Code should execute sequentially if run in a Jupyter notebook

# Fiscal Risk and Government Debt¶

## Contents¶

In addition to what’s in Anaconda, this lecture will need the following libraries

In [1]:
!pip install quantecon


## Overview¶

This lecture studies government debt in an AMSS economy [AMSSeppala02] of the type described in Optimal Taxation without State-Contingent Debt

We study the behavior of government debt as time $t \rightarrow + \infty$

We use these techniques

• simulations
• a regression coefficient from the tail of a long simulation that allows us to verify that the asymptotic mean of government debt solves a fiscal-risk minimization problem
• an approximation to the mean of an ergodic distribution of government debt
• an approximation to the rate of convergence to an ergodic distribution of government debt

We apply tools applicable to more general incomplete markets economies that are presented on pages 648 - 650 in section III.D of [BEGS17] (BEGS)

We study an [AMSSeppala02] economy with three Markov states driving government expenditures

• In a previous lecture, we showed that with only two Markov states, it is possible that eventually endogenous interest rate fluctuations support complete markets allocations and Ramsey outcomes
• The presence of three states prevents the full spanning that eventually prevails in the two-state example featured in Fiscal Insurance via Fluctuating Interest Rates

The lack of full spanning means that the ergodic distribution of the par value of government debt is nontrivial, in contrast to the situation in Fiscal Insurance via Fluctuating Interest Rates where the ergodic distribution of the par value is concentrated on one point

Nevertheless, [BEGS17] (BEGS) establish for general settings that include ours, the Ramsey planner steers government assets to a level that comes as close as possible to providing full spanning in a precise a sense defined by BEGS that we describe below

We use code constructed in a previous lecture

Warning: Key equations in [BEGS17] section III.D carry typos that we correct below

## The Economy¶

As in Optimal Taxation without State-Contingent Debt and Optimal Taxation with State-Contingent Debt, we assume that the representative agent has utility function

$$u(c,n) = {\frac{c^{1-\sigma}}{1-\sigma}} - {\frac{n^{1+\gamma}}{1+\gamma}}$$

We work directly with labor supply instead of leisure

We assume that

$$c_t + g_t = n_t$$

The Markov state $s_t$ takes three values, namely, $0,1,2$

The initial Markov state is $0$

The Markov transition matrix is $(1/3) I$ where $I$ is a $3 \times 3$ identity matrix, so the $s_t$ process is IID

Government expenditures $g(s)$ equal $.1$ in Markov state $0$, $.2$ in Markov state $1$, and $.3$ in Markov state $2$

We set preference parameters

\begin{aligned} \beta & = .9 \cr \sigma & = 2 \cr \gamma & = 2 \end{aligned}

The following Python code sets up the economy

In [2]:
import numpy as np

class CRRAutility:

def __init__(self,
β=0.9,
σ=2,
γ=2,
π=0.5*np.ones((2, 2)),
G=np.array([0.1, 0.2]),
Θ=np.ones(2),
transfers=False):

self.β, self.σ, self.γ = β, σ, γ
self.π, self.G, self.Θ, self.transfers = π, G, Θ, transfers

# Utility function
def U(self, c, n):
σ = self.σ
if σ == 1.:
U = np.log(c)
else:
U = (c**(1 - σ) - 1) / (1 - σ)
return U - n**(1 + self.γ) / (1 + self.γ)

# Derivatives of utility function
def Uc(self, c, n):
return c**(-self.σ)

def Ucc(self, c, n):
return -self.σ * c**(-self.σ - 1)

def Un(self, c, n):
return -n**self.γ

def Unn(self, c, n):
return -self.γ * n**(self.γ - 1)


### First and Second Moments¶

We’ll want first and second moments of some key random variables below

The following code computes these moments; the code is recycled from Fiscal Insurance via Fluctuating Interest Rates

In [3]:
def mean(x, s):
'''Returns mean for x given initial state'''
x = np.array(x)
return x @ u.π[s]

def variance(x, s):
x = np.array(x)
return x**2 @ u.π[s] - mean(x, s)**2

def covariance(x, y, s):
x, y = np.array(x), np.array(y)
return x * y @ u.π[s] - mean(x, s) * mean(y, s)


## Long Simulation¶

To generate a long simulation we use the following code

We begin by showing the code that we used in earlier lectures on the AMSS model

Here it is

In [4]:
import numpy as np
from scipy.optimize import root
from quantecon import MarkovChain

class SequentialAllocation:

'''
Class that takes CESutility or BGPutility object as input returns
planner's allocation as a function of the multiplier on the
implementability constraint μ.
'''

def __init__(self, model):

# Initialize from model object attributes
self.β, self.π, self.G = model.β, model.π, model.G
self.mc, self.Θ = MarkovChain(self.π), model.Θ
self.S = len(model.π)  # Number of states
self.model = model

# Find the first best allocation
self.find_first_best()

def find_first_best(self):
'''
Find the first best allocation
'''
model = self.model
S, Θ, G = self.S, self.Θ, self.G
Uc, Un = model.Uc, model.Un

def res(z):
c = z[:S]
n = z[S:]
return np.hstack([Θ * Uc(c, n) + Un(c, n), Θ * n - c - G])

res = root(res, 0.5 * np.ones(2 * S))

if not res.success:
raise Exception('Could not find first best')

self.cFB = res.x[:S]
self.nFB = res.x[S:]

