In [1]:
import numpy as np
import pandas as pd
import linearsolve as ls
import matplotlib.pyplot as plt
plt.style.use('classic')
%matplotlib inline
linearsolve Doeslinearsolve defines a class - linearsolve.model - with several functions for approximating, solving, and simulating dynamic stochastic general equilibrium (DSGE) models. The equilibrium conditions for most DSGE models can be expressed as a vector function $f$:
where 0 is an $n\times 1$ vector of zeros, $X_t$ is an $n\times 1$ vector of endogenous variables, and $\epsilon_{t+1}$ is an $m\times 1$ vector of exogenous structural shocks to the model. $E_tX_{t+1}$ denotes the expecation of the $t+1$ endogenous variables based on the information available to agents in the model as of time period $t$.
linearsolve.model has methods for computing linear and log-linear approximations of the model in equation (1) and methods for solving and simulating the linear model.
This package defines a linearsolve.model class. An instance of the linearsolve.model has the following methods:
approximate_and_solve(log_linear=True): Approximates and solves the model by combining the previous two methods.approximated(self,round=True,precision=4): Returns a string containing the approximation to the equilibrium conditions of the model.check_ss(): Test whether the computed steady state satisfies the model's equilbrium conditions.compute_ss(,guess=None,method='fsolve',options={}): Computes the steady state of the nonlinear model.impulse(self,T=51,t0=1,shocks=None,percent=False,diff=True): Computes impulse responses for shocks to each endogenous state variable.linear_approximation(self,steady_state=None): Computes a linear approximation of the model.log_linear_approximation(self,steady_state=None): Computes a log-linear approximation of the model.set_ss(self,steady_state): Sets the steady state .ss attribute of the instance.solve_klein(self,a=None,b=None): Solves the linear model using Klein's solution method.solved(self,round=True,precision=4): Returns a string containing the solution to the approximated model.stoch_sim(self,T=51,drop_first=300,cov_mat=None,seed=None,percent=False,diff=True): Computes a stochastic simulation of the model.Here I demonstrate how how relatively straightforward it is to appoximate, solve, and simulate a DSGE model using linearsolve. In the example that follows, I describe the procedure more carefully.
In the block of code that immediately follows, I input the model, solve for the steady state, compute the log-linear approximation of the equilibirum conditions, and compute some impulse responses following a shock to technology $A_t$.
In [2]:
# Input model parameters
parameters = pd.Series()
parameters['alpha'] = .35
parameters['beta'] = 0.99
parameters['delta'] = 0.025
parameters['rhoa'] = .9
parameters['sigma'] = 1.5
parameters['A'] = 1
# Funtion that evaluates the equilibrium conditions
def equilibrium_equations(variables_forward,variables_current,parameters):
# Parameters
p = parameters
# Variables
fwd = variables_forward
cur = variables_current
# Household Euler equation
euler_eqn = p.beta*fwd.c**-p.sigma*(p.alpha*cur.a*fwd.k**(p.alpha-1)+1-p.delta) - cur.c**-p.sigma
# Goods market clearing
market_clearing = cur.c + fwd.k - (1-p.delta)*cur.k - cur.a*cur.k**p.alpha
# Exogenous technology
# technology_proc = cur.a**p.rhoa - fwd.a
technology_proc = p.rhoa*np.log(cur.a) - np.log(fwd.a)
# Stack equilibrium conditions into a numpy array
return np.array([
euler_eqn,
market_clearing,
technology_proc
])
# Initialize the model
model = ls.model(equations = equilibrium_equations,
n_states=2,
var_names=['a','k','c'],
shock_names=['e_a','e_k'],
parameters = parameters)
# Compute the steady state numerically
guess = [1,1,1]
model.compute_ss(guess)
# model.ss
# Find the log-linear approximation around the non-stochastic steady state and solve
model.approximate_and_solve()
# Compute impulse responses and plot
model.impulse(T=41,t0=5,shocks=None)
fig = plt.figure(figsize=(12,4))
ax1 =fig.add_subplot(1,2,1)
model.irs['e_a'][['a','k','c']].plot(lw='5',alpha=0.5,grid=True,ax=ax1).legend(loc='upper right',ncol=3)
ax2 =fig.add_subplot(1,2,2)
model.irs['e_a'][['e_a','a']].plot(lw='5',alpha=0.5,grid=True,ax=ax2).legend(loc='upper right',ncol=2)
Out[2]:
Consider the equilibrium conditions for a basic RBC model without labor:
\begin{align} C_t^{-\sigma} & = \beta E_t \left[C_{t+1}^{-\sigma}(\alpha A_{t+1} K_{t+1}^{\alpha-1} + 1 - \delta)\right]\\ Y_t & = A_t K_t^{\alpha}\\ I_t & = K_{t+1} - (1-\delta)K_t\\ Y_t & = C_t + I_t\\ \log A_t & = \rho_a \log A_{t-1} + \epsilon_t \end{align}In the nonstochastic steady state, we have:
\begin{align} K & = \left(\frac{\alpha A}{1/\beta+\delta-1}\right)^{\frac{1}{1-\alpha}}\\ Y & = AK^{\alpha}\\ I & = \delta K\\ C & = Y - I \end{align}Given values for the parameters $\beta$, $\sigma$, $\alpha$, $\delta$, and $A$, steady state values of capital, output, investment, and consumption are easily computed.
