P B R^{-1} B^\top \tilde P + Q = 0 . $$ -Our equation differs in two ways: our $P$ is negative definite, and we discount. +Our equation differs in two ways: our $P$ is negative definite when the effective curvatures are positive definite, and we discount. Writing $P = -\tilde P$ and collecting terms shows that our equation is the standard one with $A$ replaced by $-(A + \tfrac\rho2 I)$. The discount rate enters exactly as it does in the "$\rho/2$ shift" familiar from continuous-time control. +When no admissible solution exists, `solve_continuous_are` usually raises an error, but not always: it can return a matrix that does not solve the equation at all. + +So our function checks both the Riccati residual and the eigenvalue condition, and raises an error if either fails. + ```{code-cell} ipython3 -def mfg_riccati(A, B, Q, Γ, ρ): +def closed_loop(A, B, Γ_eff, P): + "The matrix governing aggregate dynamics, Ẋ = (B Γ_eff^{-1} B' P - A) X." + return B @ np.linalg.solve(Γ_eff, B.T @ P) - A + +def mfg_riccati(A, B, Q, Γ, ρ, tol=1e-8): """ - Solve P B Γ^{-1} B' P = Q + ρ P + P A + A' P for the negative - definite stabilizing solution P. + Solve P B Γ^{-1} B' P = Q + ρ P + P A + A' P for the solution P whose + closed-loop matrix B Γ^{-1} B' P - A has eigenvalues with real parts + below ρ/2, and raise ValueError if there is none. Passing Q + Θ_X and Γ + Θ_A gives the equilibrium of the mean field game; passing Q and Γ gives the single agent's value function curvature. """ n = A.shape[0] - return -solve_continuous_are(-(A + ρ/2 * np.eye(n)), B, Q, Γ) - -def closed_loop(A, B, Γ_eff, P): - "The matrix governing aggregate dynamics, Ẋ = (B Γ_eff^{-1} B' P - A) X." - return B @ np.linalg.solve(Γ_eff, B.T) @ P - A + try: + P = -solve_continuous_are(-(A + ρ/2 * np.eye(n)), B, Q, Γ) + except (np.linalg.LinAlgError, ValueError): + raise ValueError("no admissible solution of the Riccati equation") + quad = P @ B @ np.linalg.solve(Γ, B.T) @ P + residual = quad - (Q + ρ*P + P @ A + A.T @ P) + scale = 1 + np.abs(Q).max() + np.abs(quad).max() + if (np.abs(residual).max() > tol*scale + or np.linalg.eigvals(closed_loop(A, B, Γ, P)).real.max() >= ρ/2): + raise ValueError("no admissible solution of the Riccati equation") + return P ``` Let's set up a two-dimensional example with complementarity in states and substitutability in actions, the configuration that {cite:t}`AlvarezArgente2026` obtain from an industry equilibrium with capital accumulation. @@ -390,21 +493,21 @@ print("individual curvature β̄₂ =\n", β2.round(4)) print("\nequilibrium matrix P =\n", P.round(4)) ``` -Let's verify that $P$ really does solve the equilibrium Riccati equation, and compare the aggregate dynamics with what a lone agent would choose. +Let's compare aggregate dynamics with individual dynamics, that is, with the dynamics of an agent's deviation from the mean. ```{code-cell} ipython3 -Λ = B @ np.linalg.solve(Γ + Θ_A, B.T) -residual = P @ Λ @ P - (Q + Θ_X + ρ*P + P @ A + A.T @ P) -print(f"Riccati residual: {np.abs(residual).max():.2e}") +D = B @ np.linalg.solve(Γ + Θ_A, B.T) -G = closed_loop(A, B, Γ, β2) # dynamics without any interaction -JG = closed_loop(A, B, Γ + Θ_A, P) # equilibrium dynamics +U_ind = closed_loop(A, B, Γ, β2) # individual dynamics +U_agg = closed_loop(A, B, Γ + Θ_A, P) # aggregate dynamics -print("\neigenvalues without interactions:", np.linalg.eigvals(G).round(4)) -print("eigenvalues in equilibrium: ", np.linalg.eigvals(JG).round(4)) +print("individual eigenvalues:", np.linalg.eigvals(U_ind).round(4)) +print("aggregate eigenvalues: ", np.linalg.eigvals(U_agg).round(4)) ``` -The equilibrium eigenvalues are closer to zero, so aggregate adjustment is slower than it would be if each agent ignored everyone else. +The aggregate eigenvalues are closer to zero, so the economy as a whole adjusts more slowly than an agent's deviation from it. + +This example has complementarity in states and substitutability in actions, and {ref}`mfg_persistence` shows that both raise the sum of the eigenvalues; here each eigenvalue moves toward zero as well. ### Checking the Hamiltonian structure @@ -416,17 +519,17 @@ Let's check both claims. BΓB = B @ np.linalg.solve(Γ, B.T) n = A.shape[0] -H = np.block([[ρ*np.eye(n) + A.T - β2 @ Λ, Θ_X + β2 @ (BΓB - Λ) @ β2], - [Λ, -A + Λ @ β2]]) +Ham = np.block([[ρ*np.eye(n) + A.T - β2 @ D, Θ_X + β2 @ (BΓB - D) @ β2], + [D, -A + D @ β2]]) -ev = np.linalg.eigvals(H) +ev = np.linalg.eigvals(Ham) print("eigenvalues of ℋ:", np.sort(ev.real).round(4)) print("paired as λ and ρ - λ:", np.allclose(np.sort(ev.real), np.sort(ρ - ev.real))) stable = np.sort(ev.real[ev.real < ρ/2]) print("\nstable half of ℋ: ", stable.round(4)) -print("closed-loop eigenvalues:", np.sort(np.linalg.eigvals(JG).real).round(4)) +print("closed-loop eigenvalues:", np.sort(np.linalg.eigvals(U_agg).real).round(4)) ``` The saddle path is the stable invariant subspace of $\mathcal H$, exactly as in {doc}`lagrangian_lqdp`. @@ -435,7 +538,7 @@ The saddle path is the stable invariant subspace of $\mathcal H$, exactly as in With $n = k = 1$ everything is explicit. -Write $q, a, b, \gamma, \theta_X, \theta_{\mathcal A}$ for the scalars. +Write $q, a, b, \gamma, \theta_X, \theta_{\mathcal A}$ for the scalars, and assume $\gamma + \theta_{\mathcal A} > 0$. The Riccati equation becomes a quadratic, and the admissible root gives the aggregate eigenvalue @@ -445,7 +548,7 @@ The Riccati equation becomes a quadratic, and the admissible root gives the aggr + \frac{b^2(q + \theta_X)}{\gamma + \theta_{\mathcal A}}} . ``` -An equilibrium exists if and only if the term under the square root is positive, +When the initial mean $X(0)$ is not zero, an equilibrium exists if and only if the term under the square root is positive, ```{math} :label: mfg_scalar_existence @@ -459,30 +562,141 @@ and the equilibrium is stable, meaning $\lambda<0$, if and only if q + \theta_X + a(a+\rho)\frac{\gamma+\theta_{\mathcal A}}{b^2} > 0 . ``` -Formula {eq}`mfg_scalar_lambda` displays the two comparative statics at a glance. +When $q + \theta_X > 0$, as in both of our examples, formula {eq}`mfg_scalar_lambda` displays the two comparative statics at a glance. More complementarity in *states* (a smaller $\theta_X$) raises $\lambda$ and makes aggregate dynamics *more* persistent: when others stay away from the steady state, each agent has less reason to return to it. More complementarity in *actions* (a smaller $\theta_{\mathcal A}$) lowers $\lambda$ and makes dynamics *less* persistent: when others adjust, each agent wants to adjust too. -Let's confirm that our solver reproduces {eq}`mfg_scalar_lambda`. +If $q + \theta_X < 0$, the second comparative static reverses, which is why the general results below assume positive effective curvatures. + +### Four regions + +Conditions {eq}`mfg_scalar_existence` and {eq}`mfg_scalar_stability` define two thresholds for the state interaction, + +$$ +\theta^{*} \equiv -q - a(a+\rho)\frac{\gamma+\theta_{\mathcal A}}{b^2}, +\qquad +\theta^{**} \equiv -q - \left(\frac\rho2+a\right)^2\frac{\gamma+\theta_{\mathcal A}}{b^2} += \theta^{*} - \left(\frac\rho2\right)^2\frac{\gamma+\theta_{\mathcal A}}{b^2} , +$$ + +so that $\theta^{**} < \theta^{*}$. + +As $\theta_X$ falls, that is, as complementarity in states increases, the economy passes through four regions: + +1. if $\theta_X > \theta^{*}$, the equilibrium is stable and $X(t)$ converges to zero +1. if $\theta_X = \theta^{*}$, then $\lambda = 0$ and $X(t)$ stays where it starts +1. if $\theta^{**} < \theta_X < \theta^{*}$, an equilibrium exists but $0 < \lambda < \rho/2$, so $X(t)$ moves away from zero, slowly enough to keep discounted payoffs finite +1. if $\theta_X \leq \theta^{**}$, there is no equilibrium + +So stability is lost before existence. + +Formulas {eq}`mfg_scalar_lambda` to {eq}`mfg_scalar_stability` depend on the two interactions only through the ratio $(q + \theta_X)/(\gamma + \theta_{\mathcal A})$. + +In the $(\theta_X, \theta_{\mathcal A})$ plane, every curve along which that ratio is constant is a ray from the point $(-q, -\gamma)$. + +So contours of the aggregate eigenvalue, the stability boundary and the existence boundary are all rays from that point. + +Let's draw them, taking a large discount rate, $\rho = 0.5$, to widen the band between the two thresholds. + +We first write the closed form, and check it against the solver. ```{code-cell} ipython3 -def scalar_lambda(ρ, a, b, q, γ, θ_X, θ_A): +def scalar_λ(ρ, a, b, q, γ, θ_X, θ_A): "Closed-form aggregate eigenvalue in the scalar case." return ρ/2 - np.sqrt((ρ/2 + a)**2 + b**2*(q + θ_X)/(γ + θ_A)) -ρ_s, a, b, q, γ = 0.05, 0.3, 1.0, 1.0, 1.0 +def half_life(λ): + "Time for a state with eigenvalue λ < 0 to close half of a gap." + return np.log(2)/(-λ) + +ρ_s, a, b, q, γ = 0.5, 0.3, 1.0, 1.0, 1.0 -print(f"{'θ_X':>6}{'θ_A':>6}{'solver':>12}{'closed form':>14}") +errors = [] for θ_X, θ_A in ((0.0, 0.0), (-0.5, 0.0), (0.0, 0.5), (-0.5, 0.5)): P_s = mfg_riccati(np.array([[a]]), np.array([[b]]), np.array([[q + θ_X]]), np.array([[γ + θ_A]]), ρ_s) λ_num = closed_loop(np.array([[a]]), np.array([[b]]), np.array([[γ + θ_A]]), P_s)[0, 0] - print(f"{θ_X:>6}{θ_A:>6}{λ_num:>12.6f}{scalar_lambda(ρ_s,a,b,q,γ,θ_X,θ_A):>14.6f}") + errors.append(abs(λ_num - scalar_λ(ρ_s, a, b, q, γ, θ_X, θ_A))) +print(f"largest gap between solver and closed form: {max(errors):.1e}") +``` + +The next figure shades the four regions and draws contours of the half-life of the aggregate state. + +```{code-cell} ipython3 +--- +mystnb: + figure: + caption: Scalar model regions and half-lives + name: fig-mfg-regions +--- +θ_X_grid = np.linspace(-2.0, 1.0, 400) +θ_A_grid = np.linspace(-0.99, 2.0, 400) +TX, TA = np.meshgrid(θ_X_grid, θ_A_grid) + +inside = (ρ_s/2 + a)**2 + b**2*(q + TX)/(γ + TA) +λ_plane = np.where(inside >= 0, ρ_s/2 - np.sqrt(np.maximum(inside, 0)), np.nan) +region = np.where(inside < 0, 2, np.where(λ_plane >= 0, 1, 0)) +hl_plane = np.where(λ_plane < 0, np.log(2)/np.abs(λ_plane), np.nan) + +def θ_X_on_contour(h, θ_A): + "The value of θ_X at which the half-life equals h, given θ_A." + λ = -np.log(2)/h + ratio = ((ρ_s/2 - λ)**2 - (ρ_s/2 + a)**2)/b**2 + return ratio*(γ + θ_A) - q + +levels = [0.75, 1, 1.5, 2, 3, 5] +colors = ['white', '#fdd0a2', '#d9d9d9'] +fig, ax = plt.subplots(figsize=(10, 5.5)) +ax.contourf(TX, TA, region, levels=[-0.5, 0.5, 1.5, 2.5], colors=colors) +cs = ax.contour(TX, TA, hl_plane, levels=levels, colors='C0', linewidths=1.2) +ax.clabel(cs, fmt='%g', fontsize=9, + manual=[(θ_X_on_contour(h, 0.6), 0.6) for h in levels]) +θ_A_line = np.linspace(-γ, 2.0, 50) +ax.plot(-q - a*(a + ρ_s)*(γ + θ_A_line)/b**2, θ_A_line, 'k', lw=1.5) +ax.axhspan(-1.2, -γ, facecolor='none', hatch='//', edgecolor='grey', lw=0) +ax.plot(-q, -γ, 'ko', ms=5) +ax.annotate(r'$(-q, -\gamma)$', (-q, -γ), xytext=(-q + 0.06, -γ + 0.06)) +handles = [Patch(facecolor='white', edgecolor='k', lw=0.5, label='stable'), + Line2D([0], [0], color='k', lw=1.5, label=r'unit root, $\lambda = 0$'), + Patch(facecolor=colors[1], label='divergent but admissible'), + Patch(facecolor=colors[2], label='no equilibrium'), + Patch(facecolor='none', hatch='//', edgecolor='grey', + label=r'excluded, $\gamma + \theta_{\mathcal{A}} \leq 0$'), + Line2D([0], [0], color='C0', lw=1.2, label='half-life contours, years')] +ax.legend(handles=handles, loc='center left', bbox_to_anchor=(1.02, 0.5)) +ax.set_xlim(-2.0, 1.0) +ax.set_ylim(-1.2, 2.0) +ax.set_xlabel(r'$\theta_X$') +ax.set_ylabel(r'$\theta_{\mathcal{A}}$') +plt.tight_layout() +plt.show() +``` + +Because the half-life contours are rays from $(-q, -\gamma)$, a move toward complementarity in states can be exactly offset by a move toward substitutability in actions. + +The black line is the stability boundary $\theta_X = \theta^{*}$, and the thin orange band between it and the existence boundary is the third region, in which equilibrium paths diverge. + +Our solver respects the boundaries: with $\theta_{\mathcal A} = 0$, it returns an admissible solution with $\lambda > 0$ inside the band and raises an error beyond it. + +```{code-cell} ipython3 +θ_star = -q - a*(a + ρ_s)*γ/b**2 +θ_2star = θ_star - (ρ_s/2)**2*γ/b**2 + +for θ_X in ((θ_star + θ_2star)/2, θ_2star - 0.02): + try: + P_s = mfg_riccati(np.array([[a]]), np.array([[b]]), + np.array([[q + θ_X]]), np.array([[γ]]), ρ_s) + λ_num = closed_loop(np.array([[a]]), np.array([[b]]), np.array([[γ]]), P_s)[0, 0] + print(f"θ_X = {θ_X:.4f}: λ = {λ_num:.4f}") + except ValueError as e: + print(f"θ_X = {θ_X:.4f}: {e}") ``` +{ref}`mfg_ex_scalar` asks you to derive {eq}`mfg_scalar_lambda`, {eq}`mfg_scalar_existence` and {eq}`mfg_scalar_stability`. + ## An industry equilibrium with capital accumulation The scalar model is not a toy. @@ -505,10 +719,14 @@ So if all other firms hold capital $K$, operating profit is proportional to Capital evolves according to $$ -dk = (i - \delta k)dt + k \sigma dW , +dk = (i - \delta k)dt + k \sigma dW . $$ -the firm buys investment goods at price $\mathcal P(I)$, where $I$ is aggregate investment, and it pays a convex adjustment cost $\psi(i)$. +The firm buys investment goods at a price $\mathcal P(I)$ that depends on aggregate investment $I$, and it pays a convex adjustment cost $\psi(i)$, so it chooses investment to maximize + +$$ +\mathbb E \int_0^\infty e^{-\rho t}\left[\Pi(k, K) - \mathcal P(I)\, i - \psi(i)\right] dt . +$$ Let $\bar i = \delta \bar k$ be steady-state investment, normalize $\mathcal P(\bar i) = 1$ and $\psi'(\bar i) = 0$, and write percentage deviations from the deterministic steady state as @@ -579,51 +797,111 @@ print(f"{'q + θ_X':>10}{-(Π_kk + Π_kK)/Π(1, 1):>20.6f}" f"{(η_c - 1)/η_c*ν_c*(1 - ν_c):>15.6f}") ``` +The effective curvature of {ref}`mfg_effective` has a concrete meaning here. + +The next figure plots the marginal return to a firm's capital, measured relative to $\bar\Pi/\bar k$, in three experiments. + +In the first, one firm changes its capital while the rest of the industry stays at $\bar k$, and the marginal return falls with slope $-q$. + +In the second, every firm changes its capital together, and the marginal return falls with the smaller slope $-(q + \theta_X)$, because a larger industry raises the demand for each firm's good. + +The third line is the marginal revenue of the whole industry when every firm holds the same capital, $\frac{d}{dk}\Pi(k, k) = \nu k^{\nu - 1}$. + +The second line is exactly $(\eta-1)/\eta$ times the third, because a firm's marginal revenue is only that fraction of the value of its marginal product, so the markup lowers both the level and the slope of the private return. + +The third line will matter when we come to the planner. + +```{code-cell} ipython3 +--- +mystnb: + figure: + caption: Marginal returns to capital + name: fig-mfg-marginal-returns +--- +Π_k = lambda k, K: ν_c*(1 - 1/η_c)*k**(ν_c*(1 - 1/η_c) - 1)*K**(ν_c/η_c) +k_grid = np.linspace(0.8, 1.2, 200) # k̄ = 1, so that Π̄ = 1 + +fig, ax = plt.subplots(figsize=(8, 4.8)) +ax.plot(k_grid, Π_k(k_grid, 1.0), lw=2, + label=r'$\Pi_k(k, \bar k)$: one firm moves, slope $-q$') +ax.plot(k_grid, Π_k(k_grid, k_grid), lw=2, + label=r'$\Pi_k(k, k)$: all firms move, slope $-(q+\theta_X)$') +ax.plot(k_grid, ν_c*k_grid**(ν_c - 1), lw=2, ls='--', + label=r'$\frac{d}{dk}\Pi(k, k)$: industry, slope $-\nu(1-\nu)$') +ax.set_xlabel(r'capital relative to steady state, $k/\bar k$') +ax.set_ylabel(r'marginal return, relative to $\bar\Pi/\bar k$') +ax.set_ylim(0.45, 0.88) +ax.legend(loc='upper right') +plt.tight_layout() +plt.show() +``` + ### Calibration Take annual units, $\delta = 0.10$, $\rho = 0.05$, an elasticity of substitution $\eta = 4$, and returns to scale $\nu = 0.7$. -For the two curvatures that the technology does not pin down we set $\gamma = 0.05$, which makes a firm that ignores the industry close half of a capital gap in three years, and $\theta_{\mathcal A} = 0.09$. +For investment costs we use a quadratic adjustment cost and a constant-elasticity supply of investment goods, + +$$ +\psi(i) = \frac{\phi}{2}\,\bar i\left(\frac{i - \bar i}{\bar i}\right)^2 , +\qquad +\mathcal P(I) = \left(\frac{I}{\bar i}\right)^{1/\varepsilon_s} , +$$ + +which satisfy the normalizations $\psi'(\bar i) = 0$ and $\mathcal P(\bar i) = 1$. + +The steady-state Euler equation $\Pi_k(\bar k, \bar k) = \rho + \delta$ gives $\bar k^{1-\nu} = \nu(\eta-1)/[\eta(\rho+\delta)]$, and substituting into the formulas for $\gamma$ and $\theta_{\mathcal A}$ gives + +$$ +\gamma = \phi\,\delta\bar k^{1-\nu} , \qquad +\theta_{\mathcal A} = \frac{\delta \bar k^{1-\nu}}{\varepsilon_s} . +$$ + +We choose $\phi = 1/7$, so that a firm that ignores the industry closes half of a capital gap in three years, and $\varepsilon_s = 35/9 \approx 3.9$, a moderately elastic supply of investment goods chosen so that $\theta_{\mathcal A} = 0.09$. -{ref}`mfg_ex5` derives both numbers from an adjustment cost function and an investment supply curve. +With these values $\gamma = 0.05$. + +{ref}`mfg_ex_capital` asks you to derive these formulas. ```{code-cell} ipython3 ρ_c, δ_c = 0.05, 0.10 -γ_c, θ_A_c = 0.05, 0.09 +φ_c, ε_s_c = 1/7, 35/9 -q_c, θ_X_c = cap_q(η_c, ν_c), cap_θ_X(η_c, ν_c) +def cap_scale(η, ν, ρ=ρ_c, δ=δ_c): + "δ k̄^{1-ν}, which converts φ and 1/ε_s into γ and θ_A." + return δ*ν*(η - 1)/(η*(ρ + δ)) -def half_life(λ): - "Time for the aggregate state to close half of a gap." - return np.log(2)/(-λ) +γ_c = φ_c*cap_scale(η_c, ν_c) +θ_A_c = cap_scale(η_c, ν_c)/ε_s_c +q_c, θ_X_c = cap_q(η_c, ν_c), cap_θ_X(η_c, ν_c) -λ_c = scalar_lambda(ρ_c, δ_c, δ_c, q_c, γ_c, θ_X_c, θ_A_c) +λ_c = scalar_λ(ρ_c, δ_c, δ_c, q_c, γ_c, θ_X_c, θ_A_c) -P_c = mfg_riccati(np.array([[δ_c]]), np.array([[δ_c]]), - np.array([[q_c + θ_X_c]]), np.array([[γ_c + θ_A_c]]), ρ_c) -λ_c_solver = closed_loop(np.array([[δ_c]]), np.array([[δ_c]]), - np.array([[γ_c + θ_A_c]]), P_c)[0, 0] +θ_star_c = -q_c - δ_c*(δ_c + ρ_c)*(γ_c + θ_A_c)/δ_c**2 # stability threshold +print(f"γ = {γ_c:.4f}, θ_A = {θ_A_c:.4f}") print(f"q = {q_c:.4f}, θ_X = {θ_X_c:.4f}, q + θ_X = {q_c + θ_X_c:.4f}") -print(f"λ from the solver = {λ_c_solver:.6f}") -print(f"λ from the closed form = {λ_c:.6f}") +print(f"stability threshold θ* = {θ_star_c:.4f}") print(f"half-life of aggregate capital = {half_life(λ_c):.2f} years") ``` +Since the technology delivers $q + \theta_X > 0$ and an upward-sloping supply of investment goods delivers $\theta_{\mathcal A} > 0$, this example is always in the first of the four regions. + +Complementarity in states would have to be five times stronger than the technology implies before the industry's capital stock stopped converging. + ### How much do the interactions matter? Because {eq}`mfg_scalar_lambda` depends on $\theta_X$ and $\theta_{\mathcal A}$ separately, we can switch each interaction off and read the answer. ```{code-cell} ipython3 -cases = {'no interactions': (0.0, 0.0), - 'state complementarity only': (θ_X_c, 0.0), +cases = {'no interactions': (0.0, 0.0), + 'state complementarity only': (θ_X_c, 0.0), 'action substitutability only': (0.0, θ_A_c), - 'equilibrium': (θ_X_c, θ_A_c), - 'planner (both doubled)': (2*θ_X_c, 2*θ_A_c)} + 'equilibrium': (θ_X_c, θ_A_c)} print(f"{'':>30}{'λ':>10}{'half-life':>12}") for label, (tx, ta) in cases.items(): - λ_case = scalar_lambda(ρ_c, δ_c, δ_c, q_c, γ_c, tx, ta) + λ_case = scalar_λ(ρ_c, δ_c, δ_c, q_c, γ_c, tx, ta) print(f"{label:>30}{λ_case:>10.4f}{half_life(λ_case):>12.2f}") ``` @@ -635,24 +913,17 @@ Substitutability in actions means that a burst of industry investment is expensi Together they stretch the half-life of the industry's capital stock from three years to five. -The planner, who internalizes both externalities, is slower still. +We will see in {ref}`mfg_planner_capital` that a planner would adjust slightly faster than the industry does, once all the terms in the planner's objective are counted. -### Micro and macro adjustment speeds +### Individual and aggregate adjustment speeds -A distinctive prediction of the model is that individual capital reverts to the mean faster than aggregate capital does, because {eq}`mfg_beta2` contains no interaction matrix. +By {eq}`mfg_decomposition`, a firm's deviation from the industry mean adjusts at the speed of a firm that ignores the industry, so the first row of the table is also the individual half-life: three years, against five for the industry, a ratio of 1.66. -```{code-cell} ipython3 -β2_c = mfg_riccati(np.array([[δ_c]]), np.array([[δ_c]]), - np.array([[q_c]]), np.array([[γ_c]]), ρ_c) -λ_micro = closed_loop(np.array([[δ_c]]), np.array([[δ_c]]), - np.array([[γ_c]]), β2_c)[0, 0] +Because state complementarity and action substitutability both slow aggregate adjustment here, individual capital reverts to the mean faster than aggregate capital does. -print(f"individual half-life = {half_life(λ_micro):.2f} years") -print(f"aggregate half-life = {half_life(λ_c):.2f} years") -print(f"ratio = {half_life(λ_c)/half_life(λ_micro):.2f}") -``` +A researcher who estimated the speed of capital adjustment from firm-level data, and used it to predict the industry's response to an industry-wide shock, would predict a half-life of three years rather than five. -A researcher who estimated the speed of capital adjustment from firm-level data, and then used it to predict how quickly the industry responds to an industry-wide