# Multiplier on the resource constraint
self.ΞFB = Uc(self.cFB, self.nFB)
self.zFB = np.hstack([self.cFB, self.nFB, self.ΞFB])

def time1_allocation(self, μ):
'''
Computes optimal allocation for time t >= 1 for a given μ
'''
model = self.model
S, Θ, G = self.S, self.Θ, self.G
Uc, Ucc, Un, Unn = model.Uc, model.Ucc, model.Un, model.Unn

def FOC(z):
c = z[:S]
n = z[S:2 * S]
Ξ = z[2 * S:]
return np.hstack([Uc(c, n) - μ * (Ucc(c, n) * c + Uc(c, n)) - Ξ,          # FOC of c
Un(c, n) - μ * (Unn(c, n) * n + Un(c, n)) + \
Θ * Ξ,  # FOC of n
Θ * n - c - G])

# Find the root of the first-order condition
res = root(FOC, self.zFB)
if not res.success:
raise Exception('Could not find LS allocation.')
z = res.x
c, n, Ξ = z[:S], z[S:2 * S], z[2 * S:]

# Compute x
I = Uc(c, n) * c + Un(c, n) * n
x = np.linalg.solve(np.eye(S) - self.β * self.π, I)

return c, n, x, Ξ

def time0_allocation(self, B_, s_0):
'''
Finds the optimal allocation given initial government debt B_ and state s_0
'''
model, π, Θ, G, β = self.model, self.π, self.Θ, self.G, self.β
Uc, Ucc, Un, Unn = model.Uc, model.Ucc, model.Un, model.Unn

# First order conditions of planner's problem
def FOC(z):
μ, c, n, Ξ = z
xprime = self.time1_allocation(μ)[2]
return np.hstack([Uc(c, n) * (c - B_) + Un(c, n) * n + β * π[s_0] @ xprime,
Uc(c, n) - μ * (Ucc(c, n) *
(c - B_) + Uc(c, n)) - Ξ,
Un(c, n) - μ * (Unn(c, n) * n +
Un(c, n)) + Θ[s_0] * Ξ,
(Θ * n - c - G)[s_0]])

# Find root
res = root(FOC, np.array(
[0, self.cFB[s_0], self.nFB[s_0], self.ΞFB[s_0]]))
if not res.success:
raise Exception('Could not find time 0 LS allocation.')

return res.x

def time1_value(self, μ):
'''
Find the value associated with multiplier μ
'''
c, n, x, Ξ = self.time1_allocation(μ)
U = self.model.U(c, n)
V = np.linalg.solve(np.eye(self.S) - self.β * self.π, U)
return c, n, x, V

def Τ(self, c, n):
'''
Computes Τ given c, n
'''
model = self.model
Uc, Un = model.Uc(c, n), model.Un(c,  n)

return 1 + Un / (self.Θ * Uc)

def simulate(self, B_, s_0, T, sHist=None):
'''
Simulates planners policies for T periods
'''
model, π, β = self.model, self.π, self.β
Uc = model.Uc

if sHist is None:
sHist = self.mc.simulate(T, s_0)

cHist, nHist, Bhist, ΤHist, μHist = np.zeros((5, T))
RHist = np.zeros(T - 1)

# Time 0
μ, cHist[0], nHist[0], _ = self.time0_allocation(B_, s_0)
ΤHist[0] = self.Τ(cHist[0], nHist[0])[s_0]
Bhist[0] = B_
μHist[0] = μ

# Time 1 onward
for t in range(1, T):
c, n, x, Ξ = self.time1_allocation(μ)
Τ = self.Τ(c, n)
u_c = Uc(c, n)
s = sHist[t]
Eu_c = π[sHist[t - 1]] @ u_c
cHist[t], nHist[t], Bhist[t], ΤHist[t] = c[s], n[s], x[s] / \
u_c[s], Τ[s]
RHist[t - 1] = Uc(cHist[t - 1], nHist[t - 1]) / (β * Eu_c)
μHist[t] = μ

return np.array([cHist, nHist, Bhist, ΤHist, sHist, μHist, RHist])

In [5]:
from scipy.optimize import fmin_slsqp

class RecursiveAllocationAMSS:

def __init__(self, model, μgrid, tol_diff=1e-4, tol=1e-4):

self.β, self.π, self.G = model.β, model.π, model.G
self.mc, self.S = MarkovChain(self.π), len(model.π)  # Number of states
self.Θ, self.model, self.μgrid = model.Θ, model, μgrid
self.tol_diff, self.tol = tol_diff, tol

# Find the first best allocation
self.solve_time1_bellman()
self.T.time_0 = True  # Bellman equation now solves time 0 problem

def solve_time1_bellman(self):
'''
Solve the time  1 Bellman equation for calibration model and
initial grid μgrid0
'''
model, μgrid0 = self.model, self.μgrid
π = model.π
S = len(model.π)

# First get initial fit from Lucas Stokey solution.
# Need to change things to be ex ante
PP = SequentialAllocation(model)
interp = interpolator_factory(2, None)

def incomplete_allocation(μ_, s_):
c, n, x, V = PP.time1_value(μ_)
return c, n, π[s_] @ x, π[s_] @ V
cf, nf, xgrid, Vf, xprimef = [], [], [], [], []
for s_ in range(S):
c, n, x, V = zip(*map(lambda μ: incomplete_allocation(μ, s_), μgrid0))
c, n = np.vstack(c).T, np.vstack(n).T
x, V = np.hstack(x), np.hstack(V)
xprimes = np.vstack([x] * S)
cf.append(interp(x, c))
nf.append(interp(x, n))
Vf.append(interp(x, V))
xgrid.append(x)
xprimef.append(interp(x, xprimes))
cf, nf, xprimef = fun_vstack(cf), fun_vstack(nf), fun_vstack(xprimef)
Vf = fun_hstack(Vf)
policies = [cf, nf, xprimef]

