Contents
In addition to what's in Anaconda, this lecture will need the following libraries:
!pip install --upgrade quantecon
This notebook formulates and computes a plan that a Stackelberg leader uses to manipulate forward-looking decisions of a Stackelberg follower that depend on continuation sequences of decisions made once and for all by the Stackelberg leader at time 0.
To facilitate computation and interpretation, we formulate things in a context that allows us to apply dynamic programming for linear-quadratic models.
From the beginning, we carry along a linear-quadratic model of duopoly in which firms face adjustment costs that make them want to forecast actions of other firms that influence future prices.
Let's start with some standard imports:
import numpy as np
import numpy.linalg as la
import quantecon as qe
from quantecon import LQ
import matplotlib.pyplot as plt
%matplotlib inline
Time is discrete and is indexed by t = 0, 1, \ldots.
Two firms produce a single good whose demand is governed by the linear inverse demand curve
p_t = a_0 - a_1 (q_{1t}+ q_{2t} )
where q_{it} is output of firm i at time t and a_0 and a_1 are both positive.
q_{10}, q_{20} are given numbers that serve as initial conditions at time 0.
By incurring a cost of change
\gamma v_{it}^2
where \gamma > 0, firm i can change its output according to
q_{it+1} = q_{it} + v_{it}
Firm i's profits at time t equal
\pi_{it} = p_t q_{it} - \gamma v_{it}^2
Firm i wants to maximize the present value of its profits
\sum_{t=0}^\infty \beta^t \pi_{it}
where \beta \in (0,1) is a time discount factor.
Each firm i=1,2 chooses a sequence \vec q_i \equiv \{q_{it+1}\}_{t=0}^\infty once and for all at time 0.
We let firm 2 be a Stackelberg leader and firm 1 be a Stackelberg follower.
The leader firm 2 goes first and chooses \{q_{2t+1}\}_{t=0}^\infty once and for all at time 0.
Knowing that firm 2 has chosen \{q_{2t+1}\}_{t=0}^\infty, the follower firm 1 goes second and chooses \{q_{1t+1}\}_{t=0}^\infty once and for all at time 0.
In choosing \vec q_2, firm 2 takes into account that firm 1 will base its choice of \vec q_1 on firm 2's choice of \vec q_2.
We can express firm 1's problem as
\max_{\vec q_1} \Pi_1(\vec q_1; \vec q_2)
where the appearance behind the semi-colon indicates that \vec q_2 is given.
Firm 1's problem induces the best response mapping
\vec q_1 = B(\vec q_2)
(Here B maps a sequence into a sequence)
The Stackelberg leader's problem is
\max_{\vec q_2} \Pi_2 (B(\vec q_2), \vec q_2)
whose maximizer is a sequence \vec q_2 that depends on the initial conditions q_{10}, q_{20} and the parameters of the model a_0, a_1, \gamma.
This formulation captures key features of the model
- Both firms make once-and-for-all choices at time 0.
- This is true even though both firms are choosing sequences of quantities that are indexed by time.
- The Stackelberg leader chooses first within time 0, knowing that the Stackelberg follower will choose second within time 0.
While our abstract formulation reveals the timing protocol and equilibrium concept well, it obscures details that must be addressed when we want to compute and interpret a Stackelberg plan and the follower's best response to it.
To gain insights about these things, we study them in more detail.
Firm 1 acts as if firm 2's sequence \{q_{2t+1}\}_{t=0}^\infty is given and beyond its control.
Firm 2 knows that firm 1 chooses second and takes this into account in choosing \{q_{2t+1}\}_{t=0}^\infty.
In the spirit of working backward, we study firm 1's problem first, taking \{q_{2t+1}\}_{t=0}^\infty as given.
We can formulate firm 1's optimum problem in terms of the Lagrangian
L=\sum_{t=0}^{\infty}\beta^{t}\{a_{0}q_{1t}-a_{1}q_{1t}^{2}-a_{1}q_{1t}q_{2t}-\gamma v_{1t}^{2}+\lambda_{t}[q_{1t}+v_{1t}-q_{1t+1}]\}
Firm 1 seeks a maximum with respect to \{q_{1t+1}, v_{1t} \}_{t=0}^\infty and a minimum with respect to \{ \lambda_t\}_{t=0}^\infty.
We approach this problem using methods described in Ljungqvist and Sargent RMT5 chapter 2, appendix A and Macroeconomic Theory, 2nd edition, chapter IX.