linearsolveTo initialize the model, we need to first set the model's parameters. We do this by creating a Pandas Series variable called parameters:
In [3]:
# Input model parameters
parameters = pd.Series()
parameters['alpha'] = .35
parameters['beta'] = 0.99
parameters['delta'] = 0.025
parameters['rhoa'] = .9
parameters['sigma'] = 1.5
parameters['A'] = 1
Next, we need to define a function that returns the equilibrium conditions of the model. The function will take as inputs two vectors: one vector of "current" variables and another of "forward-looking" or one-period-ahead variables. The function will return an array that represents the equilibirum conditions of the model. We'll enter each equation with all variables moved to one side of the equals sign. For example, here's how we'll enter the produciton fucntion:
production_function = technology_current*capital_current**alpha - output_curent
Here the variable production_function stores the production function equation set equal to zero. We can enter the equations in almost any way we want. For example, we could also have entered the production function this way:
production_function = 1 - output_curent/technology_current/capital_current**alpha
One more thing to consider: the natural log in the equation describing the evolution of total factor productivity will create problems for the solution routine later on. So rewrite the equation as:
\begin{align} A_{t+1} & = A_{t}^{\rho_a}e^{\epsilon_{t+1}}\\ \end{align}So the complete system of equations that we enter into the program looks like:
\begin{align} C_t^{-\sigma} & = \beta E_t \left[C_{t+1}^{-\sigma}(\alpha Y_{t+1} /K_{t+1}+ 1 - \delta)\right]\\ Y_t & = A_t K_t^{\alpha}\\ I_t & = K_{t+1} - (1-\delta)K_t\\ Y_t & = C_t + I_t\\ A_{t+1} & = A_{t}^{\rho_a}e^{\epsilon_{t+1}} \end{align}Now let's define the function that returns the equilibrium conditions:
In [4]:
def equilibrium_equations(variables_forward,variables_current,parameters):
# Parameters
p = parameters
# Variables
fwd = variables_forward
cur = variables_current
# Household Euler equation
euler_eqn = p.beta*fwd.c**-p.sigma*(p.alpha*fwd.y/fwd.k+1-p.delta) - cur.c**-p.sigma
# Production function
production_fuction = cur.a*cur.k**p.alpha - cur.y
# Capital evolution
capital_evolution = fwd.k - (1-p.delta)*cur.k - cur.i
# Goods market clearing
market_clearing = cur.c + cur.i - cur.y
# Exogenous technology
technology_proc = cur.a**p.rhoa- fwd.a
# Stack equilibrium conditions into a numpy array
return np.array([
euler_eqn,
production_fuction,
capital_evolution,
market_clearing,
technology_proc
])
Notice that inside the function we have to define the variables of the model form the elements of the input vectors variables_forward and variables_current. It is essential that the predetermined or state variables are ordered first.
To initialize the model, we need to specify the number of state variables in the model, the names of the endogenous varaibles in the same order used in the equilibrium_equations function, and the names of the exogenous shocks to the model.
In [5]:
# Initialize the model
rbc = ls.model(equations = equilibrium_equations,
n_states=2,
var_names=['a','k','c','y','i'],
shock_names=['e_a','e_k'],
parameters=parameters)
The solution routine solves the model as if there were a separate exogenous shock for each state variable and that's why I initialized the model with two exogenous shocks e_a and e_k even though the RBC model only has one exogenous shock.