shock, would be too fast by two thirds. +That is the missing intercept of {ref}`mfg_identification` at work. The gap is a pure interaction effect: the same technology and the same adjustment costs generate both numbers. @@ -660,7 +931,29 @@ The gap is a pure interaction effect: the same technology and the same adjustmen How does the industry's adjustment speed depend on the two technological parameters? -Both enter only through $q + \theta_X = \frac{\eta-1}{\eta}\nu(1-\nu)$, which rises with $\eta$ and is maximized at $\nu = 1/2$. +Because $\gamma$ and $\theta_{\mathcal A}$ are measured relative to profits, they depend on $\eta$ and $\nu$ through $\bar k$, so we hold the structural parameters $\phi$ and $\varepsilon_s$ fixed rather than $\gamma$ and $\theta_{\mathcal A}$. + +That embeds an assumption of our calibration: the cost of a given percentage change in investment is proportional to the scale of investment $\bar i$. + +Then + +$$ +\frac{q + \theta_X}{\gamma + \theta_{\mathcal A}} = \frac{(1-\nu)(\rho+\delta)}{\delta(\phi + 1/\varepsilon_s)} , +$$ + +in which $\eta$ does not appear: the effective curvature and the investment costs, both measured relative to profits, are proportional to $(\eta-1)/\eta$. + +So by {eq}`mfg_scalar_lambda`, market power has no effect on the industry's adjustment speed. + +The invariance depends on the scaling of adjustment costs: with a cost $\frac c2(i - \bar i)^2$ and $c$ fixed, adjustment costs relative to profits would grow with the steady state, which rises with $\eta$, and the industry's half-life would rise from 4.5 years at $\eta = 1.5$ to 5.3 years at $\eta = 12$. + +It does affect a single firm's, because + +$$ +\frac{q}{\gamma} = \frac{\rho+\delta}{\phi\,\delta}\left[(1-\nu) + \frac{\nu}{\eta}\right] +$$ + +falls as $\eta$ rises: with more elastic demand, a firm's revenue is closer to linear in its own capital, so its marginal return falls more slowly as it expands. ```{code-cell} ipython3 --- @@ -673,65 +966,53 @@ mystnb: ν_grid = np.linspace(0.05, 0.995, 200) def macro_micro(η, ν): - "Aggregate and individual half-lives as functions of (η, ν)." + "Aggregate and individual half-lives, holding φ and ε_s fixed." + γ, θ_A = φ_c*cap_scale(η, ν), cap_scale(η, ν)/ε_s_c q, θ_X = cap_q(η, ν), cap_θ_X(η, ν) - λ_agg = scalar_lambda(ρ_c, δ_c, δ_c, q, γ_c, θ_X, θ_A_c) - λ_ind = scalar_lambda(ρ_c, δ_c, δ_c, q, γ_c, 0.0, 0.0) + λ_agg = scalar_λ(ρ_c, δ_c, δ_c, q, γ, θ_X, θ_A) + λ_ind = scalar_λ(ρ_c, δ_c, δ_c, q, γ, 0.0, 0.0) return half_life(λ_agg), half_life(λ_ind) fig, axes = plt.subplots(1, 2, figsize=(11, 4.2)) hl_agg, hl_ind = np.array([macro_micro(η, ν_c) for η in η_grid]).T -axes[0].plot(η_grid, hl_agg, lw=2, label='industry') -axes[0].plot(η_grid, hl_ind, lw=2, ls='--', label='single firm') +axes[0].plot(η_grid, hl_agg, lw=2, label='aggregate (industry)') +axes[0].plot(η_grid, hl_ind, lw=2, ls='--', label='individual (firm)') axes[0].set_xlabel('elasticity of substitution $\\eta$') +axes[0].set_title('varying $\\eta$, with $\\nu = 0.7$') hl_agg, hl_ind = np.array([macro_micro(η_c, ν) for ν in ν_grid]).T -axes[1].plot(ν_grid, hl_agg, lw=2, label='industry') -axes[1].plot(ν_grid, hl_ind, lw=2, ls='--', label='single firm') +axes[1].plot(ν_grid, hl_agg, lw=2, label='aggregate (industry)') +axes[1].plot(ν_grid, hl_ind, lw=2, ls='--', label='individual (firm)') +axes[1].set_title('varying $\\nu$, with $\\eta = 4$') axes[1].axhline(np.log(2)/δ_c, color='k', lw=1, alpha=0.6) axes[1].set_xlabel('returns to scale $\\nu$') -axes[1].annotate('$\\ln 2/\\delta$', (0.12, np.log(2)/δ_c - 0.45)) +axes[1].annotate('$\\ln 2/\\delta$', (0.08, np.log(2)/δ_c - 0.35)) for ax in axes: ax.set_ylabel('half-life in years') - ax.legend() + ax.legend(loc='lower right') plt.tight_layout() plt.show() ``` -More market power, meaning a smaller $\eta$, makes the industry slower: it weakens the own-capital curvature relative to the interaction, and the left panel shows the industry half-life rising as $\eta$ falls. - -The right panel contains a sharper result. - -As $\nu \to 1$ the effective curvature $q + \theta_X$ vanishes, and {eq}`mfg_scalar_lambda` collapses to $\lambda \to -\delta$: the industry's capital stock then returns to its steady state only through depreciation, with aggregate investment not responding at all. - -{ref}`mfg_ex6` asks you to verify this limit and to explain it. - -### Where are we in the four regions? - -{ref}`mfg_ex1` divides the scalar model into four regions using the thresholds $-\theta^{*}$ and $-\theta^{**}$. +The left panel shows both results: the industry's half-life does not move with $\eta$, while a single firm's rises from about 2.3 to 3.4 years as competition intensifies. -Since the technology delivers $q + \theta_X > 0$ and an upward sloping investment supply delivers $\theta_{\mathcal A} > 0$, this example is always in the first region. +So the gap between individual and aggregate adjustment is widest when market power is greatest. -The thresholds show how much room there is to spare. +The right panel shows that both slow down as returns to scale rise. -```{code-cell} ipython3 -mθ_star = q_c + δ_c*(δ_c + ρ_c)*(γ_c + θ_A_c)/δ_c**2 -mθ_2star = q_c + (ρ_c/2 + δ_c)**2*(γ_c + θ_A_c)/δ_c**2 +As $\nu$ approaches 1 the effective curvature $q + \theta_X$ vanishes, and $\lambda$ approaches $-\delta$: the industry's capital stock then returns to its steady state almost entirely through depreciation, and aggregate investment barely responds. -print(f"complementarity in the calibration, -θ_X = {-θ_X_c:.4f}") -print(f"instability threshold, -θ* = {mθ_star:.4f}") -print(f"nonexistence threshold, -θ** = {mθ_2star:.4f}") -``` +{numref}`fig-mfg-marginal-returns` shows why: as $\nu$ approaches 1, the marginal return $\Pi_k(k, k)$ becomes flat along the diagonal, so an industry-wide capital gap creates almost no incentive to invest differently. -Complementarity would have to be five times stronger than the technology implies before the industry's capital stock stopped converging. +{ref}`mfg_ex_market_power` asks you to derive these results. ### Transition paths Finally, let's trace out the industry's response to a capital stock that starts ten percent above its steady state. -Aggregate investment follows from {eq}`mfg_aggregate_action`, which in the scalar case gives $\mathcal A(t) = \frac{\lambda + \delta}{\delta}X(t)$. +Aggregate investment follows from the aggregate law of motion: $\dot X = \delta(\mathcal A - X) = \lambda X$ gives $\mathcal A(t) = \frac{\lambda + \delta}{\delta}X(t)$. ```{code-cell} ipython3 --- @@ -744,9 +1025,9 @@ t_c = np.linspace(0, 25, 300) X0_c = 0.10 fig, axes = plt.subplots(1, 2, figsize=(11, 4.2)) -for label in ('no interactions', 'equilibrium', 'planner (both doubled)'): +for label in ('no interactions', 'equilibrium'): tx, ta = cases[label] - λ_case = scalar_lambda(ρ_c, δ_c, δ_c, q_c, γ_c, tx, ta) + λ_case = scalar_λ(ρ_c, δ_c, δ_c, q_c, γ_c, tx, ta) path = X0_c*np.exp(λ_case*t_c) axes[0].plot(t_c, 100*path, lw=2, label=label) axes[1].plot(t_c, 100*(λ_case + δ_c)/δ_c*path, lw=2, label=label) @@ -765,159 +1046,355 @@ The right panel is the industry's investment response, and it is where the two i A firm that ignored the industry would cut investment on impact by thirteen percent of its steady-state level; in equilibrium the cut is under four percent, and it is undone much more slowly. -The costate $\beta_1(t) + \bar\beta_2 x$ that supports these paths is the marginal value of installed capital, that is, Tobin's $q$. +The costate $\beta_1(t) + \bar\beta_2 x$ that supports these paths is the marginal value of installed capital, measured as a deviation from its steady-state value; divided by the price of investment goods, it is Tobin's marginal q, not to be confused with the curvature $q$. -{doc}`optimal_growth_uncertainty` studies that object in a setting where the constraint $i \geq 0$ binds and the value function is not differentiable everywhere, which is exactly the nonlinearity that the quadratic approximation here sets aside. +{doc}`optimal_growth_uncertainty` studies the marginal value of capital when the irreversibility constraint $i \geq 0$ sometimes binds, and shows that the value function remains differentiable even then; the quadratic approximation here sets that nonlinearity aside. +(mfg_identification)= +## Aggregate shocks and identification -## Persistence +Now add a shock that hits everyone at once. -How do the interactions change aggregate adjustment when $n > 1$? +Let $\mathcal J$ be a compensated jump process, the difference between two independent Poisson processes with equal intensities, so that its increments have mean zero, and suppose -Stronger complementarity of either kind makes the stabilizing solution $P$ less negative definite. +$$ +dx = (B\alpha - Ax)dt + \Sigma^{1/2}dW + \Upsilon \, d\mathcal J . +$$ -That ordering is not enough to sign the change in every eigenvalue of the closed-loop matrix, but it does sign the change in their *sum*. +A common Brownian motion would work the same way. -Stronger complementarity in states raises the trace of the closed-loop matrix, while stronger complementarity in actions lowers it. +With common noise the value function must keep track of the aggregate state as well as the individual state, so it becomes $v(x, X)$. -The trace measures the rate at which a set of initial aggregate states contracts in volume as each point follows its equilibrium path, so in a stable equilibrium the first force slows the collapse toward the steady state and the second speeds it up. +Nevertheless, matching coefficients in the recursive HJB equation delivers a result worth emphasizing. -```{code-cell} ipython3 -print(f"{'scale on Θ_X':>14}{'trace':>10} eigenvalues") -for scale in (0.0, 0.5, 1.0, 1.5): - P_s = mfg_riccati(A, B, Q + scale*Θ_X, Γ + Θ_A, ρ) - JG_s = closed_loop(A, B, Γ + Θ_A, P_s) - print(f"{scale:>14}{np.trace(JG_s):>10.4f} {np.linalg.eigvals(JG_s).real.round(4)}") +```{prf:proposition} +:label: mfg_prop_common_noise + +With common noise, the matrix $P$ governing aggregate dynamics solves the *same* equilibrium Riccati equation as before. ``` -Raising the scale makes $\Theta_X$ more negative, that is, complementarity stronger, and the trace rises as claimed. +This is a certainty-equivalence result of the kind familiar from linear quadratic control. -{ref}`mfg_ex2` asks you to investigate whether *each* eigenvalue must move in the same direction. +The decomposition {eq}`mfg_decomposition` continues to hold, with $\Upsilon \, d\mathcal J$ added to the law of motion of $X$, and the common shock cancels from the law of motion of an agent's deviation $z$. -## The planner +### The missing intercept -A utilitarian planner internalizes the effect of each agent's state and action on the corresponding cross-sectional averages. +Researchers often estimate how agents respond to shocks from panel data with time fixed effects, which control for aggregate conditions. -When $\Theta_X$ and $\Theta_{\mathcal A}$ are symmetric, differentiating the planner's objective doubles each interaction term. +Such regressions identify how an agent responds to her position relative to others, but not the common response of all agents to an aggregate shift, which the time effects absorb; this is the "missing intercept" {cite}`Wolf2023`. -```{prf:proposition} -:label: mfg_prop_planner +The decomposition {eq}`mfg_decomposition` makes the point exact. -The planner's allocation coincides with the decentralized equilibrium of an economy whose interaction matrices are $2\Theta_X$ and $2\Theta_{\mathcal A}$, so the planner's Riccati equation is +Writing an agent's action in terms of her own state and the mean gives $$ -P^{*} B(\Gamma + 2\Theta_{\mathcal A})^{-1}B^\top P^{*} -= Q + 2\Theta_X + \rho P^{*} + P^{*}A + A^\top P^{*} . +\alpha_{it} = U_{\alpha,z}\, x_{it} + \left(U_{\alpha,X} - U_{\alpha,z}\right) X_t , +\qquad +U_{\alpha,X} \equiv (\Gamma+\Theta_{\mathcal A})^{-1}B^\top P , +\qquad +U_{\alpha,z} \equiv \Gamma^{-1}B^\top\bar\beta_2 . $$ + +A regression of actions on states with time effects estimates the slope $U_{\alpha,z}$, which contains no interaction matrix, while the time effects absorb $(U_{\alpha,X} - U_{\alpha,z})X_t$, which contains all of them. + +```{prf:proposition} +:label: mfg_prop_identification + +Data on $x(t) - X(t)$ and $\alpha(t)-\mathcal A(t)$ contain no information about $\Theta_X$ or $\Theta_{\mathcal A}$. ``` -Internalizing state complementarity makes the allocation *more* persistent, while internalizing action complementarity makes it *less* persistent. +```{note} +Removing time effects from micro data removes exactly the variation that identifies strategic interaction. +``` -When both are present the comparison is ambiguous, as {ref}`mfg_ex3` explores. +To see this in the capital example, consider two industries that are identical except for the strength of complementarity in states: one has $\theta_X = 0$, and the other has twice the calibrated amount. + +Let's compute the two policy coefficients for each. ```{code-cell} ipython3 -P_planner = mfg_riccati(A, B, Q + 2*Θ_X, Γ + 2*Θ_A, ρ) -JG_planner = closed_loop(A, B, Γ + 2*Θ_A, P_planner) +def capital_policy(θ_X, θ_A=θ_A_c): + "Coefficients of investment on X and on z in the capital example." + one = np.array([[1.0]]) + P_s = mfg_riccati(δ_c*one, δ_c*one, (q_c + θ_X)*one, (γ_c + θ_A)*one, ρ_c)[0, 0] + β2_s = mfg_riccati(δ_c*one, δ_c*one, q_c*one, γ_c*one, ρ_c)[0, 0] + return δ_c*P_s/(γ_c + θ_A), δ_c*β2_s/γ_c -print("equilibrium eigenvalues:", np.linalg.eigvals(JG).real.round(4)) -print("planner eigenvalues: ", np.linalg.eigvals(JG_planner).real.round(4)) +industries = (('θ_X = 0', 0.0), ('θ_X = 2 × calibrated', 2*θ_X_c)) +print(f"{'':>22}{'slope on z':>12}{'slope on X':>12}") +for label, θ_X_use in industries: + U_αX, U_αz = capital_policy(θ_X_use) + print(f"{label:>22}{U_αz:>12.4f}{U_αX:>12.4f}") ``` -Let's see what this means for the path of the aggregate state. +The slope on a firm's deviation from the mean is the same in the two industries, and only the slope on the industry mean differs. + +Now let's simulate panels of 400 firms from the two industries, using the same shocks for both. + +Both start at their steady state with some dispersion across firms, and both are hit at $t = 2$ by a common shock that lowers every firm's capital by ten percent. ```{code-cell} ipython3 --- mystnb: figure: - caption: Path of the aggregate state - name: fig-mfg-planner-path + caption: Simulated panels from two industries + name: fig-mfg-panels --- -X0 = np.array([1.0, 0.5]) -times = np.linspace(0, 12, 200) +rng = np.random.default_rng(0) +N, T, dt, σ = 400, 20.0, 0.02, 0.05 +steps = int(T/dt) +dW = rng.normal(size=(steps, N))*np.sqrt(dt) # the same draws for both +x0 = 0.05*rng.normal(size=N) +t_sim = np.arange(steps)*dt -paths = {'no interactions': G, 'equilibrium': JG, 'planner': JG_planner} - -fig, ax = plt.subplots(figsize=(8, 4.5)) -for label, M in paths.items(): - traj = np.array([expm(M*t) @ X0 for t in times]) - ax.plot(times, traj[:, 0], lw=2, label=label) -ax.set_xlabel('$t$') -ax.set_ylabel('first component of $X(t)$') -ax.legend() +fig, axes = plt.subplots(1, 2, figsize=(11, 4.2)) +deviations = [] +for k, (label, θ_X_sim) in enumerate(industries): + U_αX, U_αz = capital_policy(θ_X_sim) + x = x0.copy() + X_path, z_path = np.empty(steps), np.empty(steps) + for s in range(steps): + if s == int(2.0/dt): + x = x - 0.10 # common shock + X = x.mean() + α = U_αX*X + U_αz*(x - X) + X_path[s], z_path[s] = X, x[0] - X + x = x + δ_c*(α - x)*dt + σ*dW[s] + deviations.append(z_path) + tex_label = label.replace('θ_X', r'$\theta_X$') + axes[0].plot(t_sim, 100*z_path, lw=2.5 - 1.5*k, ls='-' if k == 0 else '--', + label=tex_label) + axes[1].plot(t_sim, 100*X_path, lw=2, label=tex_label) + +axes[0].set_title("one firm's deviation from the industry mean") +axes[1].set_title('industry mean') +for ax in axes: + ax.set_xlabel('years') + ax.set_ylabel('percent') + ax.legend() plt.tight_layout() plt.show() + +print("demeaned paths identical:", np.allclose(*deviations)) ``` -## Multiproduct price setting with Kimball demand +The left panel plots one firm's deviation from its industry mean in each economy, and the two lines coincide exactly, so a regression on demeaned data would return identical estimates for the two industries. -Our second example is multidimensional, and it delivers a surprise. +The industry means in the right panel differ: after the common shock, the industry with stronger complementarity returns to its steady state more slowly. -{cite:t}`AlvarezArgente2026` study a continuum of "stores", each selling $n$ products with constant marginal costs $z_j$. +The small wiggles in the industry means come from averaging the idiosyncratic shocks of only 400 firms. -Products within a store are aggregated by a CES price index with elasticity $\bar\eta_d$, and stores are aggregated by a symmetric Kimball aggregator {cite}`Kimball1995`. +The demeaned paths coincide by construction, because the slope on $z$ does not involve $\theta_X$, so the simulation illustrates {prf:ref}`mfg_prop_identification` rather than testing it: it shows what an econometrician would see. -Let $\eta_D(y)$ be the elasticity of a store's demand with respect to its relative price, and let $\bar\eta_D > 1$ and $\bar\eta_D'$ denote its level and its derivative at the symmetric point. +### What aggregate data reveal -The derivative $\bar\eta_D'$ is the **superelasticity** of demand, and it is the reason Kimball demand is so widely used in models of price setting: a positive superelasticity means that a store which raises its prices above the average faces a more elastic demand, which discourages it from moving away from the crowd. +Aggregate data, by contrast, do carry information about strategic interaction, and can be combined with micro data to recover it. -That is strategic complementarity in prices, and it is the mechanism that {cite:t}`KlenowWillis2016` and much of the subsequent literature rely on for real rigidity. +Collect the four reduced-form matrices that link the aggregate state, the deviations from it and the actions: -Writing $x$ and $X$ for log deviations of a store's prices and of the average store's prices from the flexible-price level $\bar p_i = \bar z_i \bar\eta_D/(\bar\eta_D-1)$, and $\bar s$ for the vector of steady-state expenditure shares, the curvature matrices are +$$ +U_{\dot X,X} = DP - A, \qquad +U_{\dot z,z} = B\Gamma^{-1}B^\top\bar\beta_2 - A, \qquad +U_{\alpha,X} = (\Gamma+\Theta_{\mathcal A})^{-1}B^\top P, \qquad +U_{\alpha,z} = \Gamma^{-1}B^\top\bar\beta_2 . +$$ -```{math} -:label: mfg_kimball_Q -\begin{aligned} -Q &= (\bar\eta_D - 1)\left[\left(\bar\eta_D - \bar\eta_d -+ \frac{\bar\eta_D'}{\bar\eta_D-1}\right)\bar s\bar s^\top -+ \bar\eta_d \operatorname{diag}(\bar s)\right] , \\ -\Theta_X &= -\bar\eta_D' \, \bar s \bar s^\top , \\ -Q + \Theta_X &= (\bar\eta_D-1)\left[(\bar\eta_D - \bar\eta_d)\bar s \bar s^\top -+ \bar\eta_d \operatorname{diag}(\bar s)\right] . -\end{aligned} -``` +```{prf:proposition} +:label: mfg_prop_identification_more -Stare at the third line. +Suppose $\Theta_{\mathcal A}$, $\Gamma$ and $\rho$ are known, that $D$ and $\Sigma$ are nonsingular, and that $\mathbb E[X(t)X(t)^\top]$ is nonsingular for some $t > 0$. -The superelasticity appears in $Q$ and in $\Theta_X$, but it *cancels* from $Q + \Theta_X$. +1. If $Q$, $A$ and $B$ are also known, a time series for $X$ identifies $\Theta_X$. +1. If $B$ is also known, time series for $X$, $z$ and $\mathcal A$ identify $Q$, $\Theta_X$ and $A$. +1. If $n = k$ and $U_{\alpha,z} - U_{\alpha,X}$ is nonsingular, time series for $X$, $z$, $\alpha$ and $\mathcal A$ identify $Q$, $\Theta_X$, $A$ and $B$. +``` -By {prf:ref}`mfg_prop_riccati`, the sum $Q+\Theta_X$ is the only channel through which either matrix reaches aggregate dynamics. +In every case $\Theta_{\mathcal A}$ must be known. -Actions are the rates of change of prices, and stores pay quadratic Rotemberg costs of changing them. +{ref}`mfg_ex_identification` walks through the recovery, and shows why $\Theta_X$ and $\Theta_{\mathcal A}$ cannot be told apart without a normalization. -The matrix $\Gamma$ lets that cost depend on which bundle of prices is changed, with negative off-diagonal elements representing economies of scope in repricing of the kind emphasized by {cite:t}`Midrigan2011` and {cite:t}`AlvarezLippi2014`. +(mfg_persistence)= +## Persistence -There is no interaction through aggregate actions, so $\Theta_{\mathcal A} = 0$, and a nonzero rate of cost inflation makes $A$ diagonal. +How do the interactions change aggregate adjustment when $n > 1$? -```{code-cell} ipython3 -def kimball(η_d, η_D, η_D_prime, s): - "Curvature and state-interaction matrices under Kimball demand." - s = np.asarray(s, dtype=float) - S = np.outer(s, s) - Q = (η_D - 1)*((η_D - η_d + η_D_prime/(η_D - 1))*S + η_d*np.diag(s)) - Θ_X = -η_D_prime*S - return Q, Θ_X +Throughout this section we follow {cite:t}`AlvarezArgente2026` and assume that $Q + \Theta_X$ and $\Gamma + \Theta_{\mathcal A}$ are positive definite in every economy we compare. -s_bar = np.array([0.5, 0.3, 0.2]) # three products, unequal shares -η_d, η_D = 6.0, 4.0 # within-store and across-store elasticities -ρ_K, π_K = 0.04, 0.02 # discount rate and cost inflation +As the scalar case showed, the comparative statics for actions can reverse sign without that assumption. -n_K = len(s_bar) -A_K = π_K*np.eye(n_K) -B_K = np.eye(n_K) -Γ_K = 20.0*(np.eye(n_K) - 0.25*(np.ones((n_K, n_K)) - np.eye(n_K))) +```{prf:proposition} +:label: mfg_prop_trace -print("Rotemberg cost matrix Γ =\n", Γ_K) -print("\neigenvalues of Γ:", np.linalg.eigvalsh(Γ_K).round(4)) +Compare two economies that differ in one interaction matrix only. + +1. More complementarity in states makes the stabilizing solution $P$ less negative definite and raises the trace of the closed-loop matrix. +1. More complementarity in actions also makes $P$ less negative definite, but lowers the trace of the closed-loop matrix. ``` -### The superelasticity and aggregate dynamics +The two kinds of interaction move $P$ in the same direction and the trace in opposite directions because they reach the closed-loop matrix $DP - A$ in different ways. -Now sweep the superelasticity across a wide range, including a negative value, and watch what changes and what does not. +States act only through $P$. + +Actions also change $D = B(\Gamma + \Theta_{\mathcal A})^{-1}B^\top$: more complementarity in actions raises $D$, so agents respond more strongly to a given marginal value, and the net effect lowers the trace. + +The trace is the sum of the eigenvalues, and also the rate at which a set of initial aggregate states contracts in volume, so in a stable equilibrium the first force slows aggregate adjustment on average and the second speeds it up. + +Let's check the first part in our two-dimensional example. ```{code-cell} ipython3 -print(f"{'η_D′':>6} {'eigenvalues of Q':>28} {'eigenvalues of Q + Θ_X':>28}") -for η_Dp in (-3.0, 0.0, 3.0, 10.0): - Q_K, Θ_K = kimball(η_d, η_D, η_Dp, s_bar) +print(f"{'scale on Θ_X':>14}{'trace':>10} eigenvalues") +for scale in (0.0, 0.5, 1.0, 1.5): + P_s = mfg_riccati(A, B, Q + scale*Θ_X, Γ + Θ_A, ρ) + U_s = closed_loop(A, B, Γ + Θ_A, P_s) + print(f"{scale:>14}{np.trace(U_s):>10.4f} {np.linalg.eigvals(U_s).real.round(4)}") +``` + +Raising the scale makes $\Theta_X$ more negative, that is, complementarity stronger, and the trace rises as claimed. + +A rise in the trace says that the sum of the eigenvalues moves, not that each of them does. + +{cite:t}`AlvarezArgente2026` give a two-dimensional example in which a strict Loewner ordering of the Riccati solutions coexists with eigenvalues that move in opposite directions. + +It sets $\rho = 0$, holds $D$ fixed, and compares two effective state curvatures $Q_1$ and $Q_2$ with $Q_2 - Q_1$ positive definite, so that the first economy has more complementarity in states. + +```{code-cell} ipython3 +A_ce = np.array([[1.0, 2.0], + [2.0, 5.0]]) +D_ce = np.array([[1.0, -3.0], + [-3.0, 10.0]]) +Q_1 = np.array([[2.0, -1.0], + [-1.0, 1.0]]) +Q_2 = np.array([[3.0, 5.0], + [5.0, 51.0]]) + +# with B = I, D = (Γ + Θ_A)^{-1} +Γ_ce = np.linalg.inv(D_ce) +P_ce = {} +for label, Q_i in (('Q_1', Q_1), ('Q_2', Q_2)): + P_ce[label] = mfg_riccati(A_ce, np.eye(2), Q_i, Γ_ce, 0.0) + U_i = closed_loop(A_ce, np.eye(2), Γ_ce, P_ce[label]) + print(f"{label}: trace {np.trace(U_i):.2f}, " + f"closed-loop eigenvalues {np.sort(np.linalg.eigvals(U_i).real).round(2)}") + +print("\neigenvalues of P_1 - P_2:", np.linalg.eigvalsh(P_ce['Q_1'] - P_ce['Q_2']).round(3)) +print("A and D commute:", np.allclose(A_ce @ D_ce, D_ce @ A_ce)) +``` + +The Riccati solutions are strictly ordered, $P_1 \succ P_2$, and the first economy has the larger trace, as the theory requires, but only one of its eigenvalues is larger: one mode becomes more persistent and the other less. + +What goes wrong is that $A$ and $D$ do not commute. + +Since both are symmetric here, that is the same as saying that $D^{-1}A$ is not symmetric, so the closed-loop matrix $DP - A$ need not be similar to a symmetric matrix. + +Two restrictions repair this. + +```{prf:proposition} +:label: mfg_prop_eigen_symmetric + +Hold $D$ fixed and suppose that $D^{-1}A$ is symmetric. + +Then more complementarity in states raises every eigenvalue of the closed-loop matrix. +``` + +The reason is that $DP - A = D(P - D^{-1}A)$ is similar to the symmetric matrix $D^{1/2}(P - D^{-1}A)D^{1/2}$, whose eigenvalues rise one by one when $P$ rises in the Loewner order. + +```{prf:proposition} +:label: mfg_prop_eigen_aI + +Suppose $A = aI$. + +Then the eigenvalues of the closed-loop matrix are real and given by + +$$ +\lambda_j = \frac\rho2 - \sqrt{\left(\frac\rho2 + a\right)^2 + \mu_j}, +$$ + +where $\mu_j$ are the eigenvalues of $D^{1/2}(Q+\Theta_X)D^{1/2}$. + +So more complementarity in states raises every $\lambda_j$, more complementarity in actions lowers every $\lambda_j$, and mode $j$ is stable if and only if $\mu_j + a(a+\rho) > 0$. +``` + +The second result is still genuinely multidimensional. + +Aggregate dynamics do decouple into scalar modes, but the coordinates that achieve this mix the eigenvectors of $Q + \Theta_X$ and $D$, and they differ from the coordinates that decouple individual dynamics, so the model as a whole does not separate into independent one-dimensional decisions. + +{ref}`mfg_ex_eigenvalues` asks you to measure how often eigenvalue-by-eigenvalue monotonicity fails in random economies, and to confirm {prf:ref}`mfg_prop_eigen_symmetric`. + +## Multiproduct price setting with Kimball demand + +Our second example is multidimensional, and it delivers a surprise. + +{cite:t}`AlvarezArgente2026` study a continuum of "stores", each selling $n$ products with constant marginal costs $c_j$. + +Products within a store are aggregated by a CES price index with elasticity $\bar\eta_d$, and stores are aggregated by a symmetric Kimball aggregator {cite}`Kimball1995`. + +Let $\eta_D(y)$ be the elasticity of a store's demand with respect to its relative price $y$, and let $\bar\eta_D > 1$ and $\bar\eta_D'$ denote its level and its derivative at the symmetric point $y = 1$. + +The derivative $\bar\eta_D'$ is the **superelasticity** of demand, and it is the reason Kimball demand is so widely used in models of price setting: a positive superelasticity means that a store which raises its prices above the average faces a more elastic demand, which discourages it from moving away from the crowd. + +That is strategic complementarity in prices, and it is the mechanism that {cite:t}`KlenowWillis2016` and much of the subsequent literature rely on for real rigidity. + +Writing $x$ and $X$ for log deviations (equivalently, to first order, percentage deviations) of a store's prices and of the average store's prices from the flexible-price level $\bar p_i = \bar c_i \bar\eta_D/(\bar\eta_D-1)$, and $\bar s$ for the vector of steady-state expenditure shares, the curvature matrices are + +```{math} +:label: mfg_kimball_Q +\begin{aligned} +Q &= (\bar\eta_D - 1)\left[\left(\bar\eta_D - \bar\eta_d ++ \frac{\bar\eta_D'}{\bar\eta_D-1}\right)\bar s\bar s^\top ++ \bar\eta_d \operatorname{diag}(\bar s)\right] , \\ +\Theta_X &= -\bar\eta_D' \, \bar s \bar s^\top , \\ +Q + \Theta_X &= (\bar\eta_D-1)\left[(\bar\eta_D - \bar\eta_d)\bar s \bar s^\top ++ \bar\eta_d \operatorname{diag}(\bar s)\right] . +\end{aligned} +``` + +Stare at the third line. + +The superelasticity appears in $Q$ and in $\Theta_X$, but it *cancels* from $Q + \Theta_X$. + +By {prf:ref}`mfg_prop_riccati`, the sum $Q+\Theta_X$ is the only channel through which either matrix reaches aggregate dynamics. + +Actions are price changes, and stores pay quadratic Rotemberg costs of changing prices. + +The matrix $\Gamma$ lets that cost depend on which bundle of prices is changed, with negative off-diagonal elements representing economies of scope in repricing of the kind emphasized by {cite:t}`Midrigan2011` and {cite:t}`AlvarezLippi2014`. + +There is no interaction through aggregate actions, so $\Theta_{\mathcal A} = 0$. + +With constant marginal costs $A = 0$, and to obtain a nontrivial drift we follow {cite:t}`AlvarezArgente2026` in letting marginal costs, and hence flexible prices, grow at a common rate $\pi$. + +Let $\bar p_i(t)$ be the trend price, which grows at rate $\pi$, and measure a price by its percentage deviation from trend, $x_i = p_i/\bar p_i - 1$, and the action by the change in the price minus the change in the trend, in units of the trend price, $\alpha_i = (\dot p_i - \dot{\bar p}_i)/\bar p_i$. + +Then $\dot x_i = \alpha_i - \pi x_i$, so $A = \pi I$: a store that leaves its price change equal to the trend change sees its gap from trend eroded at rate $\pi$. + +```{code-cell} ipython3 +def kimball(η_d, η_D, η_D_prime, s): + "Curvature and state-interaction matrices under Kimball demand." + s = np.asarray(s, dtype=float) + ss = np.outer(s, s) + Q = (η_D - 1)*((η_D - η_d + η_D_prime/(η_D - 1))*ss + η_d*np.diag(s)) + Θ_X = -η_D_prime*ss + return Q, Θ_X + +s_bar = np.array([0.5, 0.3, 0.2]) # three products, unequal shares +η_d, η_D = 6.0, 4.0 # within-store and across-store elasticities +ρ_K, π_K = 0.04, 0.02 # discount rate and cost inflation + +n_K = len(s_bar) +A_K = π_K*np.eye(n_K) +B_K = np.eye(n_K) +Γ_K = 20.0*(np.eye(n_K) - 0.25*(np.ones((n_K, n_K)) - np.eye(n_K))) + +print("Rotemberg cost matrix Γ =\n", Γ_K) +print("\neigenvalues of Γ:", np.linalg.eigvalsh(Γ_K).round(4)) +``` + +### The superelasticity and aggregate dynamics + +Now sweep the superelasticity across a wide range, including a negative value, and watch what changes and what does not. + +```{code-cell} ipython3 +print(f"{'η_D′':>6} {'eigenvalues of Q':>28} {'eigenvalues of Q + Θ_X':>28}") +for η_Dp in (-3.0, 0.0, 3.0, 10.0): + Q_K, Θ_K = kimball(η_d, η_D, η_Dp, s_bar) print(f"{η_Dp:>6} {str(np.linalg.eigvalsh(Q_K).round(3)):>28}" f" {str(np.linalg.eigvalsh(Q_K + Θ_K).round(3)):>28}") ``` @@ -939,12 +1416,25 @@ for η_Dp in (-3.0, 0.0, 3.0, 10.0): The left block of numbers is identical in every row, and the right block is not. -The superelasticity changes how an individual store behaves without changing how the industry behaves. +The superelasticity changes how an individual store behaves without changing how the industry behaves, just as market power did in the capital accumulation example. -Almost all of the movement is in a single mode, because $\Theta_X$ is rank one and points in the direction $\bar s$, which is the store's own share-weighted price index. +Almost all of the movement is in a single mode, because $\Theta_X$ is rank one and acts only through the store's price index $\bar s^\top x$. The remaining modes, which involve relative prices within the store, barely move. +Because $A_K = \pi I$, the aggregate eigenvalues are also given in closed form by {prf:ref}`mfg_prop_eigen_aI`, with $\mu_j$ the eigenvalues of $(\Gamma + \Theta_{\mathcal A})^{-1}(Q + \Theta_X)$, which equal those of $D^{1/2}(Q + \Theta_X)D^{1/2}$ when $B = I$. + +Let's confirm that the formula matches the solver. + +```{code-cell} ipython3 +Q_K, Θ_K = kimball(η_d, η_D, 3.0, s_bar) +P_K = mfg_riccati(A_K, B_K, Q_K + Θ_K, Γ_K, ρ_K) +ev_solver = np.sort(np.linalg.eigvals(closed_loop(A_K, B_K, Γ_K, P_K)).real) +μ = np.linalg.eigvals(np.linalg.solve(Γ_K, Q_K + Θ_K)).real +ev_formula = np.sort(ρ_K/2 - np.sqrt((ρ_K/2 + π_K)**2 + μ)) +print(f"largest gap between formula and solver: {np.abs(ev_solver - ev_formula).max():.1e}") +``` + ### Static complementarity It is worth seeing how much static complementarity we are varying. @@ -958,6 +1448,10 @@ Maximizing the static profit function gives the best response $x^{*}(X) = -Q^{-1 \kappa \equiv \frac{\bar\eta_D'}{(\bar\eta_D-1)\bar\eta_D} . ``` +We call the row sum of this matrix, $\kappa/(1+\kappa)$, **static pass-through**: it is the response of a store's prices to a uniform one percent rise in its competitors' prices. + +The next cell confirms the formula and reports pass-through for each superelasticity. + ```{code-cell} ipython3 print(f"{'η_D′':>6}{'pass-through':>14} best response matrix, first row") for η_Dp in (-3.0, 0.0, 3.0, 10.0): @@ -968,39 +1462,19 @@ for η_Dp in (-3.0, 0.0, 3.0, 10.0): assert np.allclose(BR, BR_closed) print(f"{η_Dp:>6}{BR.sum(axis=1)[0]:>14.4f} {BR[0].round(4)}") -print("\nΘ_X symmetric: ", np.allclose(Θ_K, Θ_K.T)) +print("\nΘ_X symmetric: ", np.allclose(Θ_K, Θ_K.T)) print("best response symmetric:", np.allclose(BR, BR.T)) ``` Static pass-through runs from $-33\%$ to $+45\%$ across these rows, and it changes sign with the superelasticity, yet every one of these economies has exactly the same aggregate dynamics. -An intuition that reads stronger static complementarity as more aggregate propagation is therefore unreliable. - -Notice also that the best response matrix is *not* symmetric, because shares differ across products, while $\Theta_X$ always is. - -Symmetry of $\Theta_X$ is what {prf:ref}`mfg_prop_planner` needs, and it survives even when the static game looks asymmetric. - -### Closed-form eigenvalues - -Because $A_K$ is a multiple of the identity and $B_K = I$, the aggregate eigenvalues have the same form as in the scalar case, one for each eigenvalue $\omega_i$ of $(\Gamma+\Theta_{\mathcal A})^{-1}(Q+\Theta_X)$: - -$$ -\lambda_i = \frac\rho2 - \sqrt{\left(\frac\rho2 + a\right)^2 + \omega_i} . -$$ - -```{code-cell} ipython3 -Q_K, Θ_K = kimball(η_d, η_D, 3.0, s_bar) -P_K = mfg_riccati(A_K, B_K, Q_K + Θ_K, Γ_K, ρ_K) -JG_K = closed_loop(A_K, B_K, Γ_K, P_K) - -ω = np.linalg.eigvals(np.linalg.solve(Γ_K, Q_K + Θ_K)).real -predicted = np.sort(ρ_K/2 - np.sqrt((ρ_K/2 + π_K)**2 + ω)) +```{note} +Stronger static complementarity need not mean more aggregate persistence. -print("eigenvalues of the closed-loop matrix:", np.sort(np.linalg.eigvals(JG_K).real).round(6)) -print("from the closed-form formula: ", predicted.round(6)) +What matters for aggregate dynamics is the effective curvature $Q + \Theta_X$, not the interaction $\Theta_X$ or the static best responses on their own. ``` -{ref}`mfg_ex8` asks what happens when inflation differs across products, so that $A$ is diagonal but not a multiple of the identity. +Notice also that the best response matrix is *not* symmetric, because shares differ across products, while $\Theta_X$ is symmetric, as promised in the description of the environment. ### Aggregate and individual price paths @@ -1025,10 +1499,10 @@ for k, η_Dp in enumerate((-3.0, 0.0, 3.0, 10.0)): Q_K, Θ_K = kimball(η_d, η_D, η_Dp, s_bar) P_K = mfg_riccati(A_K, B_K, Q_K + Θ_K, Γ_K, ρ_K) β2_K = mfg_riccati(A_K, B_K, Q_K, Γ_K, ρ_K) - U_XX = closed_loop(A_K, B_K, Γ_K, P_K) - U_zz = closed_loop(A_K, B_K, Γ_K, β2_K) - agg = np.array([s_bar @ expm(U_XX*t) @ shock for t in t_K]) - dev = np.array([s_bar @ expm(U_zz*t) @ shock for t in t_K]) + U_agg_K = closed_loop(A_K, B_K, Γ_K, P_K) + U_ind_K = closed_loop(A_K, B_K, Γ_K, β2_K) + agg = np.array([s_bar @ expm(U_agg_K*t) @ shock for t in t_K]) + dev = np.array([s_bar @ expm(U_ind_K*t) @ shock for t in t_K]) label = f"$\\bar\\eta_D' = {η_Dp}$" # decreasing line widths, so that four coincident curves remain visible axes[0].plot(t_K, 100*agg, lw=6 - 1.5*k, label=label) @@ -1048,179 +1522,305 @@ The four curves in the left panel lie exactly on top of one another. The four curves in the right panel do not: the higher the superelasticity, the faster a store closes a gap between its own prices and the industry's. -Micro and macro price flexibility are governed by different objects, and only the micro one responds to the superelasticity. +Individual and aggregate price flexibility are governed by different objects, and only the individual one responds to the superelasticity. -### The planner does care +(mfg_planner)= +## The planner -Doubling the interaction gives the planner the effective curvature $Q + 2\Theta_X = (Q + \Theta_X) - \bar\eta_D' \bar s\bar s^\top$, which *does* depend on the superelasticity. +Is the equilibrium speed of adjustment efficient? -```{code-cell} ipython3 -print(f"{'η_D′':>6} {'eig(Q + 2Θ_X)':>26} {'planner eigenvalues':>32}") -for η_Dp in (-3.0, 0.0, 3.0, 10.0, 20.0): - Q_K, Θ_K = kimball(η_d, η_D, η_Dp, s_bar) - M = Q_K + 2*Θ_K - P_p = mfg_riccati(A_K, B_K, M, Γ_K, ρ_K) - ev = np.sort(np.linalg.eigvals(closed_loop(A_K, B_K, Γ_K, P_p)).real) - flag = "" if np.linalg.eigvalsh(M).min() > 0 else " <- not positive definite" - print(f"{η_Dp:>6} {str(np.linalg.eigvalsh(M).round(3)):>26}" - f" {str(ev.round(4)):>32}{flag}") -``` - -So the equilibrium's insensitivity to the superelasticity is special to the equilibrium. +To answer, we compare it with the allocation of a utilitarian planner, who chooses an action $\alpha(x,t)$ for every agent and date to maximize the discounted value of the cross-sectional average of the period return, subject to the Kolmogorov forward equation. -The gap between what a planner would do and what the industry does widens as the superelasticity rises, and in the last row the planner's effective curvature has stopped being positive definite and the returned matrix no longer stabilizes the system. +A second-order approximation to a general return also contains terms in the aggregates alone, -{ref}`mfg_ex7` locates the threshold exactly, and gives it an interpretation in terms of static pass-through. +$$ +-\tfrac12 X^\top\Lambda_X X - \tfrac12\mathcal A^\top\Lambda_{\mathcal A}\mathcal A , +$$ +with $\Lambda_X$ and $\Lambda_{\mathcal A}$ symmetric. -## Aggregate shocks and identification +An individual takes $X$ and $\mathcal A$ as given, so these terms do not affect her choices or the decentralized equilibrium, and we left them out of {eq}`mfg_return`. -Now add a shock that hits everyone at once. - -Let $\mathcal J$ be a compensated jump process and suppose +The planner cannot ignore them, and its period objective is $$ -dx = (B\alpha - Ax)dt + \Sigma^{1/2}dW + \Upsilon \, d\mathcal J . +-\tfrac12 \int x^\top Q x \, m(x) dx - X^\top\left(\Theta_X + \tfrac12\Lambda_X\right) X + -\tfrac12 \int \alpha(x)^\top \Gamma \alpha(x) \, m(x) dx + - \mathcal A^\top\left(\Theta_{\mathcal A} + \tfrac12\Lambda_{\mathcal A}\right)\mathcal A , $$ -With common noise the value function must keep track of the aggregate state as well as the individual state, so it becomes $v(x, X)$. +where $X = \int x \, m(x) dx$ and $\mathcal A = \int \alpha(x) m(x) dx$. -Nevertheless, matching coefficients in the recursive HJB equation delivers a result worth emphasizing. +The planner internalizes both sides of each interaction. + +Moving one agent's state changes her own payoff and, through $X$, everybody else's, so the derivative of $-X^\top\Theta_X X$ with respect to the density contributes $-X^\top(\Theta_X + \Theta_X^\top)x$, twice what an individual takes into account. + +The aggregate-only term adds $-X^\top\Lambda_X x$, and the same reasoning applies to actions. ```{prf:proposition} -:label: mfg_prop_common_noise +:label: mfg_prop_planner -With common noise, the matrix $P$ governing aggregate dynamics solves the *same* equilibrium Riccati equation as before. +Whenever the planner's problem has a finite optimum, the planner's allocation coincides with the decentralized equilibrium of an economy whose interaction matrices are $\Theta_X + \Theta_X^\top + \Lambda_X$ and $\Theta_{\mathcal A} + \Theta_{\mathcal A}^\top + \Lambda_{\mathcal A}$. + +With symmetric interactions these are $2\Theta_X + \Lambda_X$ and $2\Theta_{\mathcal A} + \Lambda_{\mathcal A}$, and the planner's Riccati equation is + +$$ +P^{\text{pl}} B(\Gamma + 2\Theta_{\mathcal A} + \Lambda_{\mathcal A})^{-1}B^\top P^{\text{pl}} += Q + 2\Theta_X + \Lambda_X + \rho P^{\text{pl}} + P^{\text{pl}}A + A^\top P^{\text{pl}} . +$$ + +A finite optimum exists if $Q + 2\Theta_X + \Lambda_X$ and $\Gamma + 2\Theta_{\mathcal A} + \Lambda_{\mathcal A}$ are positive definite. ``` -This is a certainty-equivalence result of the kind familiar from linear quadratic control. +Because each interaction matrix appears together with its transpose, the planner never needs them to be symmetric. -It also yields a clean decomposition. +Symmetry is a restriction we need for the decentralized equilibrium, not for the planner. -Writing $z = x - X$ for an agent's deviation from the cross-sectional mean, +The planner's curvatures have a reading that complements the one in {ref}`mfg_effective`. -```{math} -:label: mfg_decomposition -\begin{aligned} -dX &= \left(\Lambda P - A\right) X dt + \Upsilon d\mathcal J , \\ -dz &= \left(B\Gamma^{-1}B^\top\bar\beta_2 - A\right) z \, dt + \Sigma^{1/2}dW , \\ -\alpha &= (\Gamma+\Theta_{\mathcal A})^{-1}B^\top P X + \Gamma^{-1}B^\top\bar\beta_2 z . -\end{aligned} -``` +When everybody moves together, so that $x = X$, the state part of the return including the aggregate-only term is -Look carefully at the second line. +$$ +-\tfrac12 X^\top Q X - X^\top\Theta_X X - \tfrac12 X^\top\Lambda_X X = -\tfrac12 X^\top(Q + 2\Theta_X + \Lambda_X)X . +$$ -The dynamics of an agent's deviation from the mean involve $\bar\beta_2$, the solution of the *single-agent* Riccati equation, and neither interaction matrix appears. +So $Q + \Theta_X$ is the slope of an agent's *marginal* return when everybody moves together, while $Q + 2\Theta_X + \Lambda_X$ is the curvature of the *total* return when everybody moves together. -The same is true of the part of the action that responds to $z$. +The equilibrium responds to the first and the planner to the second, and the same