# Create xgrid
x = np.vstack(xgrid).T
xbar = [x.min(0).max(), x.max(0).min()]
xgrid = np.linspace(xbar[0], xbar[1], len(μgrid0))
self.xgrid = xgrid

# Now iterate on Bellman equation
T = BellmanEquation(model, xgrid, policies, tol=self.tol)
diff = 1
while diff > self.tol_diff:
PF = T(Vf)

Vfnew, policies = self.fit_policy_function(PF)
diff = np.abs((Vf(xgrid) - Vfnew(xgrid)) / Vf(xgrid)).max()

print(diff)
Vf = Vfnew

# store value function policies and Bellman Equations
self.Vf = Vf
self.policies = policies
self.T = T

def fit_policy_function(self, PF):
'''
Fits the policy functions
'''
S, xgrid = len(self.π), self.xgrid
interp = interpolator_factory(3, 0)
cf, nf, xprimef, Tf, Vf = [], [], [], [], []
for s_ in range(S):
PFvec = np.vstack([PF(x, s_) for x in self.xgrid]).T
Vf.append(interp(xgrid, PFvec[0, :]))
cf.append(interp(xgrid, PFvec[1:1 + S]))
nf.append(interp(xgrid, PFvec[1 + S:1 + 2 * S]))
xprimef.append(interp(xgrid, PFvec[1 + 2 * S:1 + 3 * S]))
Tf.append(interp(xgrid, PFvec[1 + 3 * S:]))
policies = fun_vstack(cf), fun_vstack(
nf), fun_vstack(xprimef), fun_vstack(Tf)
Vf = fun_hstack(Vf)
return Vf, policies

def Τ(self, c, n):
'''
Computes Τ given c and n
'''
model = self.model
Uc, Un = model.Uc(c, n), model.Un(c, n)

return 1 + Un / (self.Θ * Uc)

def time0_allocation(self, B_, s0):
'''
Finds the optimal allocation given initial government debt B_ and
state s_0
'''
PF = self.T(self.Vf)
z0 = PF(B_, s0)
c0, n0, xprime0, T0 = z0[1:]
return c0, n0, xprime0, T0

def simulate(self, B_, s_0, T, sHist=None):
'''
Simulates planners policies for T periods
'''
model, π = self.model, self.π
Uc = model.Uc
cf, nf, xprimef, Tf = self.policies

if sHist is None:
sHist = simulate_markov(π, s_0, T)

cHist, nHist, Bhist, xHist, ΤHist, THist, μHist = np.zeros((7, T))
# time 0
cHist[0], nHist[0], xHist[0], THist[0] = self.time0_allocation(B_, s_0)
ΤHist[0] = self.Τ(cHist[0], nHist[0])[s_0]
Bhist[0] = B_
μHist[0] = self.Vf[s_0](xHist[0])

# time 1 onward
for t in range(1, T):
s_, x, s = sHist[t - 1], xHist[t - 1], sHist[t]
c, n, xprime, T = cf[s_, :](x), nf[s_, :](
x), xprimef[s_, :](x), Tf[s_, :](x)

Τ = self.Τ(c, n)[s]
u_c = Uc(c, n)
Eu_c = π[s_, :] @ u_c

μHist[t] = self.Vf[s](xprime[s])

cHist[t], nHist[t], Bhist[t], ΤHist[t] = c[s], n[s], x / Eu_c, Τ
xHist[t], THist[t] = xprime[s], T[s]
return np.array([cHist, nHist, Bhist, ΤHist, THist, μHist, sHist, xHist])

class BellmanEquation:
'''
Bellman equation for the continuation of the Lucas-Stokey Problem
'''

def __init__(self, model, xgrid, policies0, tol, maxiter=1000):

self.β, self.π, self.G = model.β, model.π, model.G
self.S = len(model.π)  # Number of states
self.Θ, self.model, self.tol = model.Θ, model, tol
self.maxiter = maxiter

self.xbar = [min(xgrid), max(xgrid)]
self.time_0 = False

self.z0 = {}
cf, nf, xprimef = policies0

for s_ in range(self.S):
for x in xgrid:
self.z0[x, s_] = np.hstack([cf[s_, :](x),
nf[s_, :](x),
xprimef[s_, :](x),
np.zeros(self.S)])

self.find_first_best()

def find_first_best(self):
'''
Find the first best allocation
'''
model = self.model
S, Θ, Uc, Un, G = self.S, self.Θ, model.Uc, model.Un, self.G

def res(z):
c = z[:S]
n = z[S:]
return np.hstack([Θ * Uc(c, n) + Un(c, n), Θ * n - c - G])

res = root(res, 0.5 * np.ones(2 * S))
if not res.success:
raise Exception('Could not find first best')

self.cFB = res.x[:S]
self.nFB = res.x[S:]
IFB = Uc(self.cFB, self.nFB) * self.cFB + \
Un(self.cFB, self.nFB) * self.nFB

self.xFB = np.linalg.solve(np.eye(S) - self.β * self.π, IFB)

self.zFB = {}
for s in range(S):
self.zFB[s] = np.hstack(
[self.cFB[s], self.nFB[s], self.π[s] @ self.xFB, 0.])

def __call__(self, Vf):
'''
Given continuation value function next period return value function this
period return T(V) and optimal policies
'''
if not self.time_0:
def PF(x, s): return self.get_policies_time1(x, s, Vf)
else:
def PF(B_, s0): return self.get_policies_time0(B_, s0, Vf)
return PF

def get_policies_time1(self, x, s_, Vf):
'''
Finds the optimal policies
'''
model, β, Θ, G, S, π = self.model, self.β, self.Θ, self.G, self.S, self.π
U, Uc, Un = model.U, model.Uc, model.Un

def objf(z):
c, n, xprime = z[:S], z[S:2 * S], z[2 * S:3 * S]

