Pricing under RI with Endogenous Feedback

This example solves a pricing problem under rational inattention with endogenous feedback using the DRIPs package.

Binder to run and modify the following code (no software is needed on the local machine).

See Afrouzi and Yang (2019) for background on the theory. Include the solver and import packages for plots and performance:

Contents

Setup

Problem

Suppose now that there is general equilibrium feedback with the degree of strategic complementarity $\alpha$: $p_{i,t}^*=(1-\alpha)q_t+\alpha p_t$ where

\[\begin{aligned} \Delta q_t&=\rho \Delta q_{t-1}+u_t,\quad u_t\sim \mathcal{N}(0,\sigma_u^2) \\ p_t&\equiv \int_0^1 p_{i,t}di \end{aligned}\]

Note that now the state space representation for $p_{i,t}^*$ is no longer exogenous and is determined in the equilibrium. However, we know that this is a Guassian process and by Wold's theorem we can decompose it to its $MA(\infty)$ representation:

\[\begin{aligned} p_{i,t}^*=\Phi(L)u_t \end{aligned}\]

where $\Phi(.)$ is a lag polynomial and $u_t$ is the shock to nominal demand. Here, we have basically guessed that the process for $p_{i,t}^*$ is determined uniquely by the history of monetary shocks which requires that rational inattention errors of firms are orthogonal (See Afrouzi (2020)). Our objective is to find $\Phi(.)$.

Since we cannot put $MA(\infty)$ processes in the computer, we approximate them with truncation. In particular, we know for stationary processes, we can arbitrarily get close to the true process by truncating $MA(\infty)$ processes to $MA(T)$ processes. Our problem here is that $p_{i,t}^*$ has a unit root and is not stationary. We can bypass this issue by re-writing the state space in the following way:

\[\begin{aligned} p_{i,t}^*=\phi(L)\tilde{u}_t,\quad \tilde{u}_t=(1-L)^{-1}u_t =\sum_{j=0}^\infty u_{t-j} \end{aligned}\]

here $\tilde{u}_{t-j}$ is the unit root of the process and basically we have differenced out the unit root from the lag polynomial, and $\phi(L)=(1-L)\Phi(L)$. Notice that since the original process was difference stationary, differencing out the unit root means that $\phi(L)$ is now in $\ell_2$, and the process can now be approximated arbitrarily precisely with truncation.

Matrix Notation

For a length of truncation $L$, let $\vec{x}_t\equiv (\tilde{u}_t,\tilde{u}_{t-1},\dots,\tilde{u}_{t-(L+1)})\in\mathbb{R}^L$. Then, note that $p_{i,t}^*\approx \mathbf{H} '\vec{x}_{t}$ where $\mathbf{H}\in \mathbb{R}^L$ is the truncated matrix analog of the lag polynominal, and is endogenous to the problem. Our objective is to find the general equilibrium $\mathbf{H}$ along with the optimal information structure that it implies.

Moreover, note that $q_t=\mathbf{H}_q'\vec{x}_t,\quad \mathbf{H}_q'=(1,\rho,\rho^2,\dots,\rho^{L-1})$

We will solve for $\phi$ by iterating over the problem. In particular, in iteration $n\geq 1$, given the guess $\mathbf{H}_{(n-1)}$, we have the following state space representation for the firm's problem

\[\begin{aligned} \vec{x}_{t}& = \underset{\mathbf{A}}{\underbrace{\left[\begin{array}{ccccc} 1 & 0 & \dots & 0 & 0\\ 1 & 0 & \dots & 0 & 0\\ 0 & 1 & \dots & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & \dots & 1 & 0 \end{array}\right]}}\, \vec{x}_{t-1} + \underset{\mathbf{Q}}{\underbrace{\left[\begin{array}{c} \sigma_u\\ 0\\ 0\\ \vdots\\ 0 \end{array}\right]}}\, u_t, \\ p_{i,t}^*&=\mathbf{H}_{(n-1)}'\vec{x}_{t} \end{aligned}\]

Then we can solve the rational inattention problem of all firms and get the new guess for $p_t^*$:

\[\begin{aligned} p_t^* & = (1-\alpha)q_t + \alpha p_t \\ & = (1-\alpha)\sum_{j=0}^\infty \alpha^j q_t^{(j)}, \\ q_{t}^{(j)}&\equiv \begin{cases} q_{t} & j=0\\ \int_{0}^{1}\mathbb{E}_{i,t}[q_{t}^{(j-1)}]di & j\geq1 \end{cases} \end{aligned}\]

where $q_t^{(j)}$ is the $j$'th order belief of firms, on average, of $q_t$. Now, we need to write these higher order beliefs in terms of the state vector. Suppose, for a given $j$, there exists $\mathbf{X}_j\in \mathbb{R}^{L\times L}$ such that $ qt^{(j)} = \mathbf{H}q'\mathbf{X}j \vec{x}t $ This clearly holds for $j=0$ with $\mathbf{X}_0=\mathbf{I}$.

Now, note that

\[\begin{aligned} q_{t}^{(j+1)} &= \int_{0}^{1}\mathbb{E}_{i,t}[q_{t}^{(j)}]di \\ &= \mathbf{H}_{q}'\mathbf{X}_{j}\int_{0}^{1}\mathbb{E}_{i,t}[\vec{x}_{t}]di \\ &= \mathbf{H}_{q}'\mathbf{X}_{j}\sum_{j=0}^{\infty}[(\mathbf{I}-\mathbf{K}_{(n)}\mathbf{Y}'_{(n)})\mathbf{A}]^{j}\mathbf{K}_{(n)}\mathbf{Y}'_{(n)}\vec{x}_{t-j} \\ &\approx\underset{\equiv\mathbf{X}_{(n)}}{\mathbf{H}_{q}'\mathbf{X}_{j}\underbrace{\left[\sum_{j=0}^{\infty}[(\mathbf{I}-\mathbf{K}_{(n)}\mathbf{Y}'_{(n)})\mathbf{A}]^{j}\mathbf{K}_{(n)}\mathbf{Y}'_{(n)}\mathbf{M}'^{j}\right]}}\vec{x}_{t}=\mathbf{H}_{q}'\mathbf{X}_{j}\mathbf{X}_{(n)}\vec{x}_{t} \end{aligned}\]

where the $(n)$ subscripts refer to the solution of the RI problem in the $(n)$'th iteration. Note that this implies $\mathbf{X}{j}=\mathbf{X}{(n)}^j,\forall j\geq 0 \Rightarrow qt^{(j)}=\mathbf{X}{(n)}^{j}\vec{x}_t $

This gives us an updated guess for $\mathbf{H}$:

\[\begin{aligned} p_t^*&=(1-\alpha)\mathbf{H}_q'\underset{\equiv \mathbf{X}_{p,(n)}}{\underbrace{\left[\sum_{j=0}^\infty \alpha^j \mathbf{X}_{(n)}^j\right]}} \vec{x}_t \\ &\Rightarrow \mathbf{H}_{(n)} = (1-\alpha)\mathbf{X}_{p,(n)}'\mathbf{H}_q \end{aligned}\]

We iterate until convergence of $\mathbf{H}_{(n)}$.