is true of actions. -```{prf:proposition} -:label: mfg_prop_identification +The comparative statics of {ref}`mfg_persistence` now apply directly, under the assumptions stated there. -Data on $x(t) - X(t)$ and $\alpha(t)-\mathcal A(t)$ contain no information about $\Theta_X$ or $\Theta_{\mathcal A}$. -``` +Internalizing complementarity in states raises the trace of the closed-loop matrix, which slows aggregate adjustment on average, while internalizing complementarity in actions does the opposite. -This is a sharp statement of the "missing intercept" problem. +The aggregate-only terms can reinforce or offset these forces. -Removing time effects from micro data -- a standard way to control for aggregate conditions -- removes exactly the variation that identifies strategic interaction. +A negative semi-definite $\Lambda_X$ or a positive semi-definite $\Lambda_{\mathcal A}$ slows adjustment further, and the opposite signs speed it up. -Aggregate data, by contrast, do carry that information, and can be combined with micro data to recover it. +So the planner may adjust more slowly or more quickly than the decentralized equilibrium, and statements about individual eigenvalues need the restrictions of {prf:ref}`mfg_prop_eigen_symmetric` or {prf:ref}`mfg_prop_eigen_aI`. -```{code-cell} ipython3 -# reduced-form matrices that an econometrician could estimate -U_XX = closed_loop(A, B, Γ + Θ_A, P) # drift of the aggregate state -U_zz = closed_loop(A, B, Γ, β2) # drift of the deviation from the mean +Our two-dimensional example has no aggregate-only terms, so we set $\Lambda_X = \Lambda_{\mathcal A} = 0$ when we compute its planner's allocation. -print("aggregate drift U_XX =\n", U_XX.round(4)) -print("\ndeviation drift U_zz =\n", U_zz.round(4)) +```{code-cell} ipython3 +P_planner = mfg_riccati(A, B, Q + 2*Θ_X, Γ + 2*Θ_A, ρ) +U_pl = closed_loop(A, B, Γ + 2*Θ_A, P_planner) -# now double the state interaction and recompute -P_alt = mfg_riccati(A, B, Q + 2*Θ_X, Γ + Θ_A, ρ) -print("\nwith Θ_X doubled:") -print(" aggregate drift changes: ", - not np.allclose(U_XX, closed_loop(A, B, Γ + Θ_A, P_alt))) -print(" deviation drift changes: ", - not np.allclose(U_zz, closed_loop(A, B, Γ, β2))) +print("individual eigenvalues:", np.sort(np.linalg.eigvals(U_ind).real).round(4)) +print("aggregate eigenvalues: ", np.sort(np.linalg.eigvals(U_agg).real).round(4)) +print("planner eigenvalues: ", np.sort(np.linalg.eigvals(U_pl).real).round(4)) ``` -{ref}`mfg_ex4` shows how to recover $\Theta_X$ from aggregate data, and why $\Theta_X$ and $\Theta_{\mathcal A}$ cannot be told apart without a normalization. +With complementarity in states and substitutability in actions, both of which the planner internalizes, every planner eigenvalue is closer to zero than the corresponding equilibrium eigenvalue, so the planner's allocation is the most persistent of the three. -## Exercises +{ref}`mfg_ex_planner` compares the planner with the equilibrium in other configurations of the interactions. -```{exercise} -:label: mfg_ex1 +(mfg_planner_capital)= +### The planner in the capital accumulation example -Formulas {eq}`mfg_scalar_existence` and {eq}`mfg_scalar_stability` divide the scalar model into four regions. +The capital accumulation example does have aggregate-only terms, and they change the answer. -Define +{cite:t}`AlvarezArgente2026` do not work out the planner for this example, so we do it here. + +A second-order expansion of the firm's return produces terms in $K$ and $I$ alone, which did not matter for the firm's problem. + +They are $$ --\theta^{*} \equiv q + a(a+\rho)\frac{\gamma+\theta_{\mathcal A}}{b^2}, +\Lambda_X = -\frac{\bar k^2 \Pi_{KK}}{\bar\Pi} = \frac{\nu}{\eta}\left(1 - \frac{\nu}{\eta}\right), \qquad --\theta^{**} \equiv q + \left(\frac\rho2+a\right)^2\frac{\gamma+\theta_{\mathcal A}}{b^2} . +\Lambda_{\mathcal A} = \frac{\bar i^{\,3}\,\mathcal P''(\bar i)}{\bar\Pi} += \theta_{\mathcal A}\,\frac{\bar i\,\mathcal P''(\bar i)}{\mathcal P'(\bar i)} , $$ -Take $\rho = 0.5$, $a = 0.3$, $b = 1$, $q = 1$, $\gamma = 1$, $\theta_{\mathcal A}=0$. +where the second comes from the term $\tfrac12 \bar i\,\mathcal P''(\bar i)(I - \bar i)^2$ in the expansion of investment spending $i\,\mathcal P(I)$. -1. Compute $\theta^{*}$ and $\theta^{**}$, and verify that $-\theta^{**} = -\theta^{*} + (\rho/2)^2(\gamma+\theta_{\mathcal A})/b^2$. -1. For values of $\theta_X$ on either side of each threshold, compute $\lambda$ and classify the outcome: stable, a unit root, divergent but admissible, or no equilibrium. -1. What does `mfg_riccati` do when no equilibrium exists? -1. Plot $\lambda$ against $\theta_X$ and mark the thresholds. -``` +Adding the state interaction twice gives a clean result, -```{solution-start} mfg_ex1 -:class: dropdown -``` +$$ +q + 2\theta_X + \Lambda_X = \nu(1-\nu) , +$$ + +which is the slope of the dashed line in {numref}`fig-mfg-marginal-returns`, the marginal revenue of the whole industry when every firm holds the same capital. + +The equilibrium's effective curvature $\frac{\eta-1}{\eta}\nu(1-\nu)$ is smaller by the factor $(\eta-1)/\eta$. + +A firm's marginal revenue is only $(\eta-1)/\eta$ of the marginal value of its output, so an industry-wide capital gap changes the private return to investing by less than it changes the value of the industry's output. + +For actions, the constant-elasticity supply curve of the calibration gives $\bar i\,\mathcal P''(\bar i)/\mathcal P'(\bar i) = 1/\varepsilon_s - 1$, so the planner's action curvature is $\gamma + (1 + 1/\varepsilon_s)\theta_{\mathcal A}$. + +That exceeds the equilibrium's $\gamma + \theta_{\mathcal A}$, because the planner recognizes that a burst of industry investment raises the price that every firm pays. + +So the two sides of the planner's problem pull in opposite directions: on states the planner is faster than the equilibrium, on actions slower. + +Let's compute the net effect, and also what we would have found had we doubled the interactions and ignored the aggregate-only terms. ```{code-cell} ipython3 -ρ_e, a_e, b_e, q_e, γ_e, θA_e = 0.5, 0.3, 1.0, 1.0, 1.0, 0.0 +def cap_Λ_X(η, ν): + "Aggregate-only state curvature in the capital accumulation example." + return ν/η*(1 - ν/η) -θ_star = -(q_e + a_e*(a_e + ρ_e)*(γ_e + θA_e)/b_e**2) -θ_ss = -(q_e + (ρ_e/2 + a_e)**2*(γ_e + θA_e)/b_e**2) +Λ_X_c = cap_Λ_X(η_c, ν_c) +Λ_A_c = (1/ε_s_c - 1)*θ_A_c -print(f"θ* = {θ_star:.4f}") -print(f"θ** = {θ_ss:.4f}") -print(f"gap = {θ_star - θ_ss:.4f}, " - f"(ρ/2)²(γ+θ_A)/b² = {(ρ_e/2)**2*(γ_e + θA_e)/b_e**2:.4f}") +print(f"Λ_X = {Λ_X_c:.4f}, Λ_A = {Λ_A_c:.4f}") +print(f"q + 2θ_X + Λ_X = {q_c + 2*θ_X_c + Λ_X_c:.4f}, ν(1 - ν) = {ν_c*(1 - ν_c):.4f}") + +planner_cases = {'no interactions': (0.0, 0.0), + 'equilibrium': (θ_X_c, θ_A_c), + 'doubling only': (2*θ_X_c, 2*θ_A_c), + 'planner': (2*θ_X_c + Λ_X_c, 2*θ_A_c + Λ_A_c)} + +print(f"\n{'':>16}{'λ':>10}{'half-life':>12}{'investment on impact':>23}") +for label, (tx, ta) in planner_cases.items(): + λ_case = scalar_λ(ρ_c, δ_c, δ_c, q_c, γ_c, tx, ta) + impact = 100*X0_c*(λ_case + δ_c)/δ_c + print(f"{label:>16}{λ_case:>10.4f}{half_life(λ_case):>12.2f}{impact:>22.2f}%") ``` +The last column is the change in industry investment, relative to its steady-state level, when the capital stock starts ten percent above its steady state. + +At this calibration the state side wins narrowly. + +The planner's half-life is 4.8 years against the equilibrium's 5.0, and the planner cuts investment by more on impact. + +Doubling the interactions without the aggregate-only terms would have given the opposite answer, a half-life above six years. + +Doubling $\theta_X$ strengthens the complementarity in states, while $\Lambda_X$ more than undoes it. + +The next figure shows the corresponding paths of investment. + ```{code-cell} ipython3 -for θ_X in (0.5, -0.5, θ_star, -1.27, -1.31): - inside = (ρ_e/2 + a_e)**2 + b_e**2*(q_e + θ_X)/(γ_e + θA_e) - if inside < 0: - print(f"θ_X = {θ_X:+.4f}: no equilibrium") - continue - λ = ρ_e/2 - np.sqrt(inside) - if λ < -1e-9: - kind = "stable, X converges to zero" - elif abs(λ) <= 1e-9: - kind = "unit root, X stays where it starts" - else: - kind = "divergent but admissible" - print(f"θ_X = {θ_X:+.4f}: λ = {λ:+.5f} {kind}") -``` - -Below $\theta^{**}$ there is no real root, and the solver reports failure rather than returning a spurious answer. +--- +mystnb: + figure: + caption: Equilibrium and planner investment paths + name: fig-mfg-planner-investment +--- +fig, ax = plt.subplots(figsize=(8, 4.5)) +for label in ('equilibrium', 'doubling only', 'planner'): + tx, ta = planner_cases[label] + λ_case = scalar_λ(ρ_c, δ_c, δ_c, q_c, γ_c, tx, ta) + ls = '--' if label == 'doubling only' else '-' + ax.plot(t_c, 100*X0_c*(λ_case + δ_c)/δ_c*np.exp(λ_case*t_c), lw=2, ls=ls, label=label) +ax.axhline(0, color='k', lw=0.8) +ax.set_xlabel('years') +ax.set_ylabel('investment, % above steady state') +ax.legend() +plt.tight_layout() +plt.show() +``` + +Two caveats are worth keeping in mind. + +First, the planner maximizes the sum of the firms' payoffs, so it counts the higher price of investment goods as a cost to firms; a planner who also counted the surplus of the suppliers of investment goods would treat part of it as a transfer. + +Second, the comparison is local. + +Within the quadratic model, the planner's different steady state comes from linear terms, which do not affect the dynamics of deviations. + +But as an approximation to the nonlinear economy, the planner's own steady state differs from the equilibrium's (at this calibration it has more capital), and an expansion around it would give somewhat different curvatures. + +{ref}`mfg_ex_planner_capital` splits the gap between the planner and the equilibrium into its state and action parts, and finds the supply elasticity at which the ranking flips. + +### The planner in the Kimball example + +{cite:t}`AlvarezArgente2026` point out that in the Kimball example doubling the interaction alone would give a planner the state curvature $Q + 2\Theta_X = (Q + \Theta_X) - \bar\eta_D' \bar s\bar s^\top$, which depends on the superelasticity, and that the aggregate-only term $\Lambda_X$ restores the cancellation. + +The reading above explains why: $Q + 2\Theta_X + \Lambda_X$ is the curvature of a store's profit when its prices equal the industry's, where its relative price is one, so the superelasticity drops out, although the elasticity $\bar\eta_D$ still matters through the markup. + +With household spending held fixed, {ref}`mfg_ex_kimball_planner` shows that the planner's state curvature is + +$$ +Q + 2\Theta_X + \Lambda_X = (\bar\eta_D - 1)\left[\bar s\bar s^\top + (2\bar\eta_d - 1)V\right], +\qquad V \equiv \operatorname{diag}(\bar s) - \bar s\bar s^\top , +$$ + +against $Q + \Theta_X = (\bar\eta_D - 1)\left[\bar\eta_D\bar s\bar s^\top + \bar\eta_d V\right]$ for the equilibrium. + +The planner's curvature is lower in the direction of a proportional increase in all of a store's prices, which changes the store's price level, and higher in directions that change only relative prices. + +Let's see what that does to the modes of aggregate dynamics, measuring how much each mode moves the price level by the cosine of the angle between its eigenvector and a vector of ones. ```{code-cell} ipython3 -try: - mfg_riccati(np.array([[a_e]]), np.array([[b_e]]), - np.array([[q_e - 1.31]]), np.array([[γ_e]]), ρ_e) -except Exception as e: - print(f"solver raises {type(e).__name__} when no equilibrium exists") +ss_bar = np.outer(s_bar, s_bar) +V = np.diag(s_bar) - ss_bar +M_act = (η_D - 1)*(ss_bar + (2*η_d - 1)*V) +Q_K, Θ_K = kimball(η_d, η_D, 3.0, s_bar) + +for label, M_use in (('equilibrium', Q_K + Θ_K), ('planner', M_act)): + U_use = closed_loop(A_K, B_K, Γ_K, mfg_riccati(A_K, B_K, M_use, Γ_K, ρ_K)) + ev, W = np.linalg.eig(U_use) + for j in np.argsort(ev.real): + w = W[:, j].real + cosine = abs(w.sum())/(np.sqrt(n_K)*np.linalg.norm(w)) + print(f"{label:>12}: eigenvalue {ev[j].real:.4f}, " + f"alignment with the price level {cosine:.2f}") ``` +The price-level mode is the fastest mode in equilibrium and the slowest for the planner: within the quadratic approximation, a cartel of stores would adjust its price level more slowly, and its relative prices more quickly, than the industry does. + +Two qualifications apply. + +This planner maximizes the stores' joint profits and ignores the households who buy from them, so it is a collusive benchmark rather than an efficient one. + +And with household spending fixed, the stores' joint profit keeps rising as all prices rise together, so the cartel has no interior optimum and the comparison is purely local. + +{ref}`mfg_ex_kimball_planner` also shows that with the profit normalization used to compute $Q$ and $\Theta_X$, which drops a factor that does not depend on the store's own prices, the planner's Riccati equation has no symmetric solution, so that planner's problem has no finite optimum. + +## Summary + +The four lessons of the lecture, with the numbers from our examples: + +* **Effective curvature.** Aggregate dynamics solve a single-agent Riccati equation in which $Q$ and $\Gamma$ become $Q + \Theta_X$ and $\Gamma + \Theta_{\mathcal A}$; in the pricing example the superelasticity cancels from $Q + \Theta_X$, so aggregate price dynamics do not depend on it. +* **Individual versus aggregate.** In the capital example a firm's deviation from the industry mean has a half-life of three years, while the industry's capital stock has a half-life of five. +* **The missing intercept.** In simulated panels from two industries that differ only in complementarity, the demeaned paths of firms are identical, and only the industry means differ. +* **Marginal versus total.** The planner's half-life in the capital example is 4.8 years, against 5.0 in equilibrium, while doubling the interactions and ignoring the aggregate-only terms would have given 6.25. + +## Exercises + +```{exercise} +:label: mfg_ex_scalar + +This exercise derives the formulas of the scalar case. + +1. Write the equilibrium Riccati equation in the scalar case, solve the quadratic, and show that the root satisfying the admissibility condition gives {eq}`mfg_scalar_lambda`. +1. Derive the existence condition {eq}`mfg_scalar_existence` and the stability condition {eq}`mfg_scalar_stability`. +1. With the parameters of the regions figure and $\theta_{\mathcal A} = 0$, plot $\lambda$ against $\theta_X$ and mark the two thresholds. +``` + +```{solution-start} mfg_ex_scalar +:class: dropdown +``` + +For part 1, with $g \equiv \gamma + \theta_{\mathcal A}$ the Riccati equation is $\frac{b^2}{g}p^2 = q + \theta_X + (\rho + 2a)p$, and the closed-loop eigenvalue is $\lambda = \frac{b^2}{g}p - a$. + +Substituting $p = g(\lambda + a)/b^2$ turns the Riccati equation into + +$$ +(\lambda + a)^2 - (\rho + 2a)(\lambda + a) = \frac{b^2(q + \theta_X)}{g} , +$$ + +a quadratic in $\lambda$ whose roots are $\frac\rho2 \pm \sqrt{(\frac\rho2 + a)^2 + b^2(q+\theta_X)/g}$. + +The two roots are symmetric about $\rho/2$, as {prf:ref}`mfg_prop_roots` requires, and only the smaller one satisfies the admissibility condition $\lambda < \rho/2$. + +For part 2, a real root exists when the term under the square root is nonnegative, and the admissible root lies strictly below $\rho/2$ when it is positive, which gives {eq}`mfg_scalar_existence`. + +The admissible root is negative when $\sqrt{(\rho/2 + a)^2 + b^2(q+\theta_X)/g} > \rho/2$, which after squaring gives {eq}`mfg_scalar_stability`. + +For part 3, we evaluate the closed form along a line of $\theta_X$ values, using the thresholds `θ_star` and `θ_2star` computed in the lecture. + ```{code-cell} ipython3 -θ_grid = np.linspace(-1.30, 0.5, 400) -inside = (ρ_e/2 + a_e)**2 + b_e**2*(q_e + θ_grid)/(γ_e + θA_e) -λ_grid = np.where(inside >= 0, ρ_e/2 - np.sqrt(np.maximum(inside, 0)), np.nan) +θ_line = np.linspace(-1.30, 0.5, 400) +inside_line = (ρ_s/2 + a)**2 + b**2*(q + θ_line)/γ +λ_line = np.where(inside_line >= 0, + ρ_s/2 - np.sqrt(np.maximum(inside_line, 0)), np.nan) fig, ax = plt.subplots(figsize=(8, 4.5)) -ax.plot(θ_grid, λ_grid, lw=2) +ax.plot(θ_line, λ_line, lw=2) ax.axhline(0, color='k', lw=0.8) -ax.axhline(ρ_e/2, color='grey', ls=':', lw=1, label=r'$\rho/2$') +ax.axhline(ρ_s/2, color='grey', ls=':', lw=1, label=r'$\rho/2$') ax.axvline(θ_star, color='r', ls='--', lw=1, label=r'$\theta^{*}$') -ax.axvline(θ_ss, color='b', ls='--', lw=1, label=r'$\theta^{**}$') +ax.axvline(θ_2star, color='b', ls='--', lw=1, label=r'$\theta^{**}$') ax.set_xlabel(r'$\theta_X$') ax.set_ylabel(r'$\lambda$') ax.legend() @@ -1236,168 +1836,161 @@ Persistence rises smoothly until $\theta^{*}$, where the aggregate state stops c ``` ```{exercise} -:label: mfg_ex2 - -The lecture showed that stronger complementarity in states raises the *trace* of the closed-loop matrix. - -Does it also raise every eigenvalue? +:label: mfg_ex_capital -Draw many random two-dimensional economies: take $B = I$, let $Q$ and $\Gamma$ be random positive definite matrices, let $\Theta_X$ be random negative definite, and set $\rho = 0.05$. +This exercise derives the calibration of the capital accumulation example. -For each draw, compare the equilibrium with interaction $\Theta_X$ to the one with $1.5\,\Theta_X$, keeping only draws in which both equilibria exist and are admissible. - -Report how often the trace rises and how often *every* eigenvalue rises. +1. Use the steady-state Euler equation $\Pi_k(\bar k, \bar k) = \rho + \delta$ to show that $\bar k^{1-\nu} = \nu(\eta-1)/[\eta(\rho+\delta)]$, and then that $\gamma = \phi\,\delta\bar k^{1-\nu}$ and $\theta_{\mathcal A} = \delta \bar k^{1-\nu}/\varepsilon_s$. +1. Show that all pairs $(\phi, \varepsilon_s)$ with the same value of $\phi + 1/\varepsilon_s$ generate identical *aggregate* dynamics, and verify numerically that they nevertheless imply different individual investment rules. +1. After reading {ref}`mfg_planner_capital`, show that these pairs also imply different planner allocations. ``` -```{solution-start} mfg_ex2 +```{solution-start} mfg_ex_capital :class: dropdown ``` -```{code-cell} ipython3 -rng = np.random.default_rng(3) -trace_rose = eig_rose = kept = 0 +For part 1, $\Pi_k(k,K) = \nu\frac{\eta-1}{\eta}k^{\nu(1-1/\eta)-1}K^{\nu/\eta}$, so on the diagonal $\Pi_k(\bar k,\bar k) = \nu\frac{\eta-1}{\eta}\bar k^{\nu-1}$. -for _ in range(400): - A_r = rng.normal(size=(2, 2))*0.3 + 0.4*np.eye(2) - B_r = np.eye(2) - M = rng.normal(size=(2, 2)); Q_r = M @ M.T + 0.5*np.eye(2) - M = rng.normal(size=(2, 2)); Γ_r = M @ M.T + 0.5*np.eye(2) - M = rng.normal(size=(2, 2)); Θ_r = -0.2 * (M @ M.T) +Setting this equal to $\rho+\delta$, which is the steady-state user cost when $\mathcal P(\bar i) = 1$ and $\psi'(\bar i) = 0$, gives the stated expression for $\bar k^{1-\nu}$. - try: - P0 = mfg_riccati(A_r, B_r, Q_r + Θ_r, Γ_r, 0.05) - P1 = mfg_riccati(A_r, B_r, Q_r + 1.5*Θ_r, Γ_r, 0.05) - except Exception: - continue +Since $\bar\Pi = \bar k^{\nu}$, $\bar i = \delta\bar k$, $\psi''(\bar i) = \phi/\bar i$ and $\mathcal P'(\bar i) = 1/(\varepsilon_s \bar i)$, - J0 = closed_loop(A_r, B_r, Γ_r, P0) - J1 = closed_loop(A_r, B_r, Γ_r, P1) - if max(np.linalg.eigvals(J0).real.max(), np.linalg.eigvals(J1).real.max()) >= 0.025: - continue +$$ +\gamma = \frac{\delta^2\bar k^2}{\bar k^{\nu}}\frac{\phi}{\delta \bar k} += \phi\,\delta\bar k^{1-\nu} , +\qquad +\theta_{\mathcal A} = \frac{\delta^2\bar k^2}{\bar k^{\nu}} +\frac{1}{\varepsilon_s \delta\bar k} = \frac{\delta\bar k^{1-\nu}}{\varepsilon_s} . +$$ - kept += 1 - trace_rose += np.trace(J1) > np.trace(J0) - 1e-10 - e0 = np.sort(np.linalg.eigvals(J0).real) - e1 = np.sort(np.linalg.eigvals(J1).real) - eig_rose += np.all(e1 >= e0 - 1e-10) +For part 2, $\gamma + \theta_{\mathcal A} = \delta\bar k^{1-\nu}(\phi + 1/\varepsilon_s)$, and by {eq}`mfg_scalar_lambda` only this sum matters for $\lambda$. -print(f"admissible draws: {kept}") -print(f"trace rose: {trace_rose}/{kept}") -print(f"every eigenvalue rose: {eig_rose}/{kept}") -``` +But the individual's Riccati equation {eq}`mfg_beta2` involves $\gamma$ alone, so it separates the pairs. -The trace result holds in every draw, as the theory says it must. +For part 3, the planner's Riccati equation involves $\gamma + 2\theta_{\mathcal A} + \Lambda_{\mathcal A} = \gamma + (1 + 1/\varepsilon_s)\theta_{\mathcal A} = \delta\bar k^{1-\nu}(\phi + 1/\varepsilon_s + 1/\varepsilon_s^2)$, so it separates them too. -Eigenvalue-by-eigenvalue monotonicity fails in a noticeable minority of cases. +The table confirms all three claims, reporting half-lives in years. -The reason is that $Q+\Theta_X$ and $B(\Gamma+\Theta_{\mathcal A})^{-1}B^\top$ need not commute, so the problem does not separate into independent one-dimensional problems, and a change that slows adjustment overall can still speed it up along some direction. +```{code-cell} ipython3 +scale = cap_scale(η_c, ν_c) +print(f"{'φ':>6}{'1/ε_s':>8}{'aggregate':>12}{'individual':>12}{'planner':>10}") +for φ, inv_ε in ((0.40, 0.00), (0.30, 0.10), (1/7, 9/35), (0.05, 0.35)): + γ_case, θ_A_case = φ*scale, inv_ε*scale + λ_agg = scalar_λ(ρ_c, δ_c, δ_c, q_c, γ_case, θ_X_c, θ_A_case) + λ_ind = scalar_λ(ρ_c, δ_c, δ_c, q_c, γ_case, 0.0, 0.0) + λ_pl = scalar_λ(ρ_c, δ_c, δ_c, q_c, γ_case, + 2*θ_X_c + Λ_X_c, (1 + inv_ε)*θ_A_case) + print(f"{φ:>6.3f}{inv_ε:>8.3f}{half_life(λ_agg):>12.4f}" + f"{half_life(λ_ind):>12.4f}{half_life(λ_pl):>10.4f}") +``` -Additional restrictions -- for instance a scalar drift matrix $A = a I$ -- restore monotonicity mode by mode. +Aggregate dynamics are identical down to the last digit, while the individual half-life ranges from about 4.4 years to under 2. + +Aggregate data alone cannot tell a congested market for capital goods from an expensive installation technology. + +Data on firms' deviations from the industry mean can, because they reveal $\gamma$ alone, which is the mirror image of {prf:ref}`mfg_prop_identification`. ```{solution-end} ``` ```{exercise} -:label: mfg_ex3 - -Compare the planner with the decentralized equilibrium. +:label: mfg_ex_market_power -Using the two-dimensional example from the lecture, compute the eigenvalues of the closed-loop matrix for the equilibrium and for the planner in four cases: +This exercise derives the results in the subsection on market power and returns to scale. -1. only complementarity in states, $\Theta_X \prec 0$ and $\Theta_{\mathcal A}=0$ -1. only substitutability in actions, $\Theta_X = 0$ and $\Theta_{\mathcal A} = 0.2 I$ -1. only complementarity in actions, $\Theta_X = 0$ and $\Theta_{\mathcal A} = -0.2 I$ -1. complementarity in both - -In which cases is the planner's allocation more persistent than the equilibrium? Explain. +1. Holding $\phi$ and $\varepsilon_s$ fixed, show that $(q + \theta_X)/(\gamma + \theta_{\mathcal A}) = (1-\nu)(\rho+\delta)/[\delta(\phi + 1/\varepsilon_s)]$ and $q/\gamma = (\rho+\delta)[(1-\nu) + \nu/\eta]/(\phi\delta)$, and explain why market power affects individual but not aggregate adjustment. +1. Show that $\lambda \to -\delta$ as $\nu \to 1$, for every $\eta > 1$, and that the response of aggregate investment to a capital gap, $(\lambda+\delta)/\delta$, goes to zero in the same limit. Explain the limit. ``` -```{solution-start} mfg_ex3 +```{solution-start} mfg_ex_market_power :class: dropdown ``` -```{code-cell} ipython3 -def eigen_pair(Θ_X_use, Θ_A_use): - "Closed-loop eigenvalues for the equilibrium and for the planner." - P_eq = mfg_riccati(A, B, Q + Θ_X_use, Γ + Θ_A_use, ρ) - P_pl = mfg_riccati(A, B, Q + 2*Θ_X_use, Γ + 2*Θ_A_use, ρ) - e_eq = np.sort(np.linalg.eigvals(closed_loop(A, B, Γ + Θ_A_use, P_eq)).real) - e_pl = np.sort(np.linalg.eigvals(closed_loop(A, B, Γ + 2*Θ_A_use, P_pl)).real) - return e_eq, e_pl +For part 1, from {eq}`mfg_capital_coeffs` and the calibration formulas, -zero = np.zeros((2, 2)) -cases = {'states, complements': (Θ_X, zero), - 'actions, substitutes': (zero, 0.2*np.eye(2)), - 'actions, complements': (zero, -0.2*np.eye(2)), - 'both complements': (Θ_X, -0.2*np.eye(2))} +$$ +\frac{q + \theta_X}{\gamma + \theta_{\mathcal A}} += \frac{\frac{\eta-1}{\eta}\nu(1-\nu)}{\frac{\delta\nu(\eta-1)}{\eta(\rho+\delta)}(\phi + 1/\varepsilon_s)} += \frac{(1-\nu)(\rho+\delta)}{\delta(\phi + 1/\varepsilon_s)} . +$$ -print(f"{'case':24}{'equilibrium':>22}{'planner':>22} planner slower?") -for label, (tx, ta) in cases.items(): - e_eq, e_pl = eigen_pair(tx, ta) - slower = bool(np.all(e_pl >= e_eq - 1e-12)) - print(f"{label:24}{str(e_eq.round(4)):>22}{str(e_pl.round(4)):>22} {slower}") -``` +Both numerator and denominator are proportional to $(\eta - 1)/\eta$, the ratio of a firm's marginal revenue to the value of its marginal product. -With only complementarity in states, the planner's allocation is more persistent. +For a single firm, $q = e(1-e)$ with $e = \nu(\eta-1)/\eta$ the elasticity of a firm's revenue with respect to its own capital, so -The planner recognizes that when one agent stays away from the steady state, others are content to stay away too, so there is less reason to hurry back. +$$ +\frac{q}{\gamma} = \frac{e(1-e)}{\phi e\delta/(\rho+\delta)} = \frac{(\rho+\delta)(1-e)}{\phi\delta} += \frac{\rho+\delta}{\phi\,\delta}\left[(1-\nu) + \frac{\nu}{\eta}\right] . +$$ -With only complementarity in *actions*, the comparison reverses and the planner adjusts faster: the planner internalizes that when one agent adjusts, others want to adjust as well. +Market power changes how fast a single firm's marginal revenue falls as it expands alone, but along the diagonal the industry's revenue is $K^\nu$ whatever $\eta$ is, and the $(\eta-1)/\eta$ factors cancel. -The case of substitutability in actions is the mirror image of the last one, and again makes the planner slower. +For part 2, $q + \theta_X = \frac{\eta-1}{\eta}\nu(1-\nu) \to 0$, and with $a = b = \delta$, -When complementarity is present in both states and actions, the two forces work against each other. +$$ +\lambda \to \frac\rho2 - \sqrt{\left(\frac\rho2 + \delta\right)^2} = -\delta . +$$ -In this calibration the state interaction dominates and the planner is still slower, but that ranking is a quantitative accident, not a theorem: in general the comparison between planner and equilibrium persistence cannot be signed. +The aggregate law of motion gives $\mathcal A = \frac{\lambda+\delta}{\delta}X$, which vanishes as $\lambda \to -\delta$, as the table confirms. + +```{code-cell} ipython3 +print(f"{'ν':>8}{'q + θ_X':>10}{'λ':>10}{'half-life':>11}{'A/X':>9}") +for ν in (0.7, 0.9, 0.99, 0.999, 0.99999): + γ_ν, θ_A_ν = φ_c*cap_scale(η_c, ν), cap_scale(η_c, ν)/ε_s_c + λ_ν = scalar_λ(ρ_c, δ_c, δ_c, cap_q(η_c, ν), γ_ν, cap_θ_X(η_c, ν), θ_A_ν) + print(f"{ν:>8}{cap_q(η_c,ν) + cap_θ_X(η_c,ν):>10.5f}{λ_ν:>10.5f}" + f"{half_life(λ_ν):>11.3f}{(λ_ν + δ_c)/δ_c:>9.4f}") +``` + +As $\nu$ approaches 1 the profit function {eq}`mfg_capital_profit` approaches one that is homogeneous of degree one in $(k, K)$ jointly, along which $\Pi_k(k,k)$ does not depend on $k$. + +An industry-wide capital gap then creates almost no incentive to invest differently, and the gap closes almost entirely through depreciation. + +The effective curvature approaches zero, the boundary of the sufficient condition $Q + \Theta_X \succ 0$ of {prf:ref}`mfg_prop_existence`, while the existence condition {eq}`mfg_scalar_existence` continues to hold with room to spare. + +The limit $\nu = 1$ itself is not an economy we can study this way: with constant returns the steady-state Euler equation has no solution, and $\bar k$ grows without bound as $\nu \to 1$. ```{solution-end} ``` ```{exercise} -:label: mfg_ex4 +:label: mfg_ex_identification -This exercise works through {prf:ref}`mfg_prop_identification` and its consequences. +This exercise works through the identification results of {ref}`mfg_identification`. -1. Verify that the drift of an agent's deviation from the cross-sectional mean is unchanged when $\Theta_X$ and $\Theta_{\mathcal A}$ change. -1. Suppose an econometrician knows $\rho$, $Q$, $A$, $B$, $\Gamma$ and $\Theta_{\mathcal A}$, and estimates the aggregate drift matrix $U_{\dot X, X}$. Show how to recover $\Theta_X$, and verify your procedure numerically. +1. Suppose an econometrician knows $\rho$, $Q$, $A$, $B$, $\Gamma$ and $\Theta_{\mathcal A}$, and estimates the aggregate drift matrix $U_{\dot X, X}$. Show how to recover $\Theta_X$, and verify your procedure numerically in the two-dimensional example. 1. In the scalar model, show that $\theta_X$ and $\theta_{\mathcal A}$ are not separately identified: construct a family of pairs that generate identical aggregate dynamics *and* identical policy coefficients. +1. Now suppose the econometrician knows only $\rho$, $\Gamma$ and $\Theta_{\mathcal A}$, but observes $X$, $z$, $\alpha$ and $\mathcal A$, so that all four reduced-form matrices are available. Replace $B$ by $\begin{bmatrix} 1 & 0.3 \\ 0 & 0.8\end{bmatrix}$ in the two-dimensional example, and recover $B$, $A$, $Q$ and $\Theta_X$. ``` -```{solution-start} mfg_ex4 +```{solution-start} mfg_ex_identification :class: dropdown ``` -For part 1, the deviation drift is $B\Gamma^{-1}B^\top\bar\beta_2 - A$, and $\bar\beta_2$ solves the single-agent Riccati equation {eq}`mfg_beta2`, in which no interaction matrix appears. - -```{code-cell} ipython3 -for label, (tx, ta) in {'baseline': (Θ_X, Θ_A), - 'very different': (3*Θ_X, -0.1*np.eye(2))}.items(): - β2_case = mfg_riccati(A, B, Q, Γ, ρ) # does not depend on tx, ta - print(f"{label:16}: U_zz =", closed_loop(A, B, Γ, β2_case).round(6).tolist()) -``` - -For part 2, invert the definition of the closed-loop matrix to get $P$, then read $\Theta_X$ off the equilibrium Riccati equation: +For part 1, invert the definition of the closed-loop matrix to get $P$, then read $\Theta_X$ off the equilibrium Riccati equation: $$ -P = \Lambda^{-1}\left(U_{\dot X, X} + A\right), +P = D^{-1}\left(U_{\dot X, X} + A\right), \qquad -\Theta_X = P\Lambda P - Q - \rho P - PA - A^\top P . +\Theta_X = PDP - Q - \rho P - PA - A^\top P . $$ +Let's apply this to the aggregate drift of the two-dimensional example. + ```{code-cell} ipython3 U_obs = closed_loop(A, B, Γ + Θ_A, P) # what the econometrician estimates -P_hat = np.linalg.solve(Λ, U_obs + A) -Θ_X_hat = P_hat @ Λ @ P_hat - Q - ρ*P_hat - P_hat @ A - A.T @ P_hat +P_hat = np.linalg.solve(D, U_obs + A) +Θ_X_hat = P_hat @ D @ P_hat - Q - ρ*P_hat - P_hat @ A - A.T @ P_hat print("true Θ_X =\n", Θ_X.round(6)) print("\nrecovered Θ_X =\n", Θ_X_hat.round(6)) print(f"\nmaximum error: {np.abs(Θ_X - Θ_X_hat).max():.2e}") ``` -For part 3, formula {eq}`mfg_scalar_lambda` shows that $\lambda$ depends on the two interactions only through the ratio $(q+\theta_X)/(\gamma+\theta_{\mathcal A})$. +For part 2, formula {eq}`mfg_scalar_lambda` shows that $\lambda$ depends on the two interactions only through the ratio $(q+\theta_X)/(\gamma+\theta_{\mathcal A})$. -Holding that ratio fixed traces out a one-parameter family of observationally equivalent economies. +Holding that ratio fixed traces out a one-parameter family of observationally equivalent economies, one of the rays in {numref}`fig-mfg-regions`. The policy coefficient cannot break the tie, because $\lambda = b\,U_{\alpha,X} - a$ ties it to $\lambda$. @@ -1407,7 +2000,7 @@ ratio = (q + (-0.4))/(γ + 0.3) # baseline θ_X = -0.4, θ_A = 0.3 print(f"{'θ_A':>7}{'θ_X':>12}{'λ':>12}{'policy coeff':>15}") for θ_A_alt in (0.3, 0.0, 0.6, 1.0): θ_X_alt = ratio*(γ + θ_A_alt) - q - λ_alt = scalar_lambda(ρ_s, a, b, q, γ, θ_X_alt, θ_A_alt) + λ_alt = scalar_λ(ρ_s, a, b, q, γ, θ_X_alt, θ_A_alt) print(f"{θ_A_alt:>7.2f}{θ_X_alt:>12.6f}{λ_alt:>12.6f}{(λ_alt + a)/b:>15.6f}") ``` @@ -1415,250 +2008,393 @@ Every row describes a different economy, with a different amount of strategic co Distinguishing them requires normalizing one interaction, or bringing outside information to bear. -```{solution-end} -``` +For part 3, the two drift matrices share $A$, so -```{exercise} -:label: mfg_ex5 +$$ +U_{\dot z,z} - U_{\dot X,X} = B\left(U_{\alpha,z} - U_{\alpha,X}\right) +$$ -This exercise derives the two curvatures that we simply assumed when calibrating the capital accumulation example. +identifies $B$. -Suppose the adjustment cost is +Then $A = B U_{\alpha,X} - U_{\dot X,X}$, $P = (B^\top)^{-1}(\Gamma + \Theta_{\mathcal A})U_{\alpha,X}$ and $\bar\beta_2 = (B^\top)^{-1}\Gamma U_{\alpha,z}$. -$$ -\psi(i) = \frac{\phi}{2}\,\bar i\left(\frac{i - \bar i}{\bar i}\right)^2 , -$$ +Finally, the single-agent Riccati equation {eq}`mfg_beta2` gives $Q$ and the equilibrium one gives $\Theta_X$. -so that $\psi'(\bar i) = 0$ and $\psi''(\bar i) = \phi/\bar i$, and suppose investment goods are supplied with constant elasticity $\varepsilon_s$, +```{code-cell} ipython3 +B_true = np.array([[1.0, 0.3], + [0.0, 0.8]]) +Γ_eq = Γ + Θ_A + +# the reduced-form matrices the econometrician estimates +β2_true = mfg_riccati(A, B_true, Q, Γ, ρ) +P_true = mfg_riccati(A, B_true, Q + Θ_X, Γ_eq, ρ) +U_XX = closed_loop(A, B_true, Γ_eq, P_true) +U_zz = closed_loop(A, B_true, Γ, β2_true) +U_αX = np.linalg.solve(Γ_eq, B_true.T @ P_true) +U_αz = np.linalg.solve(Γ, B_true.T @ β2_true) + +# recovery +B_hat = (U_zz - U_XX) @ np.linalg.inv(U_αz - U_αX) +A_hat = B_hat @ U_αX - U_XX +P_hat = np.linalg.solve(B_hat.T, Γ_eq @ U_αX) +β2_hat = np.linalg.solve(B_hat.T, Γ @ U_αz) +Q_hat = (β2_hat @ B_hat @ np.linalg.solve(Γ, B_hat.T) @ β2_hat + - ρ*β2_hat - β2_hat @ A_hat - A_hat.T @ β2_hat) +D_hat = B_hat @ np.linalg.solve(Γ_eq, B_hat.T) +Θ_X_hat = P_hat @ D_hat @ P_hat - Q_hat - ρ*P_hat - P_hat @ A_hat - A_hat.T @ P_hat + +for name, est, true in (('B', B_hat, B_true), ('A', A_hat, A), + ('Q', Q_hat, Q), ('Θ_X', Θ_X_hat, Θ_X)): + print(f"maximum error in {name:>3}: {np.abs(est - true).max():.1e}") +``` + +Every object is recovered to machine precision, but only because $\Theta_{\mathcal A}$ was known: part 2 shows that it cannot be recovered along with the rest. -$$ -\mathcal P(I) = \left(\frac{I}{\bar i}\right)^{1/\varepsilon_s} . -$$ +```{solution-end} +``` -1. Use the steady-state Euler equation $\Pi_k(\bar k, \bar k) = \rho + \delta$ to show that $\bar k^{1-\nu} = \nu(\eta-1)/[\eta(\rho+\delta)]$, and then that +```{exercise} +:label: mfg_ex_eigenvalues -$$ -\gamma = \phi\,\delta\bar k^{1-\nu} , \qquad -\theta_{\mathcal A} = \frac{\delta \bar k^{1-\nu}}{\varepsilon_s} . -$$ +{prf:ref}`mfg_prop_trace` says that stronger complementarity in states raises the *trace* of the closed-loop matrix, and the example of {cite:t}`AlvarezArgente2026` shows that it need not raise every eigenvalue. + +How common is that? -1. Which $(\phi, \varepsilon_s)$ reproduce the calibration $\gamma = 0.05$ and $\theta_{\mathcal A} = 0.09$? -1. Show that all pairs with the same value of $\phi + 1/\varepsilon_s$ generate identical *aggregate* dynamics, and verify numerically that they nevertheless imply different individual investment rules and different planner allocations. +1. Draw many random two-dimensional economies: take $B = I$ and $\Theta_{\mathcal A} = 0$, let $Q$ and $\Gamma$ be random positive definite matrices, let $\Theta_X$ be random negative definite, and set $\rho = 0.05$. For each draw, compare the equilibrium with interaction $\Theta_X$ to the one with $1.5\,\Theta_X$, keeping only draws in which $Q + 1.5\,\Theta_X$ is positive definite. Report how often the trace rises and how often *every* eigenvalue rises. +1. Repeat the experiment with $A = \Gamma^{-1}S$ for a random symmetric $S$, so that $D^{-1}A = S$ is symmetric, and confirm {prf:ref}`mfg_prop_eigen_symmetric`. ``` -```{solution-start} mfg_ex5 +```{solution-start} mfg_ex_eigenvalues :class: dropdown ``` -For part 1, $\Pi_k(k,K) = \nu\frac{\eta-1}{\eta}k^{\nu(1-1/\eta)-1}K^{\nu/\eta}$, so on the diagonal $\Pi_k(\bar k,\bar k) = \nu\frac{\eta-1}{\eta}\bar k^{\nu-1}$. +We write one function for both experiments, taking the rule that draws $A$ as an argument. -Setting this equal to $\rho+\delta$, which is the steady-state user cost when $\mathcal P(\bar i) = 1$ and $\psi'(\bar i) = 0$, gives the stated expression for $\bar k^{1-\nu}$. +```{code-cell} ipython3 +def compare_draws(draw_A, n_draws=400, seed=3, ρ=0.05): + """ + Compare equilibria with interactions Θ_X and 1.5 Θ_X in random + two-dimensional economies with B = I and Θ_A = 0. draw_A(rng, Γ) + returns the drift matrix. Returns the number of draws kept, the number + in which the trace rose and the number in which every eigenvalue rose. + """ + rng = np.random.default_rng(seed) + I2 = np.eye(2) + kept = trace_rose = eig_rose = 0 + for _ in range(n_draws): + M = rng.normal(size=(2, 2)); Q_r = M @ M.T + 0.5*I2 + M = rng.normal(size=(2, 2)); Γ_r = M @ M.T + 0.5*I2 + M = rng.normal(size=(2, 2)); Θ_r = -0.2*(M @ M.T) + A_r = draw_A(rng, Γ_r) + if np.linalg.eigvalsh(Q_r + 1.5*Θ_r).min() <= 0: + continue + J0 = closed_loop(A_r, I2, Γ_r, mfg_riccati(A_r, I2, Q_r + Θ_r, Γ_r, ρ)) + J1 = closed_loop(A_r, I2, Γ_r, mfg_riccati(A_r, I2, Q_r + 1.5*Θ_r, Γ_r, ρ)) + e0 = np.sort(np.linalg.eigvals(J0).real) + e1 = np.sort(np.linalg.eigvals(J1).real) + kept += 1 + trace_rose += np.trace(J1) > np.trace(J0) - 1e-10 + eig_rose += np.all(e1 >= e0 - 1e-10) + return kept, trace_rose, eig_rose + +def general_drift(rng, Γ): + "A random drift matrix with no particular structure." + return rng.normal(size=(2, 2))*0.3 + 0.4*np.eye(2) + +def symmetric_drift(rng, Γ): + "A random drift matrix with Γ A symmetric, so that D^{-1} A is symmetric." + M = rng.normal(size=(2, 2)) + return np.linalg.solve(Γ, 0.3*(M + M.T) + 0.4*np.eye(2)) + +for label, draw_A in (('general A', general_drift), ('D⁻¹A symmetric', symmetric_drift)): + kept, trace_rose, eig_rose = compare_draws(draw_A) + print(f"{label:>16}: draws kept {kept}, trace rose {trace_rose}, " + f"every eigenvalue rose {eig_rose}") +``` -Since $\bar\Pi = \bar k^{\nu}$ and $\bar i = \delta\bar k$, +The trace result holds in every draw, as the theory says it must. -$$ -\gamma = \frac{\delta^2\bar k^2}{\bar k^{\nu}}\frac{\phi}{\delta \bar k} -= \phi\,\delta\bar k^{1-\nu} , -\qquad -\theta_{\mathcal A} = \frac{\delta^2\bar k^2}{\bar k^{\nu}} -\frac{1}{\varepsilon_s \delta\bar k} = \frac{\delta\bar k^{1-\nu}}{\varepsilon_s} . -$$ +With a general drift matrix, eigenvalue-by-eigenvalue monotonicity fails in a minority of draws. -```{code-cell} ipython3 -scale_c = δ_c*ν_c*(η_c - 1)/(η_c*(ρ_c + δ_c)) # δ k̄^{1-ν} +The obstacle is that $D^{-1}A$ need not be symmetric (here $A$ itself is not symmetric), so the closed-loop matrix $DP - A$ need not be similar to a symmetric matrix, and a change that slows adjustment overall can still speed it up along some direction. -print(f"δ k̄^(1-ν) = {scale_c:.4f}") -print(f"φ implied by γ = {γ_c}: {γ_c/scale_c:.4f}") -print(f"ε_s implied by θ_A = {θ_A_c}: {scale_c/θ_A_c:.4f}") +When $D^{-1}A$ is symmetric, every eigenvalue rises in every draw, as {prf:ref}`mfg_prop_eigen_symmetric` says. + +Note that $Q + \Theta_X$ and $D$ need not commute in either experiment: by {prf:ref}`mfg_prop_eigen_aI`, that kind of non-commutativity does not by itself break monotonicity. + +```{solution-end} ``` -The calibration corresponds to a moderately convex adjustment cost, $\phi = 0.14$, and an investment supply elasticity of about $3.9$. +```{exercise} +:label: mfg_ex_planner + +Compare the planner with the decentralized equilibrium in the two-dimensional example from the lecture, with no aggregate-only terms. -For part 3, $\gamma + \theta_{\mathcal A} = \delta\bar k^{1-\nu}(\phi + 1/\varepsilon_s)$, and by {eq}`mfg_scalar_lambda` only this sum matters for $\lambda$. +Compute the eigenvalues of the closed-loop matrix for the equilibrium and for the planner in four cases: only complementarity in states ($\Theta_X$ as in the lecture, $\Theta_{\mathcal A}=0$); only substitutability in actions ($\Theta_X = 0$, $\Theta_{\mathcal A} = 0.2 I$); only complementarity in actions ($\Theta_X = 0$, $\Theta_{\mathcal A} = -0.2 I$); and complementarity in both. -But the individual's Riccati equation {eq}`mfg_beta2` involves $\gamma$ alone, and the planner's involves $\gamma + 2\theta_{\mathcal A}$, so both separate the pairs. +In which cases is the planner's allocation more persistent than the equilibrium? Explain. +``` + +```{solution-start} mfg_ex_planner +:class: dropdown +``` + +We compute both sets of eigenvalues in each case. ```{code-cell} ipython3 -print(f"{'φ':>6}{'1/ε_s':>8}{'aggregate':>12}{'individual':>12}{'planner':>10}") -for φ, inv_ε in ((0.40, 0.00), (0.30, 0.10), (1/7, 0.257143), (0.05, 0.35)): - γ_case, θ_A_case = φ*scale_c, inv_ε*scale_c - λ_agg = scalar_lambda(ρ_c, δ_c, δ_c, q_c, γ_case, θ_X_c, θ_A_case) - λ_ind = scalar_lambda(ρ_c, δ_c, δ_c, q_c, γ_case, 0.0, 0.0) - λ_pl = scalar_lambda(ρ_c, δ_c, δ_c, q_c, γ_case, 2*θ_X_c, 2*θ_A_case) - print(f"{φ:>6.3f}{inv_ε:>8.3f}{λ_agg:>12.6f}{λ_ind:>12.6f}{λ_pl:>10.4f}") +def eigen_pair(Θ_X_use, Θ_A_use): + "Closed-loop eigenvalues for the equilibrium and for the planner." + P_eq = mfg_riccati(A, B, Q + Θ_X_use, Γ + Θ_A_use, ρ) + P_pl = mfg_riccati(A, B, Q + 2*Θ_X_use, Γ + 2*Θ_A_use, ρ) + e_eq = np.sort(np.linalg.eigvals(closed_loop(A, B, Γ + Θ_A_use, P_eq)).real) + e_pl = np.sort(np.linalg.eigvals(closed_loop(A, B, Γ + 2*Θ_A_use, P_pl)).real) + return e_eq, e_pl + +zero = np.zeros((2, 2)) +planner_tests = {'states, complements': (Θ_X, zero), + 'actions, substitutes': (zero, 0.2*np.eye(2)), + 'actions, complements': (zero, -0.2*np.eye(2)), + 'both complements': (Θ_X, -0.2*np.eye(2))} + +print(f"{'case':24}{'equilibrium':>22}{'planner':>22} planner slower?") +for label, (tx, ta) in planner_tests.items(): + e_eq, e_pl = eigen_pair(tx, ta) + slower = bool(np.all(e_pl >= e_eq - 1e-12)) + print(f"{label:24}{str(e_eq.round(4)):>22}{str(e_pl.round(4)):>22} {slower}") ``` -Aggregate dynamics are identical down to the last digit, while individual behavior ranges from sluggish to almost frictionless. +With only complementarity in states, the planner's allocation is more persistent. + +The planner recognizes that when one agent stays away from the steady state, others are content to stay away too, so there is less reason to hurry back. + +With only substitutability in actions the planner is also slower, because it recognizes that a burst of adjustment by everyone at once is costly. + +With only complementarity in actions the comparison reverses and the planner adjusts faster: it internalizes that when one agent adjusts, others want to adjust as well. + +When complementarity is present in both states and actions, the two forces work against each other. -An econometrician with only aggregate data cannot tell a