Vprime = np.empty(S)
for s in range(S):
Vprime[s] = Vf[s](xprime[s])

return -π[s_] @ (U(c, n) + β * Vprime)

def cons(z):
c, n, xprime, T = z[:S], z[S:2 * S], z[2 * S:3 * S], z[3 * S:]
u_c = Uc(c, n)
Eu_c = π[s_] @ u_c
return np.hstack([
x * u_c / Eu_c - u_c * (c - T) - Un(c, n) * n - β * xprime,
Θ * n - c - G])

if model.transfers:
bounds = [(0., 100)] * S + [(0., 100)] * S + \
[self.xbar] * S + [(0., 100.)] * S
else:
bounds = [(0., 100)] * S + [(0., 100)] * S + \
[self.xbar] * S + [(0., 0.)] * S
out, fx, _, imode, smode = fmin_slsqp(objf, self.z0[x, s_],
f_eqcons=cons, bounds=bounds,
full_output=True, iprint=0,
acc=self.tol, iter=self.maxiter)

if imode > 0:
raise Exception(smode)

self.z0[x, s_] = out
return np.hstack([-fx, out])

def get_policies_time0(self, B_, s0, Vf):
'''
Finds the optimal policies
'''
model, β, Θ, G = self.model, self.β, self.Θ, self.G
U, Uc, Un = model.U, model.Uc, model.Un

def objf(z):
c, n, xprime = z[:-1]

return -(U(c, n) + β * Vf[s0](xprime))

def cons(z):
c, n, xprime, T = z
return np.hstack([
-Uc(c, n) * (c - B_ - T) - Un(c, n) * n - β * xprime,
(Θ * n - c - G)[s0]])

if model.transfers:
bounds = [(0., 100), (0., 100), self.xbar, (0., 100.)]
else:
bounds = [(0., 100), (0., 100), self.xbar, (0., 0.)]
out, fx, _, imode, smode = fmin_slsqp(objf, self.zFB[s0], f_eqcons=cons,
bounds=bounds, full_output=True, iprint=0)

if imode > 0:
raise Exception(smode)

return np.hstack([-fx, out])

In [6]:
from scipy.interpolate import UnivariateSpline

class interpolate_wrapper:

def __init__(self, F):
self.F = F

def __getitem__(self, index):
return interpolate_wrapper(np.asarray(self.F[index]))

def reshape(self, *args):
self.F = self.F.reshape(*args)
return self

def transpose(self):
self.F = self.F.transpose()

def __len__(self):
return len(self.F)

def __call__(self, xvec):
x = np.atleast_1d(xvec)
shape = self.F.shape
if len(x) == 1:
fhat = np.hstack([f(x) for f in self.F.flatten()])
return fhat.reshape(shape)
else:
fhat = np.vstack([f(x) for f in self.F.flatten()])
return fhat.reshape(np.hstack((shape, len(x))))

class interpolator_factory:

def __init__(self, k, s):
self.k, self.s = k, s

def __call__(self, xgrid, Fs):
shape, m = Fs.shape[:-1], Fs.shape[-1]
Fs = Fs.reshape((-1, m))
F = []
xgrid = np.sort(xgrid)  # Sort xgrid
for Fhat in Fs:
F.append(UnivariateSpline(xgrid, Fhat, k=self.k, s=self.s))
return interpolate_wrapper(np.array(F).reshape(shape))

def fun_vstack(fun_list):

Fs = [IW.F for IW in fun_list]
return interpolate_wrapper(np.vstack(Fs))

def fun_hstack(fun_list):

Fs = [IW.F for IW in fun_list]
return interpolate_wrapper(np.hstack(Fs))

def simulate_markov(π, s_0, T):

sHist = np.empty(T, dtype=int)
sHist[0] = s_0
S = len(π)
for t in range(1, T):
sHist[t] = np.random.choice(np.arange(S), p=π[sHist[t - 1]])

return sHist


Next, we show the code that we use to generate a very long simulation starting from initial government debt equal to $-.5$

Here is a graph of a long simulation of 102000 periods

In [7]:
import matplotlib.pyplot as plt
%matplotlib inline

μ_grid = np.linspace(-0.09, 0.1, 100)

log_example = CRRAutility(π=(1 / 3) * np.ones((3, 3)),
G=np.array([0.1, 0.2, .3]),
Θ=np.ones(3))

log_example.transfers = True                        # Government can use transfers
log_sequential = SequentialAllocation(log_example)  # Solve sequential problem
log_bellman = RecursiveAllocationAMSS(log_example, μ_grid,
tol=1e-12, tol_diff=1e-10)

T = 102000  # Set T to 102000 periods

sim_seq_long = log_sequential.simulate(0.5, 0, T)
sHist_long = sim_seq_long[-3]
sim_bel_long = log_bellman.simulate(0.5, 0, T, sHist_long)

titles = ['Government Debt', 'Tax Rate']

fig, axes = plt.subplots(2, 1, figsize=(10, 8))

for ax, title, id in zip(axes.flatten(), titles, [2, 3]):
ax.plot(sim_seq_long[id], '-k', sim_bel_long[id], '-.b', alpha=0.5)
ax.set(title=title)
ax.grid()

axes[0].legend(('Complete Markets', 'Incomplete Markets'))
plt.tight_layout()
plt.show()

