14. The Aiyagari Model#

GPU

This lecture was built using hardware that has access to a GPU.

To run this lecture on Google Colab, click on the “play” icon top right, select Colab, and set the runtime environment to include a GPU.

To run this lecture on your own machine, you need to install Google JAX.

14.1. Overview#

In this lecture, we describe the structure of a class of models that build on work by Truman Bewley [Bew77].

We begin by discussing an example of a Bewley model due to Rao Aiyagari [Aiy94].

The model features

  • Heterogeneous agents

  • A single exogenous vehicle for borrowing and lending

  • Limits on amounts individual agents may borrow

The Aiyagari model has been used to investigate many topics, including

  • precautionary savings and the effect of liquidity constraints [Aiy94]

  • risk sharing and asset pricing [HL96]

  • the shape of the wealth distribution [BBZ15]

14.1.1. References#

The primary reference for this lecture is [Aiy94].

A textbook treatment is available in chapter 18 of [LS18].

A less sophisticated version of this lecture (without JAX) can be found here.

14.1.2. Preliminaries#

We use the following imports

import matplotlib.pyplot as plt
import numpy as np
import jax
import jax.numpy as jnp
from collections import namedtuple

Let’s check the GPU we are running

!nvidia-smi
Thu Dec 21 00:43:53 2023       
+-----------------------------------------------------------------------------+
| NVIDIA-SMI 470.182.03   Driver Version: 470.182.03   CUDA Version: 12.1     |
|-------------------------------+----------------------+----------------------+
| GPU  Name        Persistence-M| Bus-Id        Disp.A | Volatile Uncorr. ECC |
| Fan  Temp  Perf  Pwr:Usage/Cap|         Memory-Usage | GPU-Util  Compute M. |
|                               |                      |               MIG M. |
|===============================+======================+======================|
|   0  Tesla V100-SXM2...  Off  | 00000000:00:1E.0 Off |                    0 |
| N/A   32C    P0    39W / 300W |      0MiB / 16160MiB |      2%      Default |
|                               |                      |                  N/A |
+-------------------------------+----------------------+----------------------+
                                                                               
+-----------------------------------------------------------------------------+
| Processes:                                                                  |
|  GPU   GI   CI        PID   Type   Process name                  GPU Memory |
|        ID   ID                                                   Usage      |
|=============================================================================|
|  No running processes found                                                 |
+-----------------------------------------------------------------------------+

We will use 64 bit floats with JAX in order to increase the precision.

jax.config.update("jax_enable_x64", True)

We will use the following function to compute stationary distributions of stochastic matrices. (For a reference to the algorithm, see p. 88 of Economic Dynamics.)

# Compute the stationary distribution of P by matrix inversion.

@jax.jit
def compute_stationary(P):
    n = P.shape[0]
    I = jnp.identity(n)
    O = jnp.ones((n, n))
    A = I - jnp.transpose(P) + O
    return jnp.linalg.solve(A, jnp.ones(n))

14.2. Firms#

Firms produce output by hiring capital and labor.

Firms act competitively and face constant returns to scale.

Since returns to scale are constant the number of firms does not matter.

Hence we can consider a single (but nonetheless competitive) representative firm.

The firm’s output is

\[ Y_t = A K_t^{\alpha} N^{1 - \alpha} \]

where

  • \( A \) and \( \alpha \) are parameters with \( A > 0 \) and \( \alpha \in (0, 1) \)

  • \( K_t \) is aggregate capital

  • \( N \) is total labor supply (which is constant in this simple version of the model)

The firm’s problem is

\[ \max_{K, N} \left\{ A K_t^{\alpha} N^{1 - \alpha} - (r + \delta) K - w N \right\} \]

The parameter \( \delta \) is the depreciation rate.

These parameters are stored in the following namedtuple.

