Solutions for http://quant-econ.net/jl/rational_expectations.html
The following solutions were put together by Chase Coleman, Spencer Lyon, Thomas Sargent and John Stachurski
In [3]:
using QuantEcon
To map a problem into a discounted optimal linear control problem, we need to define
state vector $x_t$ and control vector $u_t$
matrices $A, B, Q, R$ that define preferences and the law of motion for the state
For the state and control vectors we choose
$$ x_t = \begin{bmatrix} y_t \\ Y_t \\ 1 \end{bmatrix}, \qquad u_t = y_{t+1} - y_{t} $$For $, B, Q, R$ we set
$$ A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \kappa_1 & \kappa_0 \\ 0 & 0 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} , \quad R = \begin{bmatrix} 0 & a_1/2 & -a_0/2 \\ a_1/2 & 0 & 0 \\ -a_0/2 & 0 & 0 \end{bmatrix}, \quad Q = \gamma / 2 $$By multiplying out you can confirm that
$x_t' R x_t + u_t' Q u_t = - r_t$
$x_{t+1} = A x_t + B u_t$
We'll use the module lqcontrol.py to solve the firm's problem at the stated parameter values
This will return an LQ policy $F$ with the interpretation $u_t = - F x_t$, or
$$ y_{t+1} - y_t = - F_0 y_t - F_1 Y_t - F_2 $$Matching parameters with $y_{t+1} = h_0 + h_1 y_t + h_2 Y_t$ leads to
$$ h_0 = -F_2, \quad h_1 = 1 - F_0, \quad h_2 = -F_1 $$Here's our solution
In [4]:
# == Model parameters == #
a0 = 100
a1 = 0.05
bet = 0.95
gamma = 10.0
# == Beliefs == #
kappa0 = 95.5
kappa1 = 0.95
# == Formulate the LQ problem == #
A = [1 0 0
0 kappa1 kappa0
0 0 1]
B = [1.0, 0.0, 0.0]
R = [0 a1/2 -a0/2
a1/2 0 0
-a0/2 0 0]
Q = 0.5 * gamma
# == Solve for the optimal policy == #
lq = LQ(Q, R, A, B, nothing, bet)
P, F, d = stationary_values(lq)
hh = h0, h1, h2 = -F[3], 1 - F[1], -F[2]
@printf("F = [%.3f, %.3f, %.3f]\n", F[1], F[2], F[3])
@printf("(h1, h2, h3) = [%.3f, %.3f, %.3f]\n", h0, h1, h2)
The implication is that
$$ y_{t+1} = 96.949 + y_t - 0.046 \, Y_t $$For the case $n > 1$, recall that $Y_t = n y_t$, which, combined with the previous equation, yields
$$ Y_{t+1} = n \left( 96.949 + y_t - 0.046 \, Y_t \right) = n 96.949 + (1 - n 0.046) Y_t $$To determine whether a $\kappa_0, \kappa_1$ pair forms the aggregate law of motion component of a rational expectations equilibrium, we can proceed as follows:
Determine the corresponding firm law of motion $y_{t+1} = h_0 + h_1 y_t + h_2 Y_t$
Test whether the associated aggregate law :$Y_{t+1} = n h(Y_t/n, Y_t)$ evaluates to $Y_{t+1} = \kappa_0 + \kappa_1 Y_t$
In the second step we can use $Y_t = n y_t = y_t$, so that $Y_{t+1} = n h(Y_t/n, Y_t)$ becomes
$$ Y_{t+1} = h(Y_t, Y_t) = h_0 + (h_1 + h_2) Y_t $$Hence to test the second step we can test $\kappa_0 = h_0$ and $\kappa_1 = h_1 + h_2$
The following code implements this test
In [3]:
candidates = ([94.0886298678, 0.923409232937],
[93.2119845412, 0.984323478873],
[95.0818452486, 0.952459076301])
for (k0, k1) in candidates
A = [1 0 0
0 k1 k0
0 0 1]
lq = LQ(Q, R, A, B, nothing, bet)
P, F, d = stationary_values(lq)
hh = h0, h1, h2 = -F[3], 1 - F[1], -F[2]
if isapprox(k0, h0) && isapprox(k1, h1 + h2)
@printf("Equilibrium pair= (%.6f, %.6f)\n", k0, k1)
@printf("(h1, h2, h3) = [%.6f, %.6f, %.6f]\n", h0, h1, h2)
end
end
The output tells us that the answer is pair (iii), which implies $(h_0, h_1, h_2) = (95.0819, 1.0000, -.0475)$
(Notice we use np.allclose to test equality of floating point numbers, since exact equality is too strict)
Regarding the iterative algorithm, one could loop from a given $(\kappa_0, \kappa_1)$ pair to the associated firm law and then to a new $(\kappa_0, \kappa_1)$ pair
This amounts to implementing the operator $\Phi$ described in the lecture
(There is in general no guarantee that this iterative process will converge to a rational expectations equilibrium)
We are asked to write the planner problem as an LQ problem
For the state and control vectors we choose
$$ x_t = \begin{bmatrix} Y_t \\ 1 \end{bmatrix}, \quad u_t = Y_{t+1} - Y_{t} $$For the LQ matrices we set
$$ A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad R = \begin{bmatrix} a_1/2 & -a_0/2 \\ -a_0/2 & 0 \end{bmatrix}, \quad Q = \gamma / 2 $$By multiplying out you can confirm that
$x_t' R x_t + u_t' Q u_t = - s(Y_t, Y_{t+1})$
$x_{t+1} = A x_t + B u_t$
By obtaining the optimal policy and using $u_t = - F x_t$ or
$$ Y_{t+1} - Y_t = -F_0 Y_t - F_1 $$we can obtain the implied aggregate law of motion via $\kappa_0 = -F_1$ and $\kappa_1 = 1-F_0$
The Python code to solve this problem is below:
In [4]:
# == Formulate the planner's LQ problem == #
A = eye(2)
B = [1.0, 0.0]
R = [a1 / 2.0 -a0 / 2.0
-a0 / 2.0 0.0]
Q = gamma / 2.0
# == Solve for the optimal policy == #
lq = LQ(Q, R, A, B, nothing, bet)
P, F, d = stationary_values(lq)
# == Print the results == #
kappa0, kappa1 = -F[2], 1 - F[1]
println("kappa0=$kappa0\tkappa1=$kappa1")
The output yields the same $(\kappa_0, \kappa_1)$ pair obtained as an equilibrium from the previous exercise
In [5]:
# == Formulate the monopolist's LQ problem == #
A = eye(2)
B = [1.0, 0.0]
R = [a1 -a0 / 2.0
-a0 / 2.0 0.0]
Q = gamma / 2.0
# == Solve for the optimal policy == #
lq = LQ(Q, R, A, B, nothing, bet)
P, F, d = stationary_values(lq)
# == Print the results == #
m0, m1 = -F[2], 1 - F[1]
println("m0=$m0\tm1=$m1")
We see that the law of motion for the monopolist is approximately $Y_{t+1} = 73.4729 + 0.9265 Y_t$
In the rational expectations case the law of motion was approximately $Y_{t+1} = 95.0818 + 0.9525 Y_t$
One way to compare these two laws of motion is by their fixed points, which give long run equilibrium output in each case
For laws of the form $Y_{t+1} = c_0 + c_1 Y_t$, the fixed point is $c_0 / (1 - c_1)$
If you crunch the numbers, you will see that the monopolist adopts a lower long run quantity than obtained by the competitive market, implying a higher market price
This is analogous to the elementary static-case results