/home/qebuild/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:24: RuntimeWarning: divide by zero encountered in reciprocal
/home/qebuild/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:29: RuntimeWarning: divide by zero encountered in power
/home/qebuild/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:232: RuntimeWarning: invalid value encountered in true_divide
/home/qebuild/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:232: RuntimeWarning: invalid value encountered in multiply

0.03826635338765925
0.0015144378246584984
0.0013387575049829455
0.0011833202399953704
0.0010600307116151308
0.0009506620325028087
0.0008518776516937746
0.0007625857030716029
0.0006819563061621401
0.0006094002926927259
0.0005443007358227137
0.0004859950035124384
0.00043383959352032413
0.00038722730861434493
0.000345595412214899
0.0003084287064063272
0.0002752590187094664
0.0002456631291600592
0.00021925988530998263
0.00019570695817042554
0.00017469751640521595
0.0001559569713071983
0.00013923987965085293
0.00012432704760933488
0.00011102285952965586
9.915283206803345e-05
8.856139174858334e-05
7.91098648574037e-05
7.067466535012738e-05
6.31456673681484e-05
5.6424746008860264e-05
5.042447143154252e-05
4.506694212534692e-05
4.028274355430257e-05
3.601001918083999e-05
3.2193642882531256e-05
2.878448111493858e-05
2.5738738819018375e-05
2.301736976750311e-05
2.0585562762952467e-05
1.841227366505203e-05
1.647009732636953e-05
1.4734148263778101e-05
1.3182214397654561e-05
1.1794654663586968e-05
1.0553942919813837e-05
9.444436170445705e-06
8.452171096119784e-06
7.564681527564076e-06
6.770836691014705e-06
6.0606991281269e-06
5.425387729296574e-06
4.856977427893397e-06
4.348382669160568e-06
3.893276412835248e-06
3.4860031510823107e-06
3.1215109737669223e-06
2.795284109545752e-06
2.503284080753522e-06
2.241904849713046e-06
2.0079207043630637e-06
1.7984473598229776e-06
1.6109043156289632e-06
1.4429883335786674e-06
1.2926350820537814e-06
1.1580014056712184e-06
1.037436438388734e-06
9.294649648188667e-07
8.3276668236914e-07
7.461586315970762e-07
6.685859440207697e-07
5.991018791164966e-07
5.36860205044418e-07
4.811036780956593e-07
4.311540879734326e-07
3.864052975497157e-07
3.4631272976818724e-07
3.103916419559902e-07
2.7820604666234584e-07
2.493665757525022e-07
2.235242330547715e-07
2.0036662752964725e-07
1.7961406916271733e-07
1.6101610597963257e-07
1.4434841355653347e-07
1.2941010571693734e-07
1.1602128543466011e-07
1.0402082434646952e-07
9.326441852343976e-08
8.362274988135493e-08
7.49799939308504e-08
6.723239527266927e-08
6.028702608399241e-08
5.406062550179954e-08
4.847860158085037e-08
4.347411399612939e-08
3.898727291456688e-08
3.496441211560151e-08
3.135744827221921e-08
2.8123291587559614e-08
2.5223328515894898e-08
2.26229541381511e-08
2.0291155429696614e-08
1.8200137067287912e-08
1.632498637211791e-08
1.464337367650618e-08
1.313528409236892e-08
1.1782776814280018e-08
1.056978340425791e-08
9.481875293010455e-09
8.506129973427988e-09
7.630960687058812e-09
6.845981534564069e-09
6.141882022205125e-09
5.510312745970958e-09
4.943790102561541e-09
4.435605224917734e-09
3.979784496747062e-09
3.570876548574359e-09
3.204046477022082e-09
2.8749568316999748e-09
2.5797181798724397e-09
2.3148435453620055e-09
2.0772039322334064e-09
1.8639959361200393e-09
1.6727033117204262e-09
1.5010698530901528e-09
1.347071344111145e-09
1.2088951024813536e-09
1.0849106904200139e-09
9.736599966038641e-10
8.738333708142846e-10
7.842541269558106e-10
7.038711045065831e-10
6.317371277104943e-10
5.670057512125597e-10
5.08915453134878e-10
4.5678390119330306e-10
4.0999973929598685e-10
3.6801375426220425e-10
3.303311273300117e-10
2.965134487651615e-10
2.6616089632731117e-10
2.3891982879379247e-10
2.1446961352153604e-10
1.9252572584463677e-10
1.7282859124437483e-10
1.5514900929464294e-10
1.392805601387578e-10
1.250371249717539e-10
1.1225226616102369e-10
1.0077473312033485e-10
9.04737312852764e-11


The long simulation apparently indicates eventual convergence to an ergodic distribution

It takes about 1000 periods to reach the ergodic distribution – an outcome that is forecast by approximations to rates of convergence that appear in [BEGS17] and that we discuss in a previous lecture

We discard the first 2000 observations of the simulation and construct the histogram of the part value of government debt

We obtain the following graph for the histogram of the last 100,000 observations on the par value of government debt

The black vertical line denotes the sample mean for the last 100,000 observations included in the histogram; the green vertical line denotes the value of $\frac{ {\mathcal B}^*}{E u_c}$, associated with the sample (presumably) from the ergodic where ${\mathcal B}^*$ is the regression coefficient described below; the red vertical line denotes an approximation by [BEGS17] to the mean of the ergodic distribution that can be precomputed before sampling from the ergodic distribution, as described below

Before moving on to discuss the histogram and the vertical lines approximating the ergodic mean of government debt in more detail, the following graphs show government debt and taxes early in the simulation, for periods 1-100 and 101 to 200 respectively

In [8]:
titles = ['Government Debt', 'Tax Rate']

fig, axes = plt.subplots(4, 1, figsize=(10, 15))

for i, id in enumerate([2, 3]):
axes[i].plot(sim_seq_long[id][:99], '-k', sim_bel_long[id][:99], '-.b', alpha=0.5)
axes[i+2].plot(range(100, 199), sim_seq_long[id][100:199], '-k',
range(100, 199), sim_bel_long[id][100:199], '-.b', alpha=0.5)
axes[i].set(title=titles[i])
axes[i+2].set(title=titles[i])
axes[i].grid()
axes[i+2].grid()

axes[0].legend(('Complete Markets', 'Incomplete Markets'))
plt.tight_layout()
plt.show()


For the short samples early in our simulated sample of 102,000 observations, fluctuations in government debt and the tax rate conceal the weak but inexorable force that the Ramsey planner puts into both series driving them toward ergodic distributions far from these early observations

• early observations are more influenced by the initial value of the par value of government debt than by the ergodic mean of the par value of government debt
• much later observations are more influenced by the ergodic mean and are independent of the initial value of the par value of government debt

## Asymptotic Mean and Rate of Convergence¶

We apply the results of [BEGS17] to interpret

• the mean of the ergodic distribution of government debt
• the rate of convergence to the ergodic distribution from an arbitrary initial government debt

We begin by computing objects required by the theory of section III.i of [BEGS17]

As in Fiscal Insurance via Fluctuating Interest Rates, we recall that [BEGS17] used a particular notation to represent what we can regard as a generalization of the AMSS model

We introduce some of the [BEGS17] notation so that readers can quickly relate notation that appears in their key formulas to the notation that we have used in previous lectures here and here

BEGS work with objects $B_t, {\mathcal B}_t, {\mathcal R}_t, {\mathcal X}_t$ that are related to notation that we used in earlier lectures by

\begin{aligned} {\mathcal R}_t & = \frac{u_{c,t}}{u_{c,t-1}} R_{t-1} = \frac{u_{c,t}}{ \beta E_{t-1} u_{c,t}} \\ B_t & = \frac{b_{t+1}(s^t)}{R_t(s^t)} \\ b_t(s^{t-1}) & = {\mathcal R}_{t-1} B_{t-1} \\ {\mathcal B}_t & = u_{c,t} B_t = (\beta E_t u_{c,t+1}) b_{t+1}(s^t) \\ {\mathcal X}_t & = u_{c,t} [g_t - \tau_t n_t] \end{aligned}

[BEGS17] call ${\mathcal X}_t$ the effective government deficit, and ${\mathcal B}_t$ the effective government debt

Equation (44) of [BEGS17] expresses the time $t$ state $s$ government budget constraint as

$${\mathcal B}(s) = {\mathcal R}_\tau(s, s_{-}) {\mathcal B}_{-} + {\mathcal X}_{\tau} (s) \tag{1}$$

where the dependence on $\tau$ is to remind us that these objects depend on the tax rate; $s_{-}$ is last period’s Markov state

BEGS interpret random variations in the right side of (1) as fiscal risks generated by

• interest-rate-driven fluctuations in time $t$ effective payments due on the government portfolio, namely, ${\mathcal R}_\tau(s, s_{-}) {\mathcal B}_{-}$, and
• fluctuations in the effective government deficit ${\mathcal X}_t$

### Asymptotic Mean¶

BEGS give conditions under which the ergodic mean of ${\mathcal B}_t$ approximately satisfies the equation

$${\mathcal B}^* = - \frac{\rm cov^{\infty}({\mathcal R}_t, {\mathcal X_t})}{\rm var^{\infty}({\mathcal R}_t)} \tag{2}$$

where the superscript $\infty$ denotes a moment taken with respect to an ergodic distribution

Formula (2) represents ${\mathcal B}^*$ as a regression coefficient of ${\mathcal X}_t$ on ${\mathcal R}_t$ in the ergodic distribution

Regression coefficient ${\mathcal B}^*$ solves a variance-minimization problem:

$${\mathcal B}^* = {\rm argmin}_{\mathcal B} {\rm var}^\infty ({\mathcal R} {\mathcal B} + {\mathcal X}) \tag{3}$$

The minimand in criterion (3) measures fiscal risk associated with a given tax-debt policy that appears on the right side of equation (1)

Expressing formula (2) in terms of our notation tells us that the ergodic mean of the par value $b$ of government debt in the AMSS model should approximately equal

$$\hat b = \frac{\mathcal B^*}{\beta E( E_t u_{c,t+1})} = \frac{\mathcal B^*}{\beta E( u_{c,t+1} )} \tag{4}$$

where mathematical expectations are taken with respect to the ergodic distribution

### Rate of Convergence¶

BEGS also derive the following approximation to the rate of convergence to ${\mathcal B}^{*}$ from an arbitrary initial condition

$$\frac{ E_t ( {\mathcal B}_{t+1} - {\mathcal B}^{*} )} { ( {\mathcal B}_{t} - {\mathcal B}^{*} )} \approx \frac{1}{1 + \beta^2 {\rm var}^\infty ({\mathcal R} )} \tag{5}$$

(See the equation above equation (47) in [BEGS17])

The remainder of this lecture is about technical material based on formulas from [BEGS17]

The topic is interpreting and extending formula (3) for the ergodic mean ${\mathcal B}^*$

### Chicken and Egg¶

Attributes of the ergodic distribution for ${\mathcal B}_t$ appear on the right side of formula (3) for the ergodic mean ${\mathcal B}^*$