congested market for capital goods from an expensive installation technology, which is the scalar version of {prf:ref}`mfg_prop_identification`. +In this calibration the state interaction dominates and the planner is still slower, but that ranking is a quantitative accident, not a theorem. ```{solution-end} ``` ```{exercise} -:label: mfg_ex6 +:label: mfg_ex_planner_capital -The right panel of the figure above suggests that something special happens as returns to scale approach constancy. +In {ref}`mfg_planner_capital` the planner adjusts slightly faster than the equilibrium, because the state side and the action side of its problem pull in opposite directions. -1. Show that $q + \theta_X \to 0$ and $\lambda \to -\delta$ as $\nu \to 1$, for every $\eta > 1$. -1. Show that the response of aggregate investment to a capital gap, $(\lambda+\delta)/\delta$, goes to zero in the same limit. -1. Explain the limit in economic terms. +1. Compute the planner's half-life when only the state side is corrected, with state curvature $q + 2\theta_X + \Lambda_X$ and action curvature $\gamma + \theta_{\mathcal A}$, and when only the action side is corrected, with $q + \theta_X$ and $\gamma + (1 + 1/\varepsilon_s)\theta_{\mathcal A}$. +1. Hold $\phi$ fixed and let the supply elasticity $\varepsilon_s$ vary. Find the elasticity at which the planner and the equilibrium have the same half-life. +1. Repeat part 2 for $\eta \in \{2, 4, 8\}$ and explain how the threshold depends on market power. ``` -```{solution-start} mfg_ex6 +```{solution-start} mfg_ex_planner_capital :class: dropdown ``` -Part 1 is immediate from {eq}`mfg_capital_coeffs`: $q + \theta_X = \frac{\eta-1}{\eta}\nu(1-\nu) \to 0$, and with $a = b = \delta$, +For part 1, we switch on one side of the planner's problem at a time. -$$ -\lambda \to \frac\rho2 - \sqrt{\left(\frac\rho2 + \delta\right)^2} = -\delta . -$$ +```{code-cell} ipython3 +sides = {'equilibrium': (θ_X_c, θ_A_c), + 'state side only': (2*θ_X_c + Λ_X_c, θ_A_c), + 'action side only': (θ_X_c, 2*θ_A_c + Λ_A_c), + 'planner': (2*θ_X_c + Λ_X_c, 2*θ_A_c + Λ_A_c)} + +for label, (tx, ta) in sides.items(): + λ_case = scalar_λ(ρ_c, δ_c, δ_c, q_c, γ_c, tx, ta) + print(f"{label:>18}: half-life {half_life(λ_case):.2f} years") +``` + +On its own, the state side would cut the half-life to 4.6 years and the action side would raise it to 5.2 years. -For part 2, the scalar version of {eq}`mfg_aggregate_action` gives $\mathcal A = \frac{\lambda+\delta}{\delta}X$, which vanishes as $\lambda \to -\delta$. +For parts 2 and 3, we write the gap between the two half-lives as a function of $1/\varepsilon_s$ and find its root. ```{code-cell} ipython3 -print(f"{'ν':>8}{'q + θ_X':>10}{'λ':>10}{'half-life':>11}{'A/X':>9}") -for ν in (0.7, 0.9, 0.99, 0.999, 0.99999): - λ_ν = scalar_lambda(ρ_c, δ_c, δ_c, cap_q(η_c, ν), γ_c, cap_θ_X(η_c, ν), θ_A_c) - print(f"{ν:>8}{cap_q(η_c,ν) + cap_θ_X(η_c,ν):>10.5f}{λ_ν:>10.5f}" - f"{half_life(λ_ν):>11.3f}{(λ_ν + δ_c)/δ_c:>9.4f}") +def planner_gap(inv_ε, η): + "Planner's half-life minus the equilibrium's, as a function of 1/ε_s." + scale_η = cap_scale(η, ν_c) + γ_η, θ_A_η = φ_c*scale_η, inv_ε*scale_η + q_η, θ_X_η = cap_q(η, ν_c), cap_θ_X(η, ν_c) + λ_eq = scalar_λ(ρ_c, δ_c, δ_c, q_η, γ_η, θ_X_η, θ_A_η) + λ_pl = scalar_λ(ρ_c, δ_c, δ_c, q_η, γ_η, + 2*θ_X_η + cap_Λ_X(η, ν_c), (1 + inv_ε)*θ_A_η) + return half_life(λ_pl) - half_life(λ_eq) + +for η in (2.0, 4.0, 8.0): + inv_star = brentq(planner_gap, 0.01, 20.0, args=(η,)) + print(f"η = {η}: half-lives coincide at ε_s = {1/inv_star:.2f}") ``` -For part 3, note that $q + \theta_X = -\bar k^2\left[\Pi_{kk} + \Pi_{kK}\right]/\bar\Pi$ measures the curvature of profit along the *diagonal*, that is, the rate at which the marginal profitability of capital falls when the whole industry expands together. +With $\eta = 4$, the planner is faster whenever the supply of investment goods is more elastic than about 2.3, which includes the calibrated 3.9, and slower when it is less elastic. -With $\nu = 1$ the profit function {eq}`mfg_capital_profit` is homogeneous of degree one in $(k, K)$ jointly, so $\Pi_k(k,k)$ does not depend on $k$ at all. +The threshold rises with $\eta$. -An industry-wide capital gap then creates no incentive to invest differently, and the gap closes only through depreciation. +The planner's advantage on the state side is $\nu(1-\nu) - \frac{\eta-1}{\eta}\nu(1-\nu) = \nu(1-\nu)/\eta$, which shrinks as market power falls. -Effective curvature is zero, the economy sits exactly on the boundary of the existence region in {eq}`mfg_scalar_existence`, and the equilibrium remains unique and stable with $\lambda = -\delta$. +So in a more competitive industry, a smaller penalty on the action side, which means a more elastic supply of investment goods, is enough to make the planner slower. ```{solution-end} ``` ```{exercise} -:label: mfg_ex7 +:label: mfg_ex_kimball_planner + +This exercise derives the planner's state curvature in the Kimball example, and checks it against the profit function itself. + +Normalize flexible prices to one, so that the CES weights equal the expenditure shares $\bar s$ and marginal costs equal $(\bar\eta_D - 1)/\bar\eta_D$, and use a Kimball demand whose elasticity at relative price $y$ is $\bar\eta_D + \bar\eta_D'(y-1)$. -In the Kimball example, the equilibrium's effective curvature $Q + \Theta_X$ is positive definite for every superelasticity, but the planner's, $Q + 2\Theta_X$, is not. +Let $\Phi(p)$ be the CES price index of a store that charges prices $p$, let $\tilde p$ be the other stores' prices, and let $\mathcal Y$ be household spending, which we hold fixed. + +A store's actual profits are proportional to $\mathcal Y/\Phi(\tilde p)$; {cite:t}`AlvarezArgente2026` drop this factor, which does not depend on the store's own prices, when they compute $Q$ and $\Theta_X$. + +Define -1. Using $Q + 2\Theta_X = (Q+\Theta_X) - \bar\eta_D'\,\bar s\bar s^\top$, show that it is positive definite if and only if -$\bar\eta_D'\,\bar s^\top(Q+\Theta_X)^{-1}\bar s < 1$. -1. Show that $(Q+\Theta_X)\mathbb{1} = (\bar\eta_D-1)\bar\eta_D\,\bar s$, where $\mathbb{1}$ is a vector of ones, and hence that the condition is $\bar\eta_D' < (\bar\eta_D-1)\bar\eta_D$. -1. Show that this is exactly the condition that static pass-through in {eq}`mfg_kimball_br` be below one half, and verify the threshold numerically. +$$ +V \equiv \operatorname{diag}(\bar s) - \bar s\bar s^\top, +\qquad +\ell(X) \equiv \frac{1}{1-\bar\eta_d}\log\sum_j \bar s_j e^{(1-\bar\eta_d)X_j} , +$$ + +so that $\ell$ is the log of the store's price index. + +1. Show that along the diagonal, where a store's prices equal the industry's, actual profit is proportional to $1 - \frac{\bar\eta_D-1}{\bar\eta_D}e^{(\bar\eta_d - 1)\ell(X)}\sum_j\bar s_j e^{-\bar\eta_d X_j}$, and that profit with the factor dropped is proportional to $e^{\ell(X)}$ times that expression. +1. Using $\nabla\ell(0) = \bar s$ and $\nabla^2\ell(0) = (1 - \bar\eta_d)V$, show that $Q + 2\Theta_X + \Lambda_X$ equals $(\bar\eta_D - 1)[\bar s\bar s^\top + (2\bar\eta_d - 1)V]$ with actual profits and $[\bar\eta_d(2\bar\eta_D - 1) - \bar\eta_D]V - \bar\eta_D\bar s\bar s^\top$ with the factor dropped. +1. Check both formulas, and the formulas for $Q$ and $\Theta_X$, against finite-difference Hessians of the profit function, for several values of the superelasticity. +1. Using $V\mathbb{1} = 0$, where $\mathbb{1}$ is a vector of ones, show that the first matrix is positive definite whenever $\bar\eta_d > 1/2$ and the second is not positive definite. Use the matrix $E$ of {eq}`mfg_E` to show that with the factor dropped the planner's Riccati equation has no symmetric solution. +1. Compare the planner's curvature with the equilibrium's, $Q + \Theta_X = (\bar\eta_D - 1)[\bar\eta_D\bar s\bar s^\top + \bar\eta_d V]$, in the direction $\mathbb{1}$ and in directions $v$ with $\bar s^\top v = 0$. ``` -```{solution-start} mfg_ex7 +```{solution-start} mfg_ex_kimball_planner :class: dropdown ``` -For part 1, if $\bar\eta_D' \leq 0$ the rank-one term is added rather than subtracted and positive definiteness is immediate. +For part 1, along the diagonal the relative price $\Phi(p)/\Phi(\tilde p)$ equals one, so the Kimball demand function is evaluated at a constant. -If $\bar\eta_D' > 0$, then for $M \succ 0$ the matrix $M - c vv^\top$ with $c>0$ is positive definite if and only if $c\, v^\top M^{-1}v < 1$, which follows from the determinant identity $\det(M - cvv^\top) = \det(M)(1 - c\,v^\top M^{-1}v)$ applied to every leading block, or directly from the Schur complement of the bordered matrix. +Revenue from the store's products is $\sum_j p_j \bar s_j (p_j/\Phi(p))^{-\bar\eta_d} = \Phi(p)^{\bar\eta_d}\sum_j \bar s_j p_j^{1-\bar\eta_d} = \Phi(p) = e^{\ell(X)}$, and costs are $\frac{\bar\eta_D-1}{\bar\eta_D}e^{\bar\eta_d\ell(X)}\sum_j\bar s_j e^{-\bar\eta_d X_j}$. -For part 2, using $\bar s^\top\mathbb{1} = 1$ in the third line of {eq}`mfg_kimball_Q`, +Dividing by $\Phi(\tilde p) = e^{\ell(X)}$ gives actual profit. -$$ -(Q+\Theta_X)\mathbb{1} -= (\bar\eta_D-1)\left[(\bar\eta_D-\bar\eta_d)\bar s + \bar\eta_d \bar s\right] -= (\bar\eta_D-1)\bar\eta_D\,\bar s . -$$ +For part 2, let $g(X) = \sum_j\bar s_j e^{-\bar\eta_d X_j}$, so that $\log g$ has gradient $-\bar\eta_d\bar s$ and Hessian $\bar\eta_d^2 V$ at zero. -So $(Q+\Theta_X)^{-1}\bar s = \mathbb{1}/[(\bar\eta_D-1)\bar\eta_D]$ and therefore +For any function $f$ with $f(0) = 0$, the Hessian of $e^{f}$ at zero is $\nabla f\nabla f^\top + \nabla^2 f$. -$$ -\bar s^\top(Q+\Theta_X)^{-1}\bar s = \frac{1}{(\bar\eta_D-1)\bar\eta_D} , -$$ +For $e^{(\bar\eta_d - 1)\ell}g$, the logarithm has gradient $(\bar\eta_d - 1)\bar s - \bar\eta_d\bar s = -\bar s$ and Hessian $(\bar\eta_d - 1)(1-\bar\eta_d)V + \bar\eta_d^2 V = (2\bar\eta_d - 1)V$, so its Hessian is $\bar s\bar s^\top + (2\bar\eta_d - 1)V$. -which turns the condition in part 1 into $\bar\eta_D' < (\bar\eta_D-1)\bar\eta_D$. +For $e^{\ell}$, the Hessian is $\bar s\bar s^\top + (1-\bar\eta_d)V$, and for $e^{\bar\eta_d\ell}g$ the logarithm has gradient zero and Hessian $\bar\eta_d(1 - \bar\eta_d)V + \bar\eta_d^2V = \bar\eta_d V$. -For part 3, that inequality says precisely that $\kappa < 1$, and $\kappa/(1+\kappa)$ is increasing in $\kappa$ with value $1/2$ at $\kappa = 1$. +Both profit functions equal $1/\bar\eta_D$ at zero, and dividing the negative Hessians by that value gives the two matrices. -Total static pass-through is $\sum_j \partial x_i^{*}/\partial X_j = \kappa/(1+\kappa)$ because the shares sum to one, so the planner's problem is well behaved exactly when a store would pass less than half of an industry-wide price increase into its own prices. +For part 3, we write the profit function and a finite-difference Hessian. ```{code-cell} ipython3 -lo, hi = 0.0, 100.0 -for _ in range(60): - mid = (lo + hi)/2 - Q_m, Θ_m = kimball(η_d, η_D, mid, s_bar) - if np.linalg.eigvalsh(Q_m + 2*Θ_m).min() > 0: - lo = mid - else: - hi = mid - -κ_star = lo/((η_D - 1)*η_D) -print(f"threshold by bisection: η_D′ = {lo:.6f}") -print(f"(η_D - 1) η_D = {(η_D - 1)*η_D:.6f}") -print(f"κ at the threshold = {κ_star:.6f}") -print(f"pass-through there = {κ_star/(1 + κ_star):.6f}") -``` +def hessian(f, v, h=1e-4): + "Central finite-difference Hessian of the scalar function f at v." + m = len(v) + E = h*np.eye(m) + hess = np.empty((m, m)) + for i in range(m): + for j in range(m): + hess[i, j] = (f(v + E[i] + E[j]) - f(v + E[i] - E[j]) + - f(v - E[i] + E[j]) + f(v - E[i] - E[j]))/(4*h**2) + return hess + +def store_profit(x, X, η_d, η_D, η_Dp, s, actual=True): + """ + Profit of a store charging log prices x when all other stores charge X, + with flexible prices normalized to one. With actual=False the factor + 𝒴/Φ(p̃), which does not depend on x, is dropped. + """ + p, p_others = np.exp(x), np.exp(X) + Φ = lambda p: (s @ p**(1 - η_d))**(1/(1 - η_d)) + y = Φ(p)/Φ(p_others) + demand = y**(η_Dp - η_D)*np.exp(-η_Dp*(y - 1)) + mc = (η_D - 1)/η_D + profit = np.sum((p - mc)*s*(p/Φ(p))**(-η_d))*demand + return profit/Φ(p_others) if actual else profit -```{solution-end} +def profit_curvatures(η_d, η_D, η_Dp, s, actual=True): + "Q, Θ_X and Λ_X from the Hessian of store profits at the steady state." + n = len(s) + f = lambda v: store_profit(v[:n], v[n:], η_d, η_D, η_Dp, s, actual) + v0 = np.zeros(2*n) + hess = -hessian(f, v0)/f(v0) + return hess[:n, :n], hess[:n, n:], hess[n:, n:] ``` -```{exercise} -:label: mfg_ex8 - -The closed-form eigenvalues in the Kimball example used $A = \pi I$. +Now we compare the finite differences with the closed forms. -Replace it by $A = \operatorname{diag}(0.00, 0.02, 0.06)$, so that the three products face different rates of cost inflation. +```{code-cell} ipython3 +M_norm = (η_d*(2*η_D - 1) - η_D)*V - η_D*ss_bar -1. Check that the closed-form formula fails. -1. Check that the trace comparative static of the *Persistence* section still holds, by scaling $\Theta_X$ and recording the trace and the eigenvalues of the closed-loop matrix. -1. Does the invariance of aggregate dynamics to the superelasticity survive? +for actual, M_closed in ((True, M_act), (False, M_norm)): + label = "actual" if actual else "factor dropped" + for η_Dp in (-3.0, 3.0, 10.0): + Q_f, Θ_f, Λ_f = profit_curvatures(η_d, η_D, η_Dp, s_bar, actual) + Q_K, Θ_K = kimball(η_d, η_D, η_Dp, s_bar) + err_QΘ = max(np.abs(Q_f - Q_K).max(), np.abs(Θ_f - Θ_K).max()) + err_M = np.abs(Q_f + 2*Θ_f + Λ_f - M_closed).max() + print(f"{label:>15}, η_D′ = {η_Dp:>5}: error in Q and Θ_X {err_QΘ:.0e}, " + f"error in Q + 2Θ_X + Λ_X {err_M:.0e}") ``` -```{solution-start} mfg_ex8 -:class: dropdown -``` +The formulas hold for every superelasticity, and neither planner matrix depends on it. -```{code-cell} ipython3 -A_het = np.diag([0.00, 0.02, 0.06]) -Q_h, Θ_h = kimball(η_d, η_D, 3.0, s_bar) +Dropping the factor leaves $Q$ and $\Theta_X$ unchanged, because the store's first-order condition holds at the steady state, but it changes $\Lambda_X$. -P_h = mfg_riccati(A_het, B_K, Q_h + Θ_h, Γ_K, ρ_K) -JG_h = closed_loop(A_het, B_K, Γ_K, P_h) +For part 4, $v^\top V v = \sum_j\bar s_j v_j^2 - (\bar s^\top v)^2 \geq 0$, with equality only when $v$ is proportional to $\mathbb{1}$, and then $v^\top\bar s\bar s^\top v > 0$. -ω_h = np.linalg.eigvals(np.linalg.solve(Γ_K, Q_h + Θ_h)).real -print("closed-loop eigenvalues:", np.sort(np.linalg.eigvals(JG_h).real).round(5)) -print("closed-form formula: ", - np.sort(ρ_K/2 - np.sqrt((ρ_K/2 + np.diag(A_het))**2 + np.sort(ω_h))).round(5)) -``` +So the actual matrix is positive definite whenever $2\bar\eta_d - 1 > 0$. + +With household spending fixed, an industry-wide price increase lowers quantities, so joint profit is concave in the level of prices. -The formula no longer holds: with $A$ not a multiple of the identity there is no single scalar shift to put inside the square root, and the eigenvectors of $A$ and of $Q+\Theta_X$ need not agree. +For the other matrix, $V\mathbb{1} = 0$ and $\bar s^\top\mathbb{1} = 1$ give $\mathbb{1}^\top M\mathbb{1} = -\bar\eta_D < 0$. -The errors are modest here because the three inflation rates are close together, but they are systematic, and they grow with the dispersion in $A$. +Dropping the factor holds quantities fixed, so raising all prices in proportion raises profit at an increasing rate. -The trace result, on the other hand, requires no such restriction. +The planner's problem has $\Gamma + 2\Theta_{\mathcal A} + \Lambda_{\mathcal A} = \Gamma$, so its version of $E$ is $M + (\pi + \rho/2)^2\Gamma$. ```{code-cell} ipython3 -print(f"{'scale on Θ_X':>14}{'trace':>10} eigenvalues") -for scale in (0.0, 0.5, 1.0, 1.5): - P_s = mfg_riccati(A_het, B_K, Q_h + scale*Θ_h, Γ_K, ρ_K) - JG_s = closed_loop(A_het, B_K, Γ_K, P_s) - print(f"{scale:>14}{np.trace(JG_s):>10.5f} " - f"{np.sort(np.linalg.eigvals(JG_s).real).round(5)}") +for label, M_closed in (('actual', M_act), ('factor dropped', M_norm)): + E_pl = M_closed + (π_K + ρ_K/2)**2*Γ_K + print(f"{label:>15}: eigenvalues of E {np.linalg.eigvalsh(E_pl).round(3)}") + try: + mfg_riccati(A_K, B_K, M_closed, Γ_K, ρ_K) + print(" "*17 + "the planner's Riccati equation has an admissible solution") + except ValueError as e: + print(" "*17 + str(e)) ``` -Raising the scale strengthens complementarity and raises the trace, exactly as in the two-dimensional example earlier. +With the factor dropped, $E$ has a negative eigenvalue, so by {prf:ref}`mfg_prop_existence` no symmetric solution exists, the planner's problem has no finite optimum, and our solver duly raises an error. + +For part 5, the equilibrium's curvature in the direction $\mathbb{1}$ is $(\bar\eta_D - 1)\bar\eta_D$, against $\bar\eta_D - 1$ for the planner, smaller by the factor $\bar\eta_D$. + +A single store that raises its prices loses customers to other stores at elasticity $\bar\eta_D$, while stores that raise their prices together keep their revenue and only save on costs. -For part 3, the invariance has nothing to do with $A$: it comes from the cancellation of $\bar\eta_D'$ in $Q + \Theta_X$, and {prf:ref}`mfg_prop_riccati` shows that only $Q+\Theta_X$ and $\Gamma+\Theta_{\mathcal A}$ enter the equilibrium Riccati equation. +In directions with $\bar s^\top v = 0$, which leave the store's price index unchanged, the equilibrium's curvature is $(\bar\eta_D - 1)\bar\eta_d\, v^\top\operatorname{diag}(\bar s)v$ and the planner's is $(\bar\eta_D - 1)(2\bar\eta_d - 1)\, v^\top\operatorname{diag}(\bar s)v$, which is larger because $\bar\eta_d > 1$. ```{code-cell} ipython3 -for η_Dp in (-3.0, 0.0, 3.0, 10.0): - Q_i, Θ_i = kimball(η_d, η_D, η_Dp, s_bar) - P_i = mfg_riccati(A_het, B_K, Q_i + Θ_i, Γ_K, ρ_K) - ev = np.sort(np.linalg.eigvals(closed_loop(A_het, B_K, Γ_K, P_i)).real) - print(f"η_D′ = {η_Dp:>5}: {ev.round(6)}") +ones = np.ones(n_K) +v = np.array([1.0, -1.0, 0.5]) +v = v - (s_bar @ v)*ones # now s̄'v = 0 +Q_K, Θ_K = kimball(η_d, η_D, 3.0, s_bar) + +for name, M_use in (('equilibrium', Q_K + Θ_K), ('planner', M_act)): + ratio_v = (v @ M_use @ v)/(v @ np.diag(s_bar) @ v) + print(f"{name:>12}: 1'M1 = {ones @ M_use @ ones:7.3f}, " + f"v'Mv / v'diag(s̄)v = {ratio_v:7.3f}") ``` +This is why the planner adjusts its price level more slowly than the industry does and its relative prices more quickly. + ```{solution-end} ``` - ## Further reading -{cite:t}`AlvarezArgente2026` develop the results in this lecture, along with the two economic examples that we implemented above. +{cite:t}`AlvarezArgente2026` develop the results in this lecture, along with the two economic examples; their appendices contain the proofs and the derivations behind both examples, and they discuss the planner in the pricing example. + +The planner in the capital example, the analysis of market power with structural parameters held fixed, the finite-difference check of the pricing planner, the observation that with their profit normalization that planner has no finite optimum, and the simulated panels are our own illustrations of their results. -They also extend the analysis to interactions through higher moments of the cross-sectional distribution. +{doc}`lq_mean_field_games_extensions` continues with cross terms between states and actions, forcing processes, and interactions through higher moments. {cite:t}`CarmonaDelarue2018` give a comprehensive probabilistic treatment of mean field games, and {cite:t}`AchdouEtAl2022` describe the numerical methods used to solve the coupled partial differential equations when the model is not linear quadratic. diff --git a/lectures/lq_mean_field_games_extensions.md b/lectures/lq_mean_field_games_extensions.md new file mode 100644 index 000000000..415394200 --- /dev/null +++ b/lectures/lq_mean_field_games_extensions.md @@ -0,0 +1,959 @@ +--- +jupytext: + text_representation: + extension: .md + format_name: myst + format_version: 0.13 +kernelspec: + display_name: Python 3 + language: python + name: python3 +--- + +(lq_mean_field_games_extensions)= +```{raw} jupyter +
+ + QuantEcon + +