Firm = namedtuple('Firm', ('A', 'N', 'α', 'β', 'δ'))

def create_firm(A=1.0,
                N=1.0,
                α=0.33,
                β=0.96,
                δ=0.05):
    return Firm(A=A, N=N, α=α, β=β, δ=δ)

From the first-order condition with respect to capital,

the firm’s inverse demand for capital is

(14.1)#\[r = A \alpha \left( \frac{N}{K} \right)^{1 - \alpha} - \delta\]
def rd(K, f):
    """
    Inverse demand curve for capital.  The interest rate associated with a
    given demand for capital K.
    """
    A, N, α, β, δ = f
    return A * α * (N / K)**(1 - α) - δ

rd = jax.jit(rd, static_argnums=(1,))

Using (14.1) and the firm’s first-order condition for labor,

we can pin down the equilibrium wage rate as a function of \( r \) as

(14.2)#\[w(r) = A (1 - \alpha) (A \alpha / (r + \delta))^{\alpha / (1 - \alpha)}\]
def r_to_w(r, f):
    """
    Equilibrium wages associated with a given interest rate r.
    """
    A, N, α, β, δ = f
    return A * (1 - α) * (A * α / (r + δ))**(α / (1 - α))

r_to_w = jax.jit(r_to_w, static_argnums=(1,))

14.3. Households#

Infinitely lived households / consumers face idiosyncratic income shocks.

A unit interval of ex-ante identical households face a common borrowing constraint.

The savings problem faced by a typical household is

\[ \max \mathbb E \sum_{t=0}^{\infty} \beta^t u(c_t) \]

subject to

\[ a_{t+1} + c_t \leq w z_t + (1 + r) a_t \quad c_t \geq 0, \quad \text{and} \quad a_t \geq -B \]

where

  • \( c_t \) is current consumption

  • \( a_t \) is assets

  • \( z_t \) is an exogenous component of labor income capturing stochastic unemployment risk, etc.

  • \( w \) is a wage rate

  • \( r \) is a net interest rate

  • \( B \) is the maximum amount that the agent is allowed to borrow

The exogenous process \( \{z_t\} \) follows a finite state Markov chain with given stochastic matrix \( P \).

In this simple version of the model, households supply labor inelastically because they do not value leisure.

Below we provide code to solve the household problem, taking \(r\) and \(w\) as fixed.

For now we assume that \(u(c) = \log(c)\).

(CRRA utility is treated in the exercises.)

14.3.1. Primitives and Operators#

This namedtuple stores the parameters that define a household asset accumulation problem and the grids used to solve it.

Household = namedtuple('Household', ('r', 'w', 'β', 'a_size', 'z_size', \
                                     'a_grid', 'z_grid', 'Π'))

def create_household(r=0.01,                      # Interest rate
                     w=1.0,                       # Wages
                     β=0.96,                      # Discount factor
                     Π=[[0.9, 0.1], [0.1, 0.9]],  # Markov chain
                     z_grid=[0.1, 1.0],           # Exogenous states
                     a_min=1e-10, a_max=20,       # Asset grid
                     a_size=200):
    
    a_grid = jnp.linspace(a_min, a_max, a_size)
    z_grid, Π = map(jnp.array, (z_grid, Π))
    Π = jax.device_put(Π)
    z_grid = jax.device_put(z_grid)
    z_size = len(z_grid)
    a_grid, z_grid, Π = jax.device_put((a_grid, z_grid, Π))
        
    return Household(r=r, w=w, β=β, a_size=a_size, z_size=z_size, \
                     a_grid=a_grid, z_grid=z_grid, Π=Π)
@jax.jit
def u(c):
    return jnp.log(c)

This is the vectorized version of the right-hand side of the Bellman equation (before maximization), which is a 3D array representing

\[ B(a, z, a') = u(wz + (1+r)a - a') + \beta \sum_{z'} v(a', z') Π(z, z') \]

for all \((a, z, a')\).

def B(v, constants, sizes, arrays):
    # Unpack
    r, w, β = constants
    a_size, z_size = sizes
    a_grid, z_grid, Π = arrays