Thus, formula (3) is not useful for estimating the mean of the ergodic in advance of actually computing the ergodic distribution

• we need to know the ergodic distribution to compute the right side of formula (3)

So the primary use of equation (3) is how it confirms that the ergodic distribution solves a fiscal-risk minimization problem

As an example, notice how we used the formula for the mean of ${\mathcal B}$ in the ergodic distribution of the special AMSS economy in Fiscal Insurance via Fluctuating Interest Rates

• first we computed the ergodic distribution using a reverse-engineering construction
• then we verified that ${\mathcal B}$ agrees with the mean of that distribution

### Approximating the Ergodic Mean¶

[BEGS17] propose an approximation to ${\mathcal B}^*$ that can be computed without first knowing the ergodic distribution

To construct the BEGS approximation to ${\mathcal B}^*$, we just follow steps set forth on pages 648 - 650 of section III.D of [BEGS17]

• notation in BEGS might be confusing at first sight, so it is important to stare and digest before computing
• there are also some sign errors in the [BEGS17] text that we’ll want to correct

Here is a step-by-step description of the [BEGS17] approximation procedure

### Step by Step¶

Step 1: For a given $\tau$ we compute a vector of values $c_\tau(s), s= 1, 2, \ldots, S$ that satisfy

$$(1-\tau) c_\tau(s)^{-\sigma} - (c_{\tau}(s) + g(s))^{\gamma} = 0$$

This is a nonlinear equation to be solved for $c_{\tau}(s), s = 1, \ldots, S$

$S=3$ in our case, but we’ll write code for a general integer $S$

Typo alert: Please note that there is a sign error in equation (42) of [BEGS17] – it should be a minus rather than a plus in the middle

• We have made the appropriate correction in the above equation

Step 2: Knowing $c_\tau(s), s=1, \ldots, S$ for a given $\tau$, we want to compute the random variables

$${\mathcal R}_\tau(s) = \frac{c_\tau(s)^{-\sigma}}{\beta \sum_{s'=1}^S c_\tau(s')^{-\sigma} \pi(s')}$$

and

$${\mathcal X}_\tau(s) = (c_\tau(s) + g(s))^{1+ \gamma} - c_\tau(s)^{1-\sigma}$$

each for $s= 1, \ldots, S$

BEGS call ${\mathcal R}_\tau(s)$ the effective return on risk-free debt and they call ${\mathcal X}_\tau(s)$ the effective government deficit

Step 3: With the preceding objects in hand, for a given ${\mathcal B}$, we seek a $\tau$ that satisfies

$${\mathcal B} = - \frac{\beta} {1-\beta} E {\mathcal X_\tau} \equiv - \frac{\beta} {1-\beta} \sum_{s} {\mathcal X}_\tau(s) \pi(s)$$

This equation says that at a constant discount factor $\beta$, equivalent government debt ${\mathcal B}$ equals the present value of the mean effective government surplus

Typo alert: there is a sign error in equation (46) of [BEGS17] –the left side should be multiplied by $-1$

• We have made this correction in the above equation

For a given ${\mathcal B}$, let a $\tau$ that solves the above equation be called $\tau(\mathcal B)$

We’ll use a Python root solver to finds a $\tau$ that this equation for a given ${\mathcal B}$

We’ll use this function to induce a function $\tau({\mathcal B})$

Step 4: With a Python program that computes $\tau(\mathcal B)$ in hand, next we write a Python function to compute the random variable

$$J({\mathcal B})(s) = \mathcal R_{\tau({\mathcal B})}(s) {\mathcal B} + {\mathcal X}_{\tau({\mathcal B})}(s) , \quad s = 1, \ldots, S$$

Step 5: Now that we have a machine to compute the random variable $J({\mathcal B})(s), s= 1, \ldots, S$, via a composition of Python functions, we can use the population variance function that we defined in the code above to construct a function ${\rm var}(J({\mathcal B}))$

We put ${\rm var}(J({\mathcal B}))$ into a function minimizer and compute

$${\mathcal B}^* = {\rm argmin}_{\mathcal B} {\rm var } (J({\mathcal B}) )$$

Step 6: Next we take the minimizer ${\mathcal B}^*$ and the Python functions for computing means and variances and compute

$${\rm rate} = \frac{1}{1 + \beta^2 {\rm var}( {\mathcal R}_{\tau({\mathcal B}^*)} )}$$

Ultimate outputs of this string of calculations are two scalars

$$({\mathcal B}^*, {\rm rate} )$$

Step 7: Compute the divisor

$$div = {\beta E u_{c,t+1}}$$

and then compute the mean of the par value of government debt in the AMSS model

$$\hat b = \frac{ {\mathcal B}^*}{div}$$

In the two-Markov-state AMSS economy in Fiscal Insurance via Fluctuating Interest Rates, $E_t u_{c,t+1} = E u_{c,t+1}$ in the ergodic distribution and we have confirmed that this formula very accurately describes a constant par value of government debt that

• supports full fiscal insurance via fluctuating interest parameters, and
• is the limit of government debt as $t \rightarrow +\infty$

In the three-Markov-state economy of this lecture, the par value of government debt fluctuates in a history-dependent way even asymptotically

In this economy, $\hat b$ given by the above formula approximates the mean of the ergodic distribution of the par value of government debt

• this is the red vertical line plotted in the histogram of the last 100,000 observations of our simulation of the par value of government debt plotted above
• the approximation is fairly accurate but not perfect
• so while the approximation circumvents the chicken and egg problem surrounding the much better approximation associated with the green vertical line, it does so by enlarging the approximation error