+``` + +# Linear Quadratic Mean Field Games: Extensions + +```{contents} Contents +:depth: 2 +``` + +## Overview + +This lecture continues {doc}`lq_mean_field_games`, and it assumes familiarity with that lecture's theory, through the split into individual and aggregate dynamics, and with its capital accumulation example. + +There we studied a continuum of agents whose quadratic payoffs depend on the cross-sectional averages of states and actions, and found that equilibrium aggregate dynamics solve the Riccati equation of a single-agent problem in which strategic interaction changes only the curvatures. + +Here we take up three extensions developed by {cite:t}`AlvarezArgente2026`: + +* **Cross terms between states and actions.** The equivalence result survives when the mixed interactions between states and actions satisfy a symmetry restriction, with the mixed interaction subtracted from the own cross term, and uniqueness can fail when they do not. +* **Forcing processes.** States that agents cannot control, such as costs, can be added without changing the Riccati equation of the controlled block, and a pricing example shows that firms pass idiosyncratic and common cost shocks through at different speeds and to different extents. +* **Higher moments.** When payoffs depend on higher moments of the cross-sectional distribution, idiosyncratic risk affects aggregate persistence. + +Let's start with some imports: + +```{code-cell} ipython3 +import numpy as np +import matplotlib.pyplot as plt +from scipy.linalg import solve_continuous_are, expm +``` + +## What we need from the previous lecture + +Throughout, lower-case letters denote an agent's own state $x$ and action $\alpha$, and upper-case or calligraphic letters denote their cross-sectional averages $X$ and $\mathcal A$. + +In {doc}`lq_mean_field_games` an agent has period return and state equation + +$$ +-\tfrac12 x^\top Qx - X^\top\Theta_X x - \tfrac12\alpha^\top\Gamma\alpha - \mathcal A^\top\Theta_{\mathcal A}\alpha , +\qquad +dx = (B\alpha - Ax)dt + \Sigma^{1/2}dW , +$$ + +and discounts at rate $\rho$. + +Given the aggregate paths, her value function is $u(x,t) = \beta_0(t) + \beta_1(t)^\top x + \tfrac12 x^\top\bar\beta_2 x$, where $\bar\beta_2$ solves the single-agent Riccati equation and involves neither interaction matrix. + +In equilibrium the aggregate state follows $\dot X = \left(B(\Gamma+\Theta_{\mathcal A})^{-1}B^\top P - A\right)X$, where the symmetric matrix $P$ solves + +$$ +P B(\Gamma+\Theta_{\mathcal A})^{-1}B^\top P = Q + \Theta_X + \rho P + PA + A^\top P , +$$ + +and the eigenvalues of the closed-loop matrix must have real parts below $\rho/2$. + +That equation comes from a state-costate system whose eigenvalues come in pairs $\lambda$ and $\rho - \lambda$, so at most half of them lie below $\rho/2$; with $B(\Gamma + \Theta_{\mathcal A})^{-1}B^\top$ positive definite, that is what makes the equilibrium unique. + +An agent's deviation from the cross-sectional mean, $z = x - X$, follows $dz = (B\Gamma^{-1}B^\top\bar\beta_2 - A)z\,dt + \Sigma^{1/2}dW$, in which no interaction matrix appears. + +So data on deviations from the mean contain no information about strategic interaction, which is the missing intercept. + +Shocks common to all agents, modeled with a compensated jump process $\mathcal J$, enter the law of motion of $X$ but leave $P$ unchanged. + +We reuse two tools: the closed form for the aggregate eigenvalue in the scalar case, + +$$ +\lambda = \frac\rho2 - \sqrt{\left(\frac\rho2 + a\right)^2 + \frac{b^2(q + \theta_X)}{\gamma + \theta_{\mathcal A}}} , +$$ + +and the calibration of the capital accumulation example. + +In that example a firm with capital $k$ earns $\Pi(k, K) = k^{\nu(1-1/\eta)}K^{\nu/\eta}$ when the rest of the industry holds $K$, pays an adjustment cost $\psi(i) = \frac\phi2\bar i\left(\frac{i - \bar i}{\bar i}\right)^2$, and buys investment goods at price $\mathcal P(I) = (I/\bar i)^{1/\varepsilon_s}$. + +In percentage deviations from the steady state it maps into the scalar model with $a = b = \delta$, $q + \theta_X = \frac{\eta-1}{\eta}\nu(1-\nu)$, $\gamma = \phi\,\delta\bar k^{1-\nu}$ and $\theta_{\mathcal A} = \delta\bar k^{1-\nu}/\varepsilon_s$, and the calibration gives half-lives of three years for a firm's deviation from the industry mean and five years for the industry. + +```{code-cell} ipython3 +def scalar_λ(ρ, a, b, q, γ, θ_X, θ_A): + "Closed-form aggregate eigenvalue in the scalar case." + return ρ/2 - np.sqrt((ρ/2 + a)**2 + b**2*(q + θ_X)/(γ + θ_A)) + +def half_life(λ): + "Time for a state with eigenvalue λ < 0 to close half of a gap." + return np.log(2)/(-λ) + +def cap_q(η, ν): + "Own-state curvature in the capital accumulation example." + return ν*(η - 1)/η**2*(η*(1 - ν) + ν) + +def cap_θ_X(η, ν): + "State interaction in the capital accumulation example." + return -(η - 1)/η*ν**2/η + +ρ_c, δ_c, η_c, ν_c = 0.05, 0.10, 4.0, 0.7 +φ_c, ε_s_c = 1/7, 35/9 + +def cap_scale(η, ν, ρ=ρ_c, δ=δ_c): + "δ k̄^{1-ν}, which converts φ and 1/ε_s into γ and θ_A." + return δ*ν*(η - 1)/(η*(ρ + δ)) + +γ_c = φ_c*cap_scale(η_c, ν_c) +θ_A_c = cap_scale(η_c, ν_c)/ε_s_c +q_c, θ_X_c = cap_q(η_c, ν_c), cap_θ_X(η_c, ν_c) + +λ_c = scalar_λ(ρ_c, δ_c, δ_c, q_c, γ_c, θ_X_c, θ_A_c) # industry +λ_micro = scalar_λ(ρ_c, δ_c, δ_c, q_c, γ_c, 0.0, 0.0) # single firm +print(f"γ = {γ_c:.3f}, θ_A = {θ_A_c:.3f}") +print(f"half-lives: firm {half_life(λ_micro):.2f} years, industry {half_life(λ_c):.2f} years") +``` + +(mfgx_cross_terms)= +## State-action cross terms + +The return in {doc}`lq_mean_field_games` was not the most general quadratic return. + +A second-order approximation to a general period return produces + +```{math} +:label: mfgx_return_general +\begin{aligned} +\tilde F(x, X, \alpha, \mathcal A) += &-\tfrac12 x^\top Q x - X^\top\Theta_X x + -\tfrac12 \alpha^\top \Gamma \alpha - \mathcal A^\top \Theta_{\mathcal A}\alpha \\ + &+ x^\top C\alpha - X^\top\Theta_{X\alpha}\alpha - x^\top\Theta_{x\mathcal A}\mathcal A \\ + &-\tfrac12 X^\top\Lambda_X X - \tfrac12\mathcal A^\top\Lambda_{\mathcal A}\mathcal A + - X^\top\Lambda_{X\mathcal A}\mathcal A . +\end{aligned} +``` + +The previous lecture used the first line throughout, and the terms in the aggregates alone in the third line when it studied the planner. + +Here we restore the second line. + +The own cross term $C$ links an agent's state and action. + +It appears when adjustment costs depend on the state as well as on the action, as with investment costs that depend on the investment *rate* {cite}`Hayashi1982,CooperHaltiwanger2006`. + +The mixed interactions appear when the average state affects the marginal return to an agent's action ($\Theta_{X\alpha}$), or the average action affects the marginal return to her state ($\Theta_{x\mathcal A}$). + +### The individual problem + +Taking the aggregate paths as given, the agent's first-order condition is now + +$$ +\alpha^{*}(x,t) = \Gamma^{-1}\left[B^\top u_x(x,t) + C^\top x - \Theta_{\mathcal A}\mathcal A(t) - \Theta_{X\alpha}^\top X(t)\right] , +$$ + +and the curvature of her value function solves + +```{math} +:label: mfgx_beta2_cross +(\bar\beta_2 B + C)\Gamma^{-1}(B^\top\bar\beta_2 + C^\top) = Q + \rho\bar\beta_2 + \bar\beta_2 A + A^\top\bar\beta_2 . +``` + +The own cross term enters the individual's curvature, but, as before, no interaction matrix does. + +Completing the square in the action, + +$$ +-\tfrac12 x^\top Qx - \tfrac12\alpha^\top\Gamma\alpha + x^\top C\alpha += -\tfrac12 x^\top(Q - C\Gamma^{-1}C^\top)x + - \tfrac12(\alpha - \Gamma^{-1}C^\top x)^\top\Gamma(\alpha - \Gamma^{-1}C^\top x), +$$ + +shows that the counterpart of $Q \succ 0$ is $Q - C\Gamma^{-1}C^\top \succ 0$, which we assume. + +### The equilibrium + +Aggregating the first-order condition gives + +$$ +(\Gamma + \Theta_{\mathcal A})\mathcal A = B^\top(\beta_1 + \bar\beta_2 X) + (C - \Theta_{X\alpha})^\top X . +$$ + +It is convenient to track $w \equiv \beta_1 + \bar\beta_2 X$, the marginal value of the state for an agent at the mean. + +Differentiating, aggregating, and using {eq}`mfgx_beta2_cross` gives the state-costate system + +```{math} +:label: mfgx_cross_system +\begin{aligned} +\dot w &= (\rho I + A^\top) w + (Q + \Theta_X) X - (C - \Theta_{x\mathcal A})\mathcal A , \\ +\dot X &= B\mathcal A - AX . +\end{aligned} +``` + +The costate equation involves $\Theta_{x\mathcal A}$, through which the average action shifts the marginal value of an agent's state. + +The aggregate action involves $\Theta_{X\alpha}$, through which the average state shifts the marginal return to her action. + +The two need not be equal, but when they are, the system keeps the Hamiltonian structure of the benchmark. + +```{prf:proposition} +:label: mfgx_prop_cross + +Suppose $\Theta_{X\alpha} = \Theta_{x\mathcal A}$, and write $\tilde C \equiv C - \Theta_{X\alpha}$. + +An equilibrium is characterized by a symmetric matrix $P$ solving + +$$ +(PB + \tilde C)(\Gamma + \Theta_{\mathcal A})^{-1}(B^\top P + \tilde C^\top) = Q + \Theta_X + \rho P + PA + A^\top P , +$$ + +with aggregate dynamics + +$$ +\dot X = \left[B(\Gamma+\Theta_{\mathcal A})^{-1}(B^\top P + \tilde C^\top) - A\right] X . +$$ + +This is the Riccati equation of a single agent whose return is + +$$ +-\tfrac12 X^\top(Q + \Theta_X)X - \tfrac12\mathcal A^\top(\Gamma + \Theta_{\mathcal A})\mathcal A + X^\top \tilde C \mathcal A . +$$ + +If $\Gamma + \Theta_{\mathcal A} \succ 0$ and $Q + \Theta_X - \tilde C(\Gamma + \Theta_{\mathcal A})^{-1}\tilde C^\top \succ 0$, an equilibrium exists and is unique. +``` + +So the equivalence result of {doc}`lq_mean_field_games` survives, with one more adjustment: the mixed interaction shifts the effective cross term from $C$ to $C - \Theta_{X\alpha}$. + +Completing the square in the effective problem shows that a cross term does two things. + +It lowers the effective state curvature to $Q + \Theta_X - \tilde C(\Gamma+\Theta_{\mathcal A})^{-1}\tilde C^\top$, and it changes the effective drift matrix from $A$ to $A - B(\Gamma+\Theta_{\mathcal A})^{-1}\tilde C^\top$. + +The comparative statics of the previous lecture apply to these transformed objects rather than to the interaction matrices themselves. + +The dynamics of an agent's deviation from the mean become + +$$ +dz = \left[B\Gamma^{-1}(B^\top\bar\beta_2 + C^\top) - A\right] z \, dt + \Sigma^{1/2} dW , +$$ + +which involve the own cross term $C$ but still no interaction matrix, so data on deviations from the mean still contain no information about strategic interaction. + +The planner is handled in the same way. + +Its effective cross term is $C - \Theta_{X\alpha} - \Theta_{x\mathcal A} - \Lambda_{X\mathcal A}$, so it needs no restriction on the mixed interactions, which enter only through their sum. + +To compute equilibria with cross terms, we extend the solver of {doc}`lq_mean_field_games`. + +The `s` argument of `solve_continuous_are` replaces $\tilde P B$ by $\tilde P B + s$ in the quadratic term of the standard Riccati equation, so setting $s = -\tilde C$ handles an effective cross term $\tilde C$. + +As in the previous lecture, the function checks the Riccati residual and the eigenvalue condition, and raises an error if either fails. + +```{code-cell} ipython3 +def closed_loop(A, B, Γ_eff, P, C=None): + "The matrix governing aggregate dynamics, Ẋ = (B Γ_eff^{-1} (B' P + C') - A) X." + K = B.T @ P if C is None else B.T @ P + C.T + return B @ np.linalg.solve(Γ_eff, K) - A + +def mfg_riccati(A, B, Q, Γ, ρ, C=None, tol=1e-8): + """ + Solve (P B + C) Γ^{-1} (B' P + C') = Q + ρ P + P A + A' P, with C = 0 by + default, for the solution P whose closed-loop matrix has eigenvalues with + real parts below ρ/2, and raise ValueError if there is none. + """ + n, k = B.shape + C_use = np.zeros((n, k)) if C is None else C + try: + P = -solve_continuous_are(-(A + ρ/2 * np.eye(n)), B, Q, Γ, s=-C_use) + except (np.linalg.LinAlgError, ValueError): + raise ValueError("no admissible solution of the Riccati equation") + quad = (P @ B + C_use) @ np.linalg.solve(Γ, B.T @ P + C_use.T) + residual = quad - (Q + ρ*P + P @ A + A.T @ P) + scale = 1 + np.abs(Q).max() + np.abs(quad).max() + if (np.abs(residual).max() > tol*scale + or np.linalg.eigvals(closed_loop(A, B, Γ, P, C)).real.max() >= ρ/2): + raise ValueError("no admissible solution of the Riccati equation") + return P +``` + +### Investment rates and Hayashi adjustment costs + +In the capital accumulation example the adjustment cost depends on investment alone. + +A common alternative makes it depend on the investment rate, + +$$ +\psi(i, k) = \frac{\phi}{2}\,\delta k\left(\frac{i}{\delta k} - 1\right)^2 , +$$ + +which is homogeneous of degree one, as in {cite:t}`Hayashi1982`, and which reduces to the cost $\psi(i)$ above when $k = \bar k$. + +Since $i - \delta k = \delta\bar k(\alpha - x)$ in percentage deviations, the normalized cost is $\frac{\gamma}{2}(\alpha - x)^2$ to second order, with the same $\gamma = \phi\,\delta\bar k^{1-\nu}$ as before. + +Expanding the square shows that the cost adds $\gamma$ to the own-state curvature and creates an own cross term, + +$$ +q \to q + \gamma , \qquad C = \gamma , +$$ + +while $Q - C\Gamma^{-1}C^\top = q$ is unchanged. + +There are no mixed interactions, so $\tilde C = \gamma$. + +Completing the square in the effective problem of {prf:ref}`mfgx_prop_cross` turns the scalar problem into one without a cross term, with the drift $a = \delta$ replaced by + +$$ +\tilde a = \delta - \frac{\delta\gamma}{\gamma+\theta_{\mathcal A}} = \frac{\delta\,\theta_{\mathcal A}}{\gamma + \theta_{\mathcal A}} +$$ + +and the effective state curvature $q + \theta_X + \gamma - \gamma^2/(\gamma+\theta_{\mathcal A}) = q + \theta_X + \gamma\theta_{\mathcal A}/(\gamma + \theta_{\mathcal A})$. + +So the aggregate eigenvalue is + +$$ +\lambda = \frac\rho2 - \sqrt{\left(\frac\rho2 + \tilde a\right)^2 ++ \frac{\delta^2}{\gamma+\theta_{\mathcal A}}\left(q + \theta_X + \frac{\gamma\,\theta_{\mathcal A}}{\gamma+\theta_{\mathcal A}}\right)} , +$$ + +and setting $\theta_X = \theta_{\mathcal A} = 0$, so that $\tilde a = 0$, gives the individual's. + +With Hayashi costs, investing at the replacement rate is free whatever the capital stock, so depreciation no longer pushes a firm back toward its steady state. + +The drift $\delta$ that sat inside the square root in the formula without cross terms disappears from the individual's problem, and it returns in the industry's problem only through the price of investment goods. + +Let's compare the half-lives with and without Hayashi costs, holding $\gamma$ at its calibrated value. + +```{code-cell} ipython3 +def λ_cross(ρ, a, b, q_eff, γ_eff, c_eff): + "Scalar aggregate eigenvalue with an effective cross term, by completing the square." + a_t = a - b*c_eff/γ_eff + q_t = q_eff - c_eff**2/γ_eff + return ρ/2 - np.sqrt((ρ/2 + a_t)**2 + b**2*q_t/γ_eff) + +λ_ind_H = λ_cross(ρ_c, δ_c, δ_c, q_c + γ_c, γ_c, γ_c) +λ_agg_H = λ_cross(ρ_c, δ_c, δ_c, q_c + γ_c + θ_X_c, γ_c + θ_A_c, γ_c) + +# the general solver agrees +one = np.array([[1.0]]) +P_H = mfg_riccati(δ_c*one, δ_c*one, (q_c + γ_c + θ_X_c)*one, + (γ_c + θ_A_c)*one, ρ_c, C=γ_c*one) +λ_agg_solver = closed_loop(δ_c*one, δ_c*one, (γ_c + θ_A_c)*one, P_H, C=γ_c*one)[0, 0] +print(f"aggregate λ: closed form {λ_agg_H:.6f}, solver {λ_agg_solver:.6f}\n") + +print(f"{'':>20}{'individual':>12}{'industry':>10}{'ratio':>8}") +for label, (λ_i, λ_a) in {'investment costs': (λ_micro, λ_c), + 'Hayashi costs': (λ_ind_H, λ_agg_H)}.items(): + print(f"{label:>20}{half_life(λ_i):>12.2f}{half_life(λ_a):>10.2f}" + f"{half_life(λ_a)/half_life(λ_i):>8.2f}") +``` + +Hayashi costs slow both the individual firm and the industry, and leave the ratio of the two half-lives about where it was. + +{ref}`mfgx_ex_hayashi` recalibrates $\phi$ to restore a three-year individual half-life, and revisits the limit of constant returns to scale. + +(mfgx_restriction)= +### When the restriction fails + +{cite:t}`AlvarezArgente2026` point out that the restriction $\Theta_{X\alpha} = \Theta_{x\mathcal A}$ has no general economic justification. + +Consider a country with capital $k$ and investment $i$ that takes world capital $K$ and world investment $I$ as given. + +Suppose its consumption is $c = f(k, K) - i$ and its period utility is $u(c)$, with $u'' < 0$ and $f_K > 0$. + +Then $\partial^2 u/\partial i\,\partial K = -u''(c) f_K > 0$, because higher world capital raises consumption and so lowers the marginal utility cost of investing. + +But $\partial^2 u/\partial k\,\partial I = 0$, because world investment does not affect the return to capital. + +So the two mixed interactions differ, and in the notation of {eq}`mfgx_return_general`, $\Theta_{X\alpha} < 0 = \Theta_{x\mathcal A}$. + +When they differ, {eq}`mfgx_cross_system` is still a linear state-costate system, but the two outer factors of the Riccati equation are no longer transposes of one another, and the matrix of the system is no longer Hamiltonian. + +Its eigenvalues then need not come in pairs $\lambda$ and $\rho - \lambda$, and the argument that delivered at most one equilibrium in {doc}`lq_mean_field_games` breaks down. + +Even with a single state, the two eigenvalues can both lie below $\rho/2$, giving a continuum of equilibria of the kind studied by {cite:t}`BenhabibFarmer1994`, or both lie above it, so that no equilibrium exists. + +To see the pairing fail, we write a function for the matrix of {eq}`mfgx_cross_system` that allows the two mixed interactions to differ. + +```{code-cell} ipython3 +def state_costate(A, B, Q_eff, Γ_eff, ρ, C1, C2): + """ + Matrix of the state-costate system for (w, X). C1 = C - Θ_xA enters the + costate equation and C2 = C - Θ_Xα enters the aggregate action. + """ + n = A.shape[0] + G_B = np.linalg.solve(Γ_eff, B.T) # (Γ + Θ_A)^{-1} B' + G_C = np.linalg.solve(Γ_eff, C2.T) # (Γ + Θ_A)^{-1} C2' + return np.block([[ρ*np.eye(n) + A.T - C1 @ G_B, Q_eff - C1 @ G_C], + [B @ G_B, B @ G_C - A]]) +``` + +We apply it to the Hayashi economy, first as it is and then with a small mixed interaction of the open-economy sign, $\Theta_{X\alpha} = -0.05$ and $\Theta_{x\mathcal A} = 0$. + +```{code-cell} ipython3 +for label, θ_Xα, θ_xA in (('restriction holds', 0.0, 0.0), + ('restriction fails', -0.05, 0.0)): + M_sc = state_costate(δ_c*one, δ_c*one, (q_c + γ_c + θ_X_c)*one, (γ_c + θ_A_c)*one, + ρ_c, (γ_c - θ_xA)*one, (γ_c - θ_Xα)*one) + ev = np.sort(np.linalg.eigvals(M_sc).real) + print(f"{label}: eigenvalues {ev.round(4)}, sum {ev.sum():.4f}, ρ = {ρ_c}") +``` + +When the restriction holds, the eigenvalues sum to $\rho$, so they are symmetric about $\rho/2$. + +When it fails, they sum to $\rho + b(\theta_{x\mathcal A} - \theta_{X\alpha})/(\gamma + \theta_{\mathcal A})$, and the open-economy sign pushes them up, toward the case in which no equilibrium exists. + +{ref}`mfgx_ex_restriction` finds parameters at which both eigenvalues lie on the same side of $\rho/2$. + +The planner is immune to all of this, because it internalizes both mixed interactions and so sees only their sum. + +(mfgx_forcing)= +## Forcing processes and pass-through + +The same machinery handles states that agents cannot control, such as a marginal cost. + +Partition the state as $x = (x^{(1)}, x^{(2)})$, where $x^{(1)}$ is controlled and $x^{(2)}$ is a forcing process with its own law of motion, + +$$ +A = \begin{bmatrix} A_{11} & A_{12} \\ 0 & A_{22}\end{bmatrix}, +\qquad +B = \begin{bmatrix} B_1 \\ 0 \end{bmatrix} . +$$ + +Now $B$ no longer has full row rank and $B\Gamma^{-1}B^\top$ is singular, so the Riccati equations are not of the standard full-rank kind, but they are block recursive. + +Following {cite:t}`AlvarezArgente2026`, we keep interactions through states only, $\Theta_{\mathcal A} = 0$. + +```{prf:proposition} +:label: mfgx_prop_forcing + +Suppose $\Theta_{\mathcal A} = 0$, that $B_1\Gamma^{-1}B_1^\top$, $Q_{11}$ and $Q_{11} + \Theta_{X,11}$ are positive definite, and that the eigenvalues of $-A_{22}$ have real parts below $\rho/2$. + +Then the upper-left block of $P$ solves the equilibrium Riccati equation of the controlled block alone, + +$$ +P_{11}B_1\Gamma^{-1}B_1^\top P_{11} = Q_{11} + \Theta_{X,11} + \rho P_{11} + P_{11}A_{11} + A_{11}^\top P_{11} , +$$ + +the remaining blocks solve linear equations, and the eigenvalues of the aggregate closed-loop matrix are those of $B_1\Gamma^{-1}B_1^\top P_{11} - A_{11}$ together with those of $-A_{22}$. +``` + +The blocks of $Q$ and $\Theta_X$ that involve only the forcing process affect the value of the problem but not the policy, so they can be chosen freely to keep both matrices positive definite. + +Here is a pricing example in the spirit of the real rigidities of {cite:t}`BallRomer1990`. + +A store's log price $x^{(1)}$ has a target that puts weight $1-\chi$ on its own log marginal cost $x^{(2)}$ and weight $\chi \in [0, 1)$ on the average price $X^{(1)}$, and the store pays a Rotemberg cost to change its price: + +$$ +-\tfrac q2\left(x^{(1)} - (1-\chi)x^{(2)} - \chi X^{(1)}\right)^2 - \tfrac\gamma2\alpha^2, +\qquad dx^{(1)} = \alpha \, dt . +$$ + +The weight $\chi$ is the static pass-through of competitors' prices, so it measures strategic complementarity, and $\Theta_{X,11} = -q\chi$. + +Terms that pair the forcing process with the average price affect neither the policy nor the dynamics, so we drop them. + +Costs follow $dx^{(2)} = -a_{22}x^{(2)}dt + \sigma dW + \upsilon \, d\mathcal J$, where $W$ is idiosyncratic and $\mathcal J$ is a common jump process. + +An idiosyncratic cost shock moves a store away from the mean, so the single-agent matrix $\bar\beta_2$ governs its response, while a common cost shock moves the industry, which $P$ governs. + +Let's trace both responses to a permanent one percent increase in costs, $a_{22} = 0$. + +```{code-cell} ipython3 +def pricing_with_costs(χ, a_22, ρ=0.04, q=1.0, γ=1.0): + "Aggregate and deviation drift matrices in the pricing example with cost shocks." + A_f = np.array([[0.0, 0.0], + [0.0, a_22]]) + B_f = np.array([[1.0], + [0.0]]) + Γ_f = np.array([[γ]]) + Q_f = q*np.array([[1.0, -(1 - χ)], + [-(1 - χ), 2.0]]) # (2, 2) element chosen freely + Θ_f = np.array([[-q*χ, 0.0], + [0.0, 0.0]]) + β2_f = mfg_riccati(A_f, B_f, Q_f, Γ_f, ρ) + P_f = mfg_riccati(A_f, B_f, Q_f + Θ_f, Γ_f, ρ) + return closed_loop(A_f, B_f, Γ_f, P_f), closed_loop(A_f, B_f, Γ_f, β2_f) +``` + +The first matrix moves the industry's average price after a common cost shock, and the second moves a store's price relative to the average after an idiosyncratic one. + +```{code-cell} ipython3 +--- +mystnb: + figure: + caption: Common and