Next, we need to compute the nonstochastic steady state of the model. The .compute_ss method can be used to compute the steady state numerically. The method's default is to use scipy's fsolve function, but other scipy root-finding functions can be used: root, broyden1, and broyden2. The optional argument options lets the user pass keywords directly to the optimization function. Check out the documentation for Scipy's nonlinear solvers here: http://docs.scipy.org/doc/scipy/reference/optimize.html
In [6]:
# Compute the steady state numerically
guess = [1,1,1,1,1]
rbc.compute_ss(guess)
print(rbc.ss)
Note that the steady state is returned as a Pandas Series. Alternatively, you could compute the steady state directly and then sent the rbc.ss attribute:
In [7]:
# Steady state solution
p = parameters
K = (p.alpha*p.A/(1/p.beta+p.delta-1))**(1/(1-p.alpha))
C = p.A*K**p.alpha - p.delta*K
Y = p.A*K**p.alpha
I = Y - C
rbc.set_ss([p.A,K,C,Y,I])
print(rbc.ss)
Now we use the .log_linear method to find the log-linear appxoximation to the model's equilibrium conditions. That is, we'll transform the nonlinear model into a linear model in which all variables are expressed as log-deviations from the steady state. Specifically, we'll compute the matrices $A$ and $B$ that satisfy:
where the vector $x_{t}$ denotes the log deviation of the endogenous variables from their steady state values.
In [8]:
# Find the log-linear approximation around the non-stochastic steady state
rbc.log_linear_approximation()
print('The matrix A:\n\n',np.around(rbc.a,4),'\n\n')
print('The matrix B:\n\n',np.around(rbc.b,4))
Finally, we need to obtain the solution to the log-linearized model. The solution is a pair of matrices $F$ and $P$ that specify:
We use the .klein method to find the solution.
In [9]:
# Solve the model
rbc.solve_klein(rbc.a,rbc.b)
# Display the output
print('The matrix F:\n\n',np.around(rbc.f,4),'\n\n')
print('The matrix P:\n\n',np.around(rbc.p,4))
One the model is solved, use the .impulse method to compute impulse responses to exogenous shocks to the state. The method creates the .irs attribute which is a dictionary with keys equal to the names of the exogenous shocks and the values are Pandas DataFrames with the computed impulse respones. You can supply your own values for the shocks, but the default is 0.01 for each exogenous shock.
In [10]:
# Compute impulse responses and plot
rbc.impulse(T=41,t0=1,shocks=None,percent=True)
print('Impulse responses to a 0.01 unit shock to A:\n\n',rbc.irs['e_a'].head())
Plotting is easy.
In [11]:
rbc.irs['e_a'][['a','k','c','y','i']].plot(lw='5',alpha=0.5,grid=True).legend(loc='upper right',ncol=2)
rbc.irs['e_a'][['e_a','a']].plot(lw='5',alpha=0.5,grid=True).legend(loc='upper right',ncol=2)
Out[11]:
Creating a stochastic simulation of the model is straightforward with the .stoch_sim method. In the following example, I create a 151 period (including t=0) simulation by first siumlating the model for 251 periods and then dropping the first 100 values. The variance of the shock to $A_t$ is set to 0.001 and the variance of the shock to $K_t$ is set to zero because there is not capital shock in the model. The seed for the numpy random number generator is set to 0.
In [12]:
rbc.stoch_sim(T=121,drop_first=100,cov_mat=np.array([[0.00763**2,0],[0,0]]),seed=0,percent=True)
rbc.simulated[['k','c','y','i']].plot(lw='5',alpha=0.5,grid=True).legend(loc='upper right',ncol=4)
rbc.simulated[['a']].plot(lw='5',alpha=0.5,grid=True).legend(ncol=4)
rbc.simulated[['e_a','e_k']].plot(lw='5',alpha=0.5,grid=True).legend(ncol=4)
Out[12]:
Consider the new-Keynesian model from Walsh (2010), chapter 8 expressed in log-linear terms:
\begin{align} y_t & = E_ty_{t+1} - \sigma^{-1} (i_t - E_t\pi_{t+1}) + g_t\\ \pi_t & = \beta E_t\pi_{t+1} + \kappa y_t + u_t\\ i_t & = \phi_x y_t + \phi_{\pi} \pi_t + v_t\\ r_t & = i_t - E_t\pi_{t+1}\\ g_{t+1} & = \rho_g g_{t} + \epsilon_{t+1}^g\\ u_{t+1} & = \rho_u u_{t} + \epsilon_{t+1}^u\\ v_{t+1} & = \rho_v v_{t} + \epsilon_{t+1}^v \end{align}where $y_t$ is the output gap (log-deviation of output from the natural rate), $\pi_t$ is the quarterly rate of inflation between $t-1$ and $t$, $i_t$ is the nominal interest rate on funds moving between period $t$ and $t+1$, $r_t$ is the real interest rate, $g_t$ is the exogenous component of demand, $u_t$ is an exogenous component of inflation, and $v_t$ is the exogenous component of monetary policy.