### Execution¶

Now let’s move on to compute things step by step

#### Step 1¶

In [9]:
u = CRRAutility(π=(1 / 3) * np.ones((3, 3)),
G=np.array([0.1, 0.2, .3]),
Θ=np.ones(3))

τ = 0.05           # Initial guess of τ (to displays calcs along the way)
S = len(u.G)       # Number of states

def solve_c(c, τ, u):
return (1 - τ) * c**(-u.σ) - (c + u.G)**u.γ

c = root(solve_c, np.ones(S), args=(τ, u)).x  # .x returns the result from root
c

Out[9]:
array([0.93852387, 0.89231015, 0.84858872])
In [10]:
root(solve_c, np.ones(S), args=(τ, u))

Out[10]:
    fjac: array([[-0.99990816, -0.00495351, -0.01261467],
[-0.00515633,  0.99985715,  0.01609659],
[-0.01253313, -0.01616015,  0.99979086]])
fun: array([ 5.61814373e-10, -4.76900741e-10,  1.17474919e-11])
message: 'The solution converged.'
nfev: 11
qtf: array([1.55568331e-08, 1.28322481e-08, 7.89913426e-11])
r: array([ 4.26943131,  0.08684775, -0.06300593, -4.71278821, -0.0743338 ,
-5.50778548])
status: 1
success: True
x: array([0.93852387, 0.89231015, 0.84858872])

#### Step 2¶

In [11]:
n = c + u.G   # compute labor supply


Remember that in our code $\pi$ is a $3 \times 3$ transition matrix

But because we are studying an IID case, $\pi$ has identical rows and we only need to compute objects for one row of $\pi$

This explains why at some places below we set $s=0$ just to pick off the first row of $\pi$ in the calculations

### Code¶

First, let’s compute ${\mathcal R}$ and ${\mathcal X}$ according to our formulas

In [12]:
def compute_R_X(τ, u, s):
c = root(solve_c, np.ones(S), args=(τ, u)).x  # Solve for vector of c's
div = u.β * (u.Uc(c[0], n[0]) * u.π[s, 0]  +  u.Uc(c[1], n[1]) * u.π[s, 1] +  u.Uc(c[2], n[2]) * u.π[s, 2])
R = c**(-u.σ) / (div)
X = (c + u.G)**(1 + u.γ) - c**(1 - u.σ)
return R, X

In [13]:
c**(-u.σ) @ u.π

Out[13]:
array([1.25997521, 1.25997521, 1.25997521])
In [14]:
u.π

Out[14]:
array([[0.33333333, 0.33333333, 0.33333333],
[0.33333333, 0.33333333, 0.33333333],
[0.33333333, 0.33333333, 0.33333333]])

We only want unconditional expectations because we are in an IID case

So we’ll set $s=0$ and just pick off expectations associated with the first row of $\pi$

In [15]:
s = 0

R, X = compute_R_X(τ, u, s)


Let’s look at the random variables ${\mathcal R}, {\mathcal X}$

In [16]:
R

Out[16]:
array([1.00116313, 1.10755123, 1.22461897])
In [17]:
mean(R, s)

Out[17]:
1.1111111111111112
In [18]:
X

Out[18]:
array([0.05457803, 0.18259396, 0.33685546])
In [19]:
mean(X, s)

Out[19]:
0.19134248445303795
In [20]:
X @ u.π

Out[20]:
array([0.19134248, 0.19134248, 0.19134248])

#### Step 3¶

In [21]:
def solve_τ(τ, B, u, s):
R, X = compute_R_X(τ, u, s)
return ((u.β - 1) / u.β) * B - X @ u.π[s]


Note that $B$ is a scalar

Let’s try out our method computing $\tau$

In [22]:
s = 0
B = 1.0

τ = root(solve_τ, .1, args=(B, u, s)).x[0]  # Very sensitive to starting value
τ

Out[22]:
0.2740159773695818

In the above cell, B is fixed at 1 and $\tau$ is to be computed as a function of B

Note that 0.2 is the initial value for $\tau$ in the root-finding algorithm

#### Step 4¶

In [23]:
def min_J(B, u, s):
τ = root(solve_τ, .5, args=(B, u, s)).x[0]  # very sensitive to initial value of τ
R, X = compute_R_X(τ, u, s)
return variance(R * B + X, s)

In [24]:
min_J(B, u, s)

Out[24]:
0.035564405653720765

#### Step 6¶

In [25]:
from scipy.optimize import minimize

B_star = minimize(min_J, .5, args=(u, s)).x[0]
B_star

Out[25]:
-1.199483167941158
In [26]:
n = c + u.G  # compute labor supply

In [27]:
div = u.β * (u.Uc(c[0], n[0]) * u.π[s, 0]  +  u.Uc(c[1], n[1]) * u.π[s, 1] +  u.Uc(c[2], n[2]) * u.π[s, 2])

In [28]:
B_hat = B_star/div
B_hat

Out[28]:
-1.0577661126390971
In [29]:
τ_star = root(solve_τ, 0.05, args=(B_star, u, s)).x[0]
τ_star

Out[29]:
0.09572916798461703
In [30]:
R_star, X_star = compute_R_X(τ_star, u, s)
R_star, X_star

Out[30]:
(array([0.9998398 , 1.10746593, 1.2260276 ]),
array([0.0020272 , 0.12464752, 0.27315299]))
In [31]:
rate = 1 / (1 + u.β**2 * variance(R_star, s))
rate

Out[31]:
0.9931353432732218
In [32]:
root(solve_c, np.ones(S), args=(τ_star, u)).x

Out[32]:
array([0.9264382 , 0.88027117, 0.83662635])
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