idiosyncratic cost pass-through + name: fig-mfgx-pass-through +--- +t_f = np.linspace(0, 6, 200) +cost_shock = np.array([0.0, 1.0]) + +fig, axes = plt.subplots(1, 2, figsize=(11, 4.2)) +for χ in (0.0, 0.5, 0.8): + U_agg, U_dev = pricing_with_costs(χ, 0.0) + agg = [(expm(U_agg*t) @ cost_shock)[0] for t in t_f] + dev = [(expm(U_dev*t) @ cost_shock)[0] for t in t_f] + axes[0].plot(t_f, agg, lw=2, label=f"$\\chi = {χ}$") + axes[1].plot(t_f, dev, lw=2, label=f"$\\chi = {χ}$") + +axes[0].set_title('average price, common cost shock') +axes[1].set_title("store's relative price, idiosyncratic cost shock") +for ax in axes: + ax.set_xlabel('years') + ax.set_ylabel('percent') + ax.legend() +plt.tight_layout() +plt.show() +``` + +A store passes an idiosyncratic cost increase through quickly but only partially, because its competitors' prices do not move: in the long run it passes through $1 - \chi$ of it. + +The industry passes an aggregate cost increase through completely, but slowly, because each store waits for the others. + +Let's compare the half-lives. + +```{code-cell} ipython3 +for χ in (0.0, 0.5, 0.8): + U_agg, U_dev = pricing_with_costs(χ, 0.0) + λ_agg_f, λ_dev_f = U_agg[0, 0], U_dev[0, 0] + print(f"χ = {χ}: store half-life {half_life(λ_dev_f):.2f} years, " + f"industry half-life {half_life(λ_agg_f):.2f} years") +``` + +With $\chi = 0.8$ a store closes half of the gap to its new target in about 0.7 years, while the industry takes about 1.6 years. + +The store's half-life is the same for every $\chi$, because a store's deviation from the average is governed by its own curvature $q$ alone. + +So how fast stores respond to their own cost shocks reveals nothing about $\chi$, which is the missing intercept of {doc}`lq_mean_field_games` in another guise. + +How far they respond is a different matter: in the long run a store passes through $1 - \chi$ of its own cost shock, because $\chi$ also enters the store's own return, through the weight on its cost. + +When a model ties the own curvature $Q$ to the interaction $\Theta_X$ in this way, micro data can be informative after all; the result of the previous lecture is that, for given $Q$, they are not. + +{ref}`mfgx_ex_forcing` verifies {prf:ref}`mfgx_prop_forcing` in this example and studies transitory cost shocks. + +(mfgx_higher_moments)= +## Coupling through higher moments + +So far agents have cared about the distribution of states only through its mean. + +{cite:t}`AlvarezArgente2026` also let payoffs depend on higher moments. + +Take a scalar state and a scalar action, set $\theta_{\mathcal A} = 0$, and write $M_r(t) = \int x^r m(x,t) dx$ for the $r$-th moment of the cross-sectional distribution. + +The period return is + +$$ +-\tfrac12 q x^2 - \left(\theta_0 + \sum_{r=1}^{\bar r}\theta_r M_r\right) x - \tfrac12\gamma\alpha^2 , +$$ + +the state follows $dx = (b\alpha - ax)dt + \sigma dW$, and the benchmark of {doc}`lq_mean_field_games` is the case $\bar r = 1$ and $\theta_0 = 0$. + +A natural notion of strategic complementarity is that a stochastically higher distribution raises the marginal return to $x$, which here means that $-\sum_r \theta_r x^r$ is nondecreasing in $x$. + +Our example below, with $\theta_1$ and $\theta_3$ negative and the other coefficients zero, satisfies it. + +The value function is still quadratic, its curvature $\bar\beta_2$ is still that of the single-agent problem, and the individual eigenvalue + +$$ +\lambda_{\text{ind}} \equiv \frac{b^2}{\gamma}\bar\beta_2 - a = \frac\rho2 - \sqrt{\left(a + \frac\rho2\right)^2 + \frac{b^2 q}{\gamma}} +$$ + +is unchanged. + +What changes is aggregation. + +The Kolmogorov forward equation implies + +$$ +\dot M_r = r\lambda_{\text{ind}} M_r + r\frac{b^2}{\gamma}\beta_1 M_{r-1} + \frac{r(r-1)\sigma^2}{2}M_{r-2} , +\qquad +\dot\beta_1 = (\rho - \lambda_{\text{ind}})\beta_1 + \theta_0 + \sum_{r=1}^{\bar r}\theta_r M_r , +$$ + +so the moments obey a finite system, but a nonlinear one, because of the products $\beta_1 M_{r-1}$. + +When $\sigma > 0$, $\lambda_{\text{ind}} < 0$, and $\theta_0 = -\sum_{r\ \mathrm{even}}\theta_r \bar M_r$, there is a steady state with $\beta_1 = 0$ in which the invariant distribution is normal with mean zero and variance $v = \sigma^2/(-2\lambda_{\text{ind}})$. + +Its odd moments vanish and its even moments are $\bar M_r = (r-1)!!\, v^{r/2}$; as $\sigma \to 0$ it collapses to a point mass at zero, which we include below as a limiting case. + +Linearizing the system around this steady state gives $\dot y = \tilde{\mathcal H} y$ for $y = (\hat\beta_1, \hat M_1, \ldots, \hat M_{\bar r})$, where hats denote deviations from the steady state. + +```{prf:proposition} +:label: mfgx_prop_moments + +Let $\bar r$ be even and define + +$$ +\bar\theta \equiv \sum_{r\ \mathrm{odd}} r\,\bar M_{r-1}\,\theta_r . +$$ + +The eigenvalues of $\tilde{\mathcal H}$ are $\tilde\lambda_r = r\lambda_{\text{ind}}$ for $r = 2, \ldots, \bar r$, together with + +$$ +\tilde\lambda_1 = \frac\rho2 - \sqrt{\left(a + \frac\rho2\right)^2 + \frac{b^2(q + \bar\theta)}{\gamma}}, +\qquad +\tilde\lambda_{\bar r+1} = \frac\rho2 + \sqrt{\left(a + \frac\rho2\right)^2 + \frac{b^2(q + \bar\theta)}{\gamma}} . +$$ + +If $q + \bar\theta + \gamma a(a+\rho)/b^2 > 0$, the steady state is locally determinate, with one unstable and $\bar r$ stable eigenvalues. +``` + +Three features stand out. + +First, the interaction coefficients enter only through $\bar\theta$, and only those on odd moments enter it, so $\bar\theta$ plays exactly the role of $\theta_X$ in the scalar formula of {doc}`lq_mean_field_games`. + +Second, $\bar\theta$ weights $\theta_3, \theta_5, \ldots$ by even moments of the invariant distribution, which grow with $\sigma^2$. + +In the benchmark, the variance of idiosyncratic shocks never enters the Riccati equations, so it has no effect on aggregate dynamics; with higher moments, idiosyncratic risk affects aggregate persistence. + +Third, coefficients on even moments do not affect the local eigenvalues, although they can affect transitional dynamics, as {ref}`mfgx_ex_moments` shows. + +Let's build $\tilde{\mathcal H}$ and check {prf:ref}`mfgx_prop_moments`, using $\bar r = 4$ and interactions through the first and third moments. + +```{code-cell} ipython3 +def moment_system(ρ, a, b, q, γ, σ, θ): + """ + Linearized matrix for (β̂₁, M̂₁, ..., M̂_r̄) around the zero-mean steady + state, with θ = (θ₁, ..., θ_r̄). Also returns the steady-state moments + M̄₀, ..., M̄_r̄ and the individual eigenvalue. + """ + r_bar = len(θ) + λ_ind = ρ/2 - np.sqrt((a + ρ/2)**2 + b**2*q/γ) + v = σ**2/(-2*λ_ind) # invariant variance + M_bar = np.zeros(r_bar + 1) + M_bar[0] = 1.0 + for r in range(2, r_bar + 1, 2): + M_bar[r] = (r - 1)*v*M_bar[r - 2] + H_tilde = np.zeros((r_bar + 1, r_bar + 1)) + H_tilde[0, 0] = ρ - λ_ind + H_tilde[0, 1:] = θ + for r in range(1, r_bar + 1): + H_tilde[r, 0] = r*M_bar[r - 1]*b**2/γ + H_tilde[r, r] = r*λ_ind + if r >= 3: # M̂₀ is identically zero + H_tilde[r, r - 2] = r*(r - 1)*σ**2/2 + return H_tilde, M_bar, λ_ind + +ρ_m, a_m, b_m, q_m, γ_m = 0.05, 0.3, 1.0, 1.0, 1.0 +θ_m = np.array([-0.3, 0.0, -0.2, 0.0]) # complementarity through M₁ and M₃ + +for σ in (0.0, 0.5, 1.0): + H_m, M_bar, λ_ind = moment_system(ρ_m, a_m, b_m, q_m, γ_m, σ, θ_m) + θ_bar = θ_m[0] + 3*M_bar[2]*θ_m[2] + root = np.sqrt((a_m + ρ_m/2)**2 + b_m**2*(q_m + θ_bar)/γ_m) + predicted = np.r_[ρ_m/2 - root, ρ_m/2 + root, [r*λ_ind for r in range(2, 5)]] + print(f"σ = {σ}: eigenvalues {np.sort(np.linalg.eigvals(H_m).real).round(4)}") + print(f" predicted {np.sort(predicted).round(4)}") +``` + +To trace the equilibrium path we keep the $\bar r$ eigenvalues below $\rho/2$, express $\hat\beta_1$ as a linear function of the moment deviations on the stable subspace, and substitute. + +That gives a law of motion for the moment deviations alone, so an initial condition specifies only those deviations, and $\hat\beta_1(0)$ follows from the saddle path. + +```{code-cell} ipython3 +def moment_dynamics(H_tilde, ρ): + "Law of motion of the moment deviations on the stable subspace of H_tilde." + ev, V = np.linalg.eig(H_tilde) + V_s = V[:, ev.real < ρ/2] + s = np.real(V_s[0] @ np.linalg.inv(V_s[1:])) # β̂₁ = s · M̂ + return H_tilde[1:, 1:] + np.outer(H_tilde[1:, 0], s) +``` + +As an initial condition we shift the invariant distribution to the right by $\Delta$, which to first order raises $M_1$ by $\Delta$ and $M_3$ by $3v\Delta$ and leaves the even moments unchanged. + +The next figure shows the path of the mean for several levels of idiosyncratic volatility. + +```{code-cell} ipython3 +--- +mystnb: + figure: + caption: Idiosyncratic risk and mean persistence + name: fig-mfgx-moments +--- +t_m = np.linspace(0, 6, 200) +Δ = 0.1 + +fig, ax = plt.subplots(figsize=(8, 4.5)) +for σ in (0.0, 0.6, 1.0, 1.3): + H_m, M_bar, λ_ind = moment_system(ρ_m, a_m, b_m, q_m, γ_m, σ, θ_m) + U_m = moment_dynamics(H_m, ρ_m) + M_hat0 = np.array([Δ, 0.0, 3*M_bar[2]*Δ, 0.0]) + path = [(expm(U_m*t) @ M_hat0)[0] for t in t_m] + θ_bar = θ_m[0] + 3*M_bar[2]*θ_m[2] + λ_1 = ρ_m/2 - np.sqrt((a_m + ρ_m/2)**2 + b_m**2*(q_m + θ_bar)/γ_m) + ax.plot(t_m, path, lw=2, + label=f"$\\sigma = {σ}$, slowest half-life {half_life(λ_1):.2f}") +ax.set_xlabel('$t$') +ax.set_ylabel('cross-sectional mean $M_1(t)$') +ax.legend() +plt.tight_layout() +plt.show() +``` + +Greater idiosyncratic volatility makes the mean more persistent here. + +A larger variance puts more weight on the third moment, through which complementarity operates, so the effective interaction $\bar\theta$ is stronger. + +## Exercises + +```{exercise} +:label: mfgx_ex_hayashi + +This exercise continues the Hayashi example. + +1. Choose $\phi$ so that a firm that ignores the industry again closes half of a capital gap in three years under Hayashi costs, and recompute the industry's half-life and the ratio of the two half-lives. +1. Holding $\phi$ from part 1 and $\varepsilon_s$ fixed, compute the aggregate eigenvalue in the limit of constant returns to scale, $\nu \to 1$, where $q + \theta_X \to 0$, once with no price effect and once with the calibrated supply elasticity. Explain why the answers differ from the limit $\lambda \to -\delta$ that holds without Hayashi costs. +``` + +```{solution-start} mfgx_ex_hayashi +:class: dropdown +``` + +For part 1, the individual's eigenvalue is $\rho/2 - \sqrt{\rho^2/4 + \delta^2 q/\gamma}$, so a target $\lambda$ requires $\gamma = \delta^2 q/[(\rho/2 - \lambda)^2 - \rho^2/4]$. + +```{code-cell} ipython3 +λ_target = -np.log(2)/3 +γ_H = δ_c**2*q_c/((ρ_c/2 - λ_target)**2 - ρ_c**2/4) + +λ_ind_H = λ_cross(ρ_c, δ_c, δ_c, q_c + γ_H, γ_H, γ_H) +λ_agg_H = λ_cross(ρ_c, δ_c, δ_c, q_c + γ_H + θ_X_c, γ_H + θ_A_c, γ_H) + +print(f"γ = {γ_H:.4f}, so φ = {γ_H/cap_scale(η_c, ν_c):.4f}") +print(f"individual half-life {half_life(λ_ind_H):.2f}, " + f"industry half-life {half_life(λ_agg_H):.2f}, " + f"ratio {half_life(λ_agg_H)/half_life(λ_ind_H):.2f}") +``` + +Matching the three-year individual half-life requires a cheaper installation technology than before, $\phi \approx 0.11$ instead of $1/7$. + +The industry's half-life is then 5.4 years, so the gap between micro and macro adjustment widens: the ratio rises from 1.66 to 1.81. + +For part 2, when $q + \theta_X = 0$ the effective state curvature in {prf:ref}`mfgx_prop_cross` is $\gamma$, and the industry's effective return is + +$$ +-\tfrac12\gamma X^2 - \tfrac12(\gamma + \theta_{\mathcal A})\mathcal A^2 + \gamma X\mathcal A += -\tfrac\gamma2(\mathcal A - X)^2 - \tfrac{\theta_{\mathcal A}}2\mathcal A^2 . +$$ + +In the limit $\nu \to 1$, the scale $\delta\bar k^{1-\nu}$ that converts $\phi$ and $1/\varepsilon_s$ into $\gamma$ and $\theta_{\mathcal A}$ tends to $\delta(\eta-1)/[\eta(\rho+\delta)]$. + +Let's compute the limit with and without a price effect. + +```{code-cell} ipython3 +φ_H = γ_H/cap_scale(η_c, ν_c) +scale_lim = cap_scale(η_c, 1.0) +γ_lim = φ_H*scale_lim +for inv_ε in (0.0, 1/ε_s_c): + θ_A_lim = inv_ε*scale_lim + λ_lim = λ_cross(ρ_c, δ_c, δ_c, γ_lim, γ_lim + θ_A_lim, γ_lim) + print(f"1/ε_s = {inv_ε:.3f}: θ_A = {θ_A_lim:.4f}, λ = {λ_lim:.5f}") +``` + +With $\theta_{\mathcal A} = 0$ the only cost is the cost of net investment, so any capital stock can be maintained for free and nothing pulls the industry back: the aggregate state has a unit root. + +Without Hayashi costs, maintaining a capital stock above its steady state requires gross investment above $\bar i$, which $\psi(i)$ penalizes, so the industry lets depreciation do the work and $\lambda \to -\delta$. + +With $\theta_{\mathcal A} > 0$ the price of investment goods penalizes gross investment away from $\bar i$, which restores some mean reversion. + +```{solution-end} +``` + +```{exercise} +:label: mfgx_ex_restriction + +This exercise shows why the restriction $\Theta_{X\alpha} = \Theta_{x\mathcal A}$ matters, even with a single state. + +Take $\rho = 0.05$, $a = 0.3$, $b = q = \gamma = 1$, $\theta_X = \theta_{\mathcal A} = 0$ and no own cross term, and use `state_costate` with $C_1 = -\theta_{x\mathcal A}$ and $C_2 = -\theta_{X\alpha}$. + +1. Show that the trace of the state-costate matrix is $\rho + b(\theta_{x\mathcal A} - \theta_{X\alpha})/(\gamma + \theta_{\mathcal A})$, so that its eigenvalues can be symmetric about $\rho/2$ only when $\theta_{X\alpha} = \theta_{x\mathcal A}$. +1. Compute the eigenvalues for $(\theta_{X\alpha}, \theta_{x\mathcal A})$ equal to $(0, 0)$, $(-2, -2)$, $(-1.5, -2.5)$ and $(-2.5, -1.5)$, and count how many have real parts below $\rho/2$. What does each count imply about the number of equilibria? +``` + +```{solution-start} mfgx_ex_restriction +:class: dropdown +``` + +For part 1, with $g = \gamma + \theta_{\mathcal A}$ the diagonal elements of the matrix are $\rho + a - b C_1/g$ and $b C_2/g - a$. + +Their sum is $\rho + b(C_2 - C_1)/g = \rho + b(\theta_{x\mathcal A} - \theta_{X\alpha})/g$. + +Eigenvalues symmetric about $\rho/2$ must sum to $\rho$, which requires $\theta_{X\alpha} = \theta_{x\mathcal A}$. + +For part 2, we compute the eigenvalues in each case. + +```{code-cell} ipython3 +ρ_e, a_e = 0.05, 0.3 + +print(f"{'(θ_Xα, θ_xA)':>14}{'eigenvalues':>34}{'trace':>8}{'below ρ/2':>11}") +for θ_Xα_e, θ_xA_e in ((0.0, 0.0), (-2.0, -2.0), (-1.5, -2.5), (-2.5, -1.5)): + M_e = state_costate(a_e*one, one, one, one, ρ_e, + -θ_xA_e*one, -θ_Xα_e*one) + ev = np.linalg.eigvals(M_e) + print(f"{str((θ_Xα_e, θ_xA_e)):>14}{str(np.round(ev, 4)):>34}" + f"{ev.real.sum():>8.3f}{np.sum(ev.real < ρ_e/2):>11}") +``` + +With no mixed interactions there is one eigenvalue on each side of $\rho/2$ and a unique equilibrium. + +At $(-2, -2)$ the restriction holds, but the sufficient condition $q - \tilde C^2/g > 0$ of {prf:ref}`mfgx_prop_cross` fails, since $1 - 4 < 0$. + +The eigenvalues are then complex with real part exactly $\rho/2$, so for $X(0) \neq 0$ there is no admissible path and no equilibrium. + +At $(-1.5, -2.5)$ both eigenvalues are below $\rho/2$, so every initial costate $w(0)$ generates an admissible path. + +Each of these paths is an equilibrium. + +Given aggregate paths that grow more slowly than $e^{\rho t/2}$, the individual's costate equation has a unique admissible solution, because its own root $\rho - \lambda_{\text{ind}}$ exceeds $\rho/2$, and the $w$ on any path of the system is that solution. + +So there is a continuum of equilibria for each initial distribution. + +At $(-2.5, -1.5)$ both eigenvalues are above $\rho/2$, and again no equilibrium exists for $X(0) \neq 0$. + +```{solution-end} +``` + +```{exercise} +:label: mfgx_ex_forcing + +This exercise works with the pricing example of {ref}`mfgx_forcing`. + +1. Take $\chi = 0.8$ and $a_{22} = 0.5$. Check that the eigenvalues of the aggregate closed-loop matrix are those of $B_1\Gamma^{-1}B_1^\top P_{11} - A_{11}$ together with $-a_{22}$, as {prf:ref}`mfgx_prop_forcing` says. Check also that changing the $(2,2)$ elements of $Q$ and $\Theta_X$ leaves the first row of $P$, and hence the pricing policy, unchanged. +1. With transitory cost shocks, $a_{22} = 0.5$, plot the industry's response to a common one percent cost increase for $\chi \in \{0, 0.5, 0.8\}$. How do the size and timing of the peak response depend on $\chi$? +``` + +```{solution-start} mfgx_ex_forcing +:class: dropdown +``` + +For part 1, we rebuild the example with the $(2,2)$ elements as arguments. + +```{code-cell} ipython3 +χ_e, a_22_e, ρ_f, q_f, γ_f = 0.8, 0.5, 0.04, 1.0, 1.0 +A_f = np.array([[0.0, 0.0], + [0.0, a_22_e]]) +B_f = np.array([[1.0], + [0.0]]) +Γ_f = np.array([[γ_f]]) + +def P_forcing(q_22, θ_22): + "Equilibrium P when the (2, 2) elements of Q and Θ_X are q_22 and θ_22." + Q_f = q_f*np.array([[1.0, -(1 - χ_e)], + [-(1 - χ_e), 0.0]]) + Q_f[1, 1] = q_22 + Θ_f = np.array([[-q_f*χ_e, 0.0], + [0.0, θ_22]]) + return mfg_riccati(A_f, B_f, Q_f + Θ_f, Γ_f, ρ_f) + +P_a, P_b = P_forcing(2.0, 0.0), P_forcing(7.0, 3.0) +λ_11 = P_a[0, 0]/γ_f # B_1 Γ^{-1} B_1' P_11 - A_11 +ev_cl = np.sort(np.linalg.eigvals(closed_loop(A_f, B_f, Γ_f, P_a)).real) +print("closed-loop eigenvalues: ", ev_cl.round(5)) +print("controlled block and -a_22:", np.sort([λ_11, -a_22_e]).round(5)) +print("first row of P unchanged:", np.allclose(P_a[0], P_b[0])) +``` + +For part 2, we plot the response of the average price. + +```{code-cell} ipython3 +t_e = np.linspace(0, 8, 400) +fig, ax = plt.subplots(figsize=(8, 4.5)) +for χ in (0.0, 0.5, 0.8): + U_agg, _ = pricing_with_costs(χ, 0.5) + agg = np.array([(expm(U_agg*t) @ np.array([0.0, 1.0]))[0] for t in t_e]) + k = agg.argmax() + ax.plot(t_e, agg, lw=2, label=f"$\\chi = {χ}$") + print(f"χ = {χ}: peak {agg[k]:.3f} percent after {t_e[k]:.2f} years") +ax.set_xlabel('years') +ax.set_ylabel('average price, percent') +ax.legend() +plt.tight_layout() +plt.show() +``` + +Stronger complementarity makes the peak response smaller and later. + +Each store raises its price more slowly because the others do, so by the time the industry's price has risen much, the cost shock has already started to fade. + +```{solution-end} +``` + +```{exercise} +:label: mfgx_ex_moments + +This exercise shows how an interaction through an even moment affects transitional dynamics without changing the local eigenvalues. + +Take $\bar r = 2$, $\theta_1 = -0.3$, $\sigma = 1$ and the other parameters of the moment example in {ref}`mfgx_higher_moments`. + +1. Verify that $\theta_2$ changes none of the eigenvalues of $\tilde{\mathcal H}$. +1. Suppose the cross-sectional variance rises by 0.1 while the mean stays at zero. Trace the path of the mean for $\theta_2 \in \{-0.5, 0, 0.5\}$ and explain what you see. +1. {cite:t}`AlvarezArgente2026` show that along the saddle path $\dot{\hat M}_1 = \tilde\lambda_1\hat M_1 + \zeta\hat M_2$ with $\zeta = b^2\theta_2/[\gamma(\tilde\lambda_1 + 2\lambda_{\text{ind}} - \rho)]$. Check this formula. +``` + +```{solution-start} mfgx_ex_moments +:class: dropdown +``` + +For part 1, we compute the eigenvalues for each $\theta_2$. + +```{code-cell} ipython3 +θ_1 = -0.3 +for θ_2 in (-0.5, 0.0, 0.5): + H_2, _, _ = moment_system(ρ_m, a_m, b_m, q_m, γ_m, 1.0, np.array([θ_1, θ_2])) + print(f"θ_2 = {θ_2:>4}: eigenvalues {np.sort(np.linalg.eigvals(H_2).real).round(5)}") +``` + +They do not move. + +At the zero-mean steady state the second moment evolves on its own, $\dot{\hat M}_2 = 2\lambda_{\text{ind}}\hat M_2$, because the term $2(b^2/\gamma)\beta_1 M_1$ in its law of motion vanishes to first order. + +For part 2, we start from $\hat M_1 = 0$ and $\hat M_2 = 0.1$. + +```{code-cell} ipython3 +t_e = np.linspace(0, 4, 200) +fig, ax = plt.subplots(figsize=(8, 4.5)) +for θ_2 in (-0.5, 0.0, 0.5): + H_2, _, _ = moment_system(ρ_m, a_m, b_m, q_m, γ_m, 1.0, np.array([θ_1, θ_2])) + U_2 = moment_dynamics(H_2, ρ_m) + path = [(expm(U_2*t) @ np.array([0.0, 0.1]))[0] for t in t_e] + ax.plot(t_e, path, lw=2, label=f"$\\theta_2 = {θ_2}$") +ax.set_xlabel('$t$') +ax.set_ylabel('cross-sectional mean $M_1(t)$') +ax.legend() +plt.tight_layout() +plt.show() +``` + +With $\theta_2 > 0$, a higher second moment lowers everybody's marginal return to $x$ through the term $-\theta_2 M_2 x$, so agents lower $x$ while the variance is high, and the mean dips before returning as the variance returns to its invariant level. + +With $\theta_2 < 0$ the mean rises instead, and with $\theta_2 = 0$ it does not move at all. + +So a shock to dispersion moves the average, even though the eigenvalues that govern local persistence are unchanged. + +For part 3, we compare the coefficients of the saddle-path law of motion with the formula. + +```{code-cell} ipython3 +for θ_2 in (-0.5, 0.5): + H_2, _, λ_ind_2 = moment_system(ρ_m, a_m, b_m, q_m, γ_m, 1.0, np.array([θ_1, θ_2])) + U_2 = moment_dynamics(H_2, ρ_m) + λ_1 = ρ_m/2 - np.sqrt((a_m + ρ_m/2)**2 + b_m**2*(q_m + θ_1)/γ_m) + ζ = b_m**2*θ_2/(γ_m*(λ_1 + 2*λ_ind_2 - ρ_m)) + print(f"θ_2 = {θ_2}: coefficients ({U_2[0, 0]:.5f}, {U_2[0, 1]:.5f}), " + f"formula ({λ_1:.5f}, {ζ:.5f})") +``` + +```{solution-end} +``` + +## Further reading + +{cite:t}`AlvarezArgente2026` develop the cross terms and the forcing processes in the appendices of their paper, and the coupling through higher moments in its main text. + +The Hayashi and pass-through examples, and the one-state analysis of the mixed interactions in {ref}`mfgx_ex_restriction`, are our own illustrations of their results.