    # Compute current consumption as array c[i, j, ip]
    a  = jnp.reshape(a_grid, (a_size, 1, 1))    # a[i]   ->  a[i, j, ip]
    z  = jnp.reshape(z_grid, (1, z_size, 1))    # z[j]   ->  z[i, j, ip]
    ap = jnp.reshape(a_grid, (1, 1, a_size))    # ap[ip] -> ap[i, j, ip]
    c = w*z + (1 + r)*a - ap

    # Calculate continuation rewards at all combinations of (a, z, ap)
    v = jnp.reshape(v, (1, 1, a_size, z_size))  # v[ip, jp] -> v[i, j, ip, jp]
    Π = jnp.reshape(Π, (1, z_size, 1, z_size))  # Π[j, jp]  -> Π[i, j, ip, jp]
    EV = jnp.sum(v * Π, axis=3)                 # sum over last index jp

    # Compute the right-hand side of the Bellman equation
    return jnp.where(c > 0, u(c) + β * EV, -jnp.inf)

B = jax.jit(B, static_argnums=(2,))

The next function computes greedy policies.

# Computes a v-greedy policy, returned as a set of indices
def get_greedy(v, constants, sizes, arrays):
    return jnp.argmax(B(v, constants, sizes, arrays), axis=2)

get_greedy = jax.jit(get_greedy, static_argnums=(2,))

We need to know rewards at a given policy for policy iteration.

The following functions computes the array \(r_{\sigma}\) which gives current rewards given policy \(\sigma\).

That is,

\[ r_{\sigma}[i, j] = r[i, j, \sigma[i, j]] \]
def compute_r_σ(σ, constants, sizes, arrays):
    # Unpack
    r, w, β = constants
    a_size, z_size = sizes
    a_grid, z_grid, Π = arrays

    # Compute r_σ[i, j]
    a = jnp.reshape(a_grid, (a_size, 1))
    z = jnp.reshape(z_grid, (1, z_size))
    ap = a_grid[σ]
    c = (1 + r)*a + w*z - ap
    r_σ = u(c)

    return r_σ

compute_r_σ = jax.jit(compute_r_σ, static_argnums=(2,))

The value \(v_{\sigma}\) of a policy \(\sigma\) is defined as

\[ v_{\sigma} = (I - \beta P_{\sigma})^{-1} r_{\sigma} \]

Here we set up the linear map \(v \rightarrow R_{\sigma} v\), where \(R_{\sigma} := I - \beta P_{\sigma}\).

In the consumption problem, this map can be expressed as

\[ (R_{\sigma} v)(a, z) = v(a, z) - \beta \sum_{z'} v(\sigma(a, z), z') Π(z, z') \]

Defining the map as above works in a more intuitive multi-index setting

(e.g. working with \(v[i, j]\) rather than flattening \(v\) to a one-dimensional array)

and avoids instantiating the large matrix \(P_{\sigma}\).

The following linear operator is also needed for policy iteration.

def R_σ(v, σ, constants, sizes, arrays):
    # Unpack
    r, w, β = constants
    a_size, z_size = sizes
    a_grid, z_grid, Π = arrays

    # Set up the array v[σ[i, j], jp]
    zp_idx = jnp.arange(z_size)
    zp_idx = jnp.reshape(zp_idx, (1, 1, z_size))
    σ = jnp.reshape(σ, (a_size, z_size, 1))
    V = v[σ, zp_idx]

    # Expand Π[j, jp] to Π[i, j, jp]
    Π = jnp.reshape(Π, (1, z_size, z_size))

    # Compute and return v[i, j] - β Σ_jp v[σ[i, j], jp] * Π[j, jp]
    return v - β * jnp.sum(V * Π, axis=2)

R_σ = jax.jit(R_σ, static_argnums=(3,))

The next function computes the lifetime value of a given policy.

# Get the value v_σ of policy σ by inverting the linear map R_σ

def get_value(σ, constants, sizes, arrays):

    r_σ = compute_r_σ(σ, constants, sizes, arrays)
    # Reduce R_σ to a function in v
    partial_R_σ = lambda v: R_σ(v, σ, constants, sizes, arrays)
    # Compute inverse v_σ = (I - β P_σ)^{-1} r_σ
    return jax.scipy.sparse.linalg.bicgstab(partial_R_σ, r_σ)[0]

get_value = jax.jit(get_value, static_argnums=(2,))

The following function is used for optimistic policy iteration.

def T_σ(v, σ, constants, sizes, arrays):
    "The σ-policy operator."