Since the model is already log-linear, there is no need to approximate the equilibrium conditions. We'll still use the .log_linear method to find the matrices $A$ and $B$, but we'll have to set the islinear option to True to avoid generating an error.
In [13]:
# Input model parameters
beta = 0.99
sigma= 1
eta = 1
omega= 0.8
kappa= (sigma+eta)*(1-omega)*(1-beta*omega)/omega
rhor = 0.9
phipi= 1.5
phiy = 0
rhog = 0.5
rhou = 0.5
rhov = 0.9
Sigma = 0.001*np.eye(3)
# This time we'll input the model parameters a list of values and we'll include a list of name strings. And
# the program will consolidate the information into a Pandas Series. FYI the parameters list is required,
# but the parameter names is optional. If omitted, the names are set to 'parameter 1', 'parameter 2', etc.
parameters = [beta,sigma,eta,omega,kappa,rhor,phipi,phiy,rhog,rhou,rhov]
parameter_names = ['beta','sigma','eta','omega','kappa','rhor','phipi','phiy','rhog','rhou','rhov']
def equilibrium_equations(variables_forward,variables_current,parameters):
# Parameters
p = parameters
# Variables
fwd = variables_forward
cur = variables_current
# Exogenous demand
g_proc = p.rhog*cur.g - fwd.g
# Exogenous inflation
u_proc = p.rhou*cur.u - fwd.u
# Exogenous monetary policy
v_proc = p.rhov*cur.v - fwd.v
# Euler equation
euler_eqn = fwd.y -1/p.sigma*(cur.i-fwd.pi) + fwd.g - cur.y
# NK Phillips curve evolution
phillips_curve = p.beta*fwd.pi + p.kappa*cur.y + fwd.u - cur.pi
# interest rate rule
interest_rule = p.phiy*cur.y+p.phipi*cur.pi + fwd.v - cur.i
# Fisher equation
fisher_eqn = cur.i - fwd.pi - cur.r
# Stack equilibrium conditions into a numpy array
return np.array([
g_proc,
u_proc,
v_proc,
euler_eqn,
phillips_curve,
interest_rule,
fisher_eqn
])
# Initialize the nk
nk = ls.model(equilibrium_equations,
n_states=3,
var_names=['g','u','v','i','r','y','pi'],
shock_names=['e_g','e_u','e_v'],
parameters=parameters,
parameter_names=parameter_names)
# Set the steady state of the nk
nk.set_ss([0,0,0,0,0,0,0])
# Find the log-linear approximation around the non-stochastic steady state
nk.linear_approximation()
# Solve the nk
nk.solve_klein(nk.a,nk.b)
In [14]:
# Compute impulse responses and plot
nk.impulse(T=11,t0=1,shocks=None)
# Create the figure and axes
fig = plt.figure(figsize=(12,12))
ax1 = fig.add_subplot(3,1,1)
ax2 = fig.add_subplot(3,1,2)
ax3 = fig.add_subplot(3,1,3)
# Plot commands
nk.irs['e_g'][['g','y','i','pi','r']].plot(lw='5',alpha=0.5,grid=True,title='Demand shock',ax=ax1).legend(loc='upper right',ncol=5)
nk.irs['e_u'][['u','y','i','pi','r']].plot(lw='5',alpha=0.5,grid=True,title='Inflation shock',ax=ax2).legend(loc='upper right',ncol=5)
nk.irs['e_v'][['v','y','i','pi','r']].plot(lw='5',alpha=0.5,grid=True,title='Interest rate shock',ax=ax3).legend(loc='upper right',ncol=5)
Out[14]:
In [15]:
nk.stoch_sim(T=151,drop_first=100,cov_mat=Sigma,seed=0)
# Create the figure and axes
fig = plt.figure(figsize=(12,8))
ax1 = fig.add_subplot(2,1,1)
ax2 = fig.add_subplot(2,1,2)
# Plot commands
nk.simulated[['y','i','pi','r']].plot(lw='5',alpha=0.5,grid=True,title='Output, inflation, and interest rates',ax=ax1).legend(ncol=4)
nk.simulated[['g','u','v']].plot(lw='5',alpha=0.5,grid=True,title='Exogenous demand, inflation, and policy',ax=ax2).legend(ncol=4,loc='lower right')
Out[15]:
In [16]:
nk.simulated[['e_g','g']].plot(lw='5',alpha=0.5,grid=True).legend(ncol=2)
nk.simulated[['e_u','u']].plot(lw='5',alpha=0.5,grid=True).legend(ncol=2)
nk.simulated[['e_v','v']].plot(lw='5',alpha=0.5,grid=True).legend(ncol=2)
Out[16]: