def F_to_K(self, F):
"""
Compute agent 2's best cost-minimizing response K, given F.
Parameters
==========
F : array_like
A self.k x self.n array
Returns
=======
K : array_like, dtype = float
P : array_like, dtype = float
"""
Q2 = self.beta * self.theta
R2 = -self.R - dot(F.T, dot(self.Q, F))
A2 = self.A - dot(self.B, F)
B2 = self.C
lq = LQ(Q2, R2, A2, B2, beta=self.beta)
P, neg_K, d = lq.stationary_values()
return -neg_K, P
def K_to_F(self, K):
"""
Compute agent 1's best value-maximizing response F, given K.
Parameters
==========
K : array_like
A self.j x self.n array
Returns
=======
F : array_like, dtype = float
P : array_like, dtype = float
"""
A1 = self.A + dot(self.C, K)
B1 = self.B
Q1 = self.Q
R1 = self.R - self.beta * self.theta * dot(K.T, K)
lq = LQ(Q1, R1, A1, B1, beta=self.beta)
P, F, d = lq.stationary_values()
return F, P
def robust_rule(self):
"""
This method solves the robust control problem by tricking it into a
stacked LQ problem, as described in chapter 2 of Hansen-Sargent's
text "Robustness." The optimal control with observed state is
u_t = - F x_t
And the value function is -x'Px
Returns
=======
F : array_like, dtype = float
The optimal control matrix from above above
P : array_like, dtype = float
The psoitive semi-definite matrix defining the value function
K : array_like, dtype = float
the worst-case shock matrix K, where :math:`w_{t+1} = K x_t` is
the worst case shock
"""
# == Simplify names == #
A, B, C, Q, R = self.A, self.B, self.C, self.Q, self.R
beta, theta = self.beta, self.theta
k, j = self.k, self.j
# == Set up LQ version == #
I = identity(j)
Z = np.zeros((k, j))
Ba = hstack([B, C])
Qa = vstack([hstack([Q, Z]), hstack([Z.T, -beta * I * theta])])
lq = LQ(Qa, R, A, Ba, beta=beta)
# == Solve and convert back to robust problem == #
P, f, d = lq.stationary_values()
F = f[:k, :]
K = -f[k:f.shape[0], :]
return F, K, P
"""
Origin: QE by John Stachurski and Thomas J. Sargent
Filename: solution_ree_ex2.py
Authors: Chase Coleman, Spencer Lyon, Thomas Sargent, John Stachurski
Solves an exercise from the rational expectations module
"""
from __future__ import print_function
import numpy as np
from lqcontrol import LQ
from solution_ree_ex1 import beta, R, Q, B
candidates = ((94.0886298678, 0.923409232937), (93.2119845412, 0.984323478873),
(95.0818452486, 0.952459076301))
for kappa0, kappa1 in candidates:
# == Form the associated law of motion == #
A = np.array([[1, 0, 0], [0, kappa1, kappa0], [0, 0, 1]])
# == Solve the LQ problem for the firm == #
lq = LQ(Q, R, A, B, beta=beta)
P, F, d = lq.stationary_values()
F = F.flatten()
h0, h1, h2 = -F[2], 1 - F[0], -F[1]
# == Test the equilibrium condition == #
if np.allclose((kappa0, kappa1), (h0, h1 + h2)):
print('Equilibrium pair =', kappa0, kappa1)
print('(h0, h1, h2) = ', h0, h1, h2)
break