    # Unpack model
    γ, w, β = constants
    a_size, z_size = sizes
    a_grid, z_grid, Π = arrays

    r_σ = compute_r_σ(σ, constants, sizes, arrays)

    # Compute the array v[σ[i, j], jp]
    zp_idx = jnp.arange(z_size)
    zp_idx = jnp.reshape(zp_idx, (1, 1, z_size))
    σ = jnp.reshape(σ, (a_size, z_size, 1))
    V = v[σ, zp_idx]

    # Convert Q[j, jp] to Q[i, j, jp]
    Π = jnp.reshape(Π, (1, z_size, z_size))

    # Calculate the expected sum Σ_jp v[σ[i, j], jp] * Q[i, j, jp]
    Ev = jnp.sum(V * Π, axis=2)

    return r_σ + β * jnp.sum(V * Π, axis=2)

T_σ = jax.jit(T_σ, static_argnums=(3,))

14.4. Solvers#

We will solve the household problem using Howard policy iteration.

def policy_iteration(household, verbose=True):
    """Howard policy iteration routine."""
    
    γ, w, β, a_size, z_size, a_grid, z_grid, Π = household
    
    constants = γ, w, β
    sizes = a_size, z_size
    arrays = a_grid, z_grid, Π

    vz = jnp.zeros(sizes)
    σ = jnp.zeros(sizes, dtype=int)
    i, error = 0, 1.0
    while error > 0:
        v_σ = get_value(σ, constants, sizes, arrays)
        σ_new = get_greedy(v_σ, constants, sizes, arrays)
        error = jnp.max(jnp.abs(σ_new - σ))
        σ = σ_new
        i = i + 1
        if verbose:
            print(f"Concluded loop {i} with error {error}.")
    return σ

We can also solve the problem using optimistic policy iteration.

def optimistic_policy_iteration(household, tol=1e-5, m=10):
    
    γ, w, β, a_size, z_size, a_grid, z_grid, Π = household
    
    constants = γ, w, β
    sizes = a_size, z_size
    arrays = a_grid, z_grid, Π

    v = jnp.zeros(sizes)
    error = tol + 1
    while error > tol:
        last_v = v
        σ = get_greedy(v, constants, sizes, arrays)
        for _ in range(m):
            v = T_σ(v, σ, constants, sizes, arrays)
        error = jnp.max(jnp.abs(v - last_v))
    return get_greedy(v, constants, sizes, arrays)

As a first example of what we can do, let’s compute and plot an optimal accumulation policy at fixed prices.

# Example prices
r = 0.03
w = 0.956

# Create an instance of Housbehold
household = create_household(r=r, w=w)
%%time

σ_star_hpi = policy_iteration(household)
Concluded loop 1 with error 101.
Concluded loop 2 with error 76.
Concluded loop 3 with error 36.
Concluded loop 4 with error 17.
Concluded loop 5 with error 12.
Concluded loop 6 with error 6.
Concluded loop 7 with error 3.
Concluded loop 8 with error 2.
Concluded loop 9 with error 1.
Concluded loop 10 with error 1.
Concluded loop 11 with error 1.
Concluded loop 12 with error 1.
Concluded loop 13 with error 1.
Concluded loop 14 with error 1.
Concluded loop 15 with error 0.
CPU times: user 917 ms, sys: 29.8 ms, total: 946 ms
Wall time: 1.27 s
%%time

σ_star = optimistic_policy_iteration(household)
CPU times: user 449 ms, sys: 69.2 ms, total: 519 ms
Wall time: 516 ms

The next plot shows asset accumulation policies at different values of the exogenous state.