Initialization

Include the package:

using DRIPs;
nothing #hide

Assign Parameters

ρ   = 0.6;        #persistence of money growth
σ_u = 0.1;        #std. deviation of shocks to money growth
α   = 0.8;        #degree of strategic complementarity
L   = 40;         #length of truncation
Hq  = ρ.^(0:L-1); #state-space rep. of Δq
nothing #hide

Specifying the primitives of the DRIP:

using LinearAlgebra;
ω   = 0.2;
β   = 0.99;
A   = [1 zeros(1,L-2) 0; Matrix(I,L-1,L-1) zeros(L-1,1)];
Q   = [σ_u; zeros(L-1,1)];
nothing #hide

A Function for Finding the Fixed Point

Let us now define a function that solves the GE problem and returns the solution in a Drip structure:

function ge_drip(ω,β,A,Q,          #primitives of drip except for H because H is endogenous
                 α,                #strategic complementarity
                 Hq,               #state space rep. of Δq
                 L;                #length of truncation
                 H0       = Hq,    #optional: initial guess for H (Hq is the true solution when α=0)
                 maxit    = 200,   #optional: max number of iterations for GE code
                 tol      = 1e-4)  #optional: tolerance for iterations
    err   = 1;
    iter  = 0;
    M     = [zeros(1,L-1) 0; Matrix(I,L-1,L-1) zeros(L-1,1)];
    while (err > tol) & (iter < maxit)
            if iter == 0
                global ge  = Drip(ω,β,A,Q,H0, w = 0.9);
            else
                global ge  = Drip(ω,β,A,Q,H0;Ω0 = ge.ss.Ω ,Σ0 = ge.ss.Σ_1,maxit=15);
            end

            XFUN(jj) = ((I-ge.ss.K*ge.ss.Y')*ge.A)^jj * (ge.ss.K*ge.ss.Y') * (M')^jj
            X = DRIPs.infinitesum(XFUN; maxit=L, start = 0);  #E[x⃗]=X×x⃗

            XpFUN(jj) = α^jj * X^(jj)
            Xp = DRIPs.infinitesum(XpFUN; maxit=L, start = 0);

            H1 = (1-α)*Xp'*Hq;
            err= 0.5*norm(H1-H0,2)/norm(H0)+0.5*err;
            H0 = H1;

            iter += 1;
            if iter == maxit
                print("GE loop hit maxit\n")
            end
            println("Iteration $iter. Difference: $err")
    end
    return(ge)
end;
nothing #hide

Solution

Solve for benchmark parameterization:

using Suppressor; # suppresses printing of function. comment to see convergence details
@time @suppress  ge = ge_drip(ω,β,A,Q,α,Hq,L); # remove suppress to see convergence log
nothing #hide
  1.956416 seconds (4.83 M allocations: 544.294 MiB, 3.98% gc time)

IRFs

Get IRFs:

geirfs = irfs(ge,T = L)

M  = [zeros(1,L-1) 0; Matrix(I,L-1,L-1) zeros(L-1,1)]; # shift matrix
dq = diagm(Hq)*geirfs.x[1,1,:];                        # change in nominal demand
Pi = (I-M)*geirfs.a[1,1,:];                            # inflation
y  = inv(I-M)*(dq-Pi);                                 # output

using Plots, LaTeXStrings; pyplot();
p1 = plot(1:L,[dq,Pi],
     label = [L"Agg. Demand Growth ($\Delta q$)" L"Inflation ($\pi$)"]);

p2 = plot(1:L,y,
     label  = L"Output ($y$)");

plot(p1,p2,
    layout     = (1,2),
    xlim       = (1,20),
    lw         = 3,
    legend     = :topright,
    legendfont = font(12),
    tickfont   = font(12),
    size       = (900,370),
    framestyle = :box)

Measure Performance

Solve and measure performance for random values of $\omega$:

using BenchmarkTools;
@suppress @benchmark ge_drip(ω,β,A,Q,α,Hq,L) setup = (ω=rand()) # solves and times the fixed point for different values of ω
BenchmarkTools.Trial: 
  memory estimate:  282.72 MiB
  allocs estimate:  34129
  --------------
  minimum time:     205.330 ms (7.93% GC)
  median time:      245.701 ms (9.30% GC)
  mean time:        268.361 ms (8.80% GC)
  maximum time:     335.964 ms (9.82% GC)
  --------------
  samples:          19
  evals/sample:     1

This page was generated using Literate.jl.