γ, w, β, a_size, z_size, a_grid, z_grid, Π = household

fig, ax = plt.subplots(figsize=(9, 9))
ax.plot(a_grid, a_grid, 'k--')  # 45 degrees
for j in range(z_size):
    lb = f'$z = {z_grid[j]:.2}$'
    ax.plot(a_grid, a_grid[σ_star[:, j]], lw=2, alpha=0.6, label=lb)
    ax.set_xlabel('current assets')
    ax.set_ylabel('next period assets')
ax.legend(loc='upper left')

plt.show()
_images/868141a977288f783a9c9aed80f160206d860e8a514c8b2350161636a4fc0ce2.png

14.4.1. Capital Supply#

To start thinking about equilibrium, we need to know how much capital households supply at a given interest rate \(r\).

This quantity can be calculated by taking the stationary distribution of assets under the optimal policy and computing the mean.

The next function implements this calculation for a given policy \(\sigma\).

First we compute the stationary distribution of \(P_{\sigma}\), which is for the bivariate Markov chain of the state \((a_t, z_t)\). Then we sum out \(z_t\) to get the marginal distribution for \(a_t\).

def compute_asset_stationary(σ, constants, sizes, arrays):

    # Unpack
    r, w, β = constants
    a_size, z_size = sizes
    a_grid, z_grid, Π = arrays

    # Construct P_σ as an array of the form P_σ[i, j, ip, jp]
    ap_idx = jnp.arange(a_size)
    ap_idx = jnp.reshape(ap_idx, (1, 1, a_size, 1))
    σ = jnp.reshape(σ, (a_size, z_size, 1, 1))
    A = jnp.where(σ == ap_idx, 1, 0)
    Π = jnp.reshape(Π, (1, z_size, 1, z_size))
    P_σ = A * Π

    # Reshape P_σ into a matrix
    n = a_size * z_size
    P_σ = jnp.reshape(P_σ, (n, n))

    # Get stationary distribution and reshape onto [i, j] grid
    ψ = compute_stationary(P_σ)
    ψ = jnp.reshape(ψ, (a_size, z_size))

    # Sum along the rows to get the marginal distribution of assets
    ψ_a = jnp.sum(ψ, axis=1)
    return ψ_a

compute_asset_stationary = jax.jit(compute_asset_stationary,
                                   static_argnums=(2,))

Let’s give this a test run.

γ, w, β, a_size, z_size, a_grid, z_grid, Π = household
constants = γ, w, β
sizes = a_size, z_size
arrays = a_grid, z_grid, Π
ψ = compute_asset_stationary(σ_star, constants, sizes, arrays)

The distribution should sum to one:

ψ.sum()
Array(1., dtype=float64)

Now we are ready to compute capital supply by households given wages and interest rates.

def capital_supply(household):
    """
    Map household decisions to the induced level of capital stock.
    """

    # Unpack
    γ, w, β, a_size, z_size, a_grid, z_grid, Π = household
    
    constants = γ, w, β
    sizes = a_size, z_size
    arrays = a_grid, z_grid, Π

    # Compute the optimal policy
    σ_star = optimistic_policy_iteration(household)
    # Compute the stationary distribution
    ψ_a = compute_asset_stationary(σ_star, constants, sizes, arrays)

    # Return K
    return float(jnp.sum(ψ_a * a_grid))

14.5. Equilibrium#

We construct a stationary rational expectations equilibrium (SREE).

In such an equilibrium

  • prices induce behavior that generates aggregate quantities consistent with the prices

  • aggregate quantities and prices are constant over time

In more detail, an SREE lists a set of prices, savings and production policies such that

  • households want to choose the specified savings policies taking the prices as given

  • firms maximize profits taking the same prices as given

  • the resulting aggregate quantities are consistent with the prices; in particular, the demand for capital equals the supply

  • aggregate quantities (defined as cross-sectional averages) are constant

In practice, once parameter values are set, we can check for an SREE by the following steps

  1. pick a proposed quantity \( K \) for aggregate capital

  2. determine corresponding prices, with interest rate \( r \) determined by (14.1) and a wage rate \( w(r) \) as given in (14.2).

  3. determine the common optimal savings policy of the households given these prices

  4. compute aggregate capital as the mean of steady state capital given this savings policy

If this final quantity agrees with \( K \) then we have a SREE. Otherwise we adjust \(K\).

These steps describe a fixed point problem which we solve below.

14.5.1. Visual inspection#

Let’s inspect visually as a first pass.

The following code draws aggregate supply and demand curves for capital.

The intersection gives equilibrium interest rates and capital.

# Create default instances
household = create_household()
firm = create_firm()

# Create a grid of r values at which to compute demand and supply of capital
num_points = 50
r_vals = np.linspace(0.005, 0.04, num_points)
%%time

# Compute supply of capital
k_vals = np.empty(num_points)
for i, r in enumerate(r_vals):
    # _replace create a new nametuple with the updated parameters
    household = household._replace(r=r, w=r_to_w(r, firm)) 
    k_vals[i] = capital_supply(household)
CPU times: user 9.62 s, sys: 2.97 s, total: 12.6 s
Wall time: 4.67 s
# Plot against demand for capital by firms

fig, ax = plt.subplots()
ax.plot(k_vals, r_vals, lw=2, alpha=0.6, label='supply of capital')
ax.plot(k_vals, rd(k_vals, firm), lw=2, alpha=0.6, label='demand for capital')
ax.grid()
ax.set_xlabel('capital')
ax.set_ylabel('interest rate')
ax.legend(loc='upper right')

plt.show()
_images/394c5b70cf25d42506bf52f46eebdec03431040b23056e577335c859c7522705.png

Here’s a plot of the excess demand function.

The equilibrium is the zero (root) of this function.

def excess_demand(K, firm, household):
    r = rd(K, firm)
    w = r_to_w(r, firm)
    household = household._replace(r=r, w=w)
    return K - capital_supply(household)
%%time

num_points = 50
k_vals = np.linspace(4, 12, num_points)
out = [excess_demand(k, firm, household) for k in k_vals]
CPU times: user 8.28 s, sys: 2.12 s, total: 10.4 s
Wall time: 3.91 s
fig, ax = plt.subplots()
ax.plot(k_vals, out, lw=2, alpha=0.6, label='excess demand')
ax.plot(k_vals, np.zeros_like(k_vals), 'k--', label="45")
ax.set_xlabel('capital')
ax.legend()
plt.show()
_images/ab5cd4ef1d26c0fef4f216064a76b22dbd7a0e14e4cd74de66f99d87cbdfec9e.png

14.5.2. Computing the equilibrium#

Now let’s compute the equilibrium

To do so, we use the bisection method, which is implemented in the next function.

def bisect(f, a, b, *args, tol=10e-2):
    """
    Implements the bisection root finding algorithm, assuming that f is a
    real-valued function on [a, b] satisfying f(a) < 0 < f(b).
    """
    lower, upper = a, b
    count = 0
    while upper - lower > tol and count < 10000:
        middle = 0.5 * (upper + lower)
        if f(middle, *args) > 0:   # root is between lower and middle
            lower, upper = lower, middle
        else:                      # root is between middle and upper
            lower, upper = middle, upper
        count += 1
    if count == 10000:
        print("Root might not be accurate")
    return 0.5 * (upper + lower)

Now we call the bisection function on excess demand.

def compute_equilibrium(firm, household):
    solution = bisect(excess_demand, 6.0, 10.0, firm, household)
    return solution
%%time

household = create_household()
firm = create_firm()
compute_equilibrium(firm, household)
CPU times: user 1.04 s, sys: 300 ms, total: 1.34 s
Wall time: 673 ms
8.09375

Notice how quickly we can compute the equilibrium capital stock using a simple method such as bisection.

14.6. Exercises#

Exercise 14.1

Using the default household and firm model, produce a graph showing the behaviour of equilibrium capital stock with the increase in \(\beta\).

Exercise 14.2

Switch to the CRRA utility function

\[ u(c) =\frac{c^{1-\gamma}}{1-\gamma} \]

and re-do the plot of demand for capital by firms against the supply of captial.

Also, recompute the equilibrium.

Use the default parameters for households and firms.

Set \(\gamma=2\).