First-order conditions for this problem are
\begin{aligned} \frac{\partial L}{\partial q_{1t}} & = a_0 - 2 a_1 q_{1t} - a_1 q_{2t} + \lambda_t - \beta^{-1}
\lambda_{t-1} = 0 , \quad t \geq 1 \cr
\frac{\partial L}{\partial v_{1t}} & = -2 \gamma v_{1t} + \lambda_t = 0 , \quad t \geq 0 \cr \end{aligned}
These first-order conditions and the constraint q_{1t+1} = q_{1t} + v_{1t} can be rearranged to take the form
\begin{aligned} v_{1t} & = \beta v_{1t+1} + \frac{\beta a_0}{2 \gamma} - \frac{\beta a_1}{\gamma} q_{1t+1} -
\frac{\beta a_1}{2 \gamma} q_{2t+1} \cr
q_{t+1} & = q_{1t} + v_{1t} \end{aligned}
We can substitute the second equation into the first equation to obtain
(q_{1t+1} - q_{1t} ) = \beta (q_{1t+2} - q_{1t+1}) + c_0 - c_1 q_{1t+1} - c_2 q_{2t+1}
where c_0 = \frac{\beta a_0}{2 \gamma}, c_1 = \frac{\beta a_1}{\gamma}, c_2 = \frac{\beta a_1}{2 \gamma}.
This equation can in turn be rearranged to become the second-order difference equation
q_{1t} + (1+\beta + c_1) q_{1t+1} - \beta q_{1t+2} = c_0 - c_2 q_{2t+1}
Equation :eq:`sstack1` is a second-order difference equation in the sequence \vec q_1 whose solution we want.
It satisfies two boundary conditions:
- an initial condition that q_{1,0}, which is given
- a terminal condition requiring that \lim_{T \rightarrow + \infty} \beta^T q_{1t}^2 < + \infty
Using the lag operators described in chapter IX of Macroeconomic Theory, Second edition (1987), difference equation :eq:`sstack1` can be written as
\beta(1 - \frac{1+\beta + c_1}{\beta} L + \beta^{-1} L^2 ) q_{1t+2} = - c_0 + c_2 q_{2t+1}
The polynomial in the lag operator on the left side can be factored as
(1 - \frac{1+\beta + c_1}{\beta} L + \beta^{-1} L^2 ) = ( 1 - \delta_1 L ) (1 - \delta_2 L)
where 0 < \delta_1 < 1 < \frac{1}{\sqrt{\beta}} < \delta_2.
Because \delta_2 > \frac{1}{\sqrt{\beta}} the operator (1 - \delta_2 L) contributes an unstable component if solved backwards but a stable component if solved forwards.
Mechanically, write
(1- \delta_2 L) = -\delta_{2} L (1 - \delta_2^{-1} L^{-1} )
and compute the following inverse operator
\left[-\delta_{2} L (1 - \delta_2^{-1} L^{-1} )\right]^{-1} = - \delta_2 (1 - {\delta_2}^{-1} )^{-1} L^{-1}
Operating on both sides of equation :eq:`sstack2` with \beta^{-1} times this inverse operator gives the follower's decision rule for setting q_{1t+1} in the feedback-feedforward form.
q_{1t+1} = \delta_1 q_{1t} - c_0 \delta_2^{-1} \beta^{-1} \frac{1}{1 -\delta_2^{-1}} + c_2 \delta_2^{-1} \beta^{-1} \sum_{j=0}^\infty \delta_2^j q_{2t+j+1} , \quad t \geq 0
The problem of the Stackelberg leader firm 2 is to choose the sequence \{q_{2t+1}\}_{t=0}^\infty to maximize its discounted profits
\sum_{t=0}^\infty \beta^t \{ (a_0 - a_1 (q_{1t} + q_{2t}) ) q_{2t} - \gamma (q_{2t+1} - q_{2t})^2 \}
subject to the sequence of constraints :eq:`sstack3` for t \geq 0.
We can put a sequence \{\theta_t\}_{t=0}^\infty of Lagrange multipliers on the sequence of equations :eq:`sstack3` and formulate the following Lagrangian for the Stackelberg leader firm 2's problem
\begin{aligned} \tilde L & = \sum_{t=0}^\infty \beta^t\{ (a_0 - a_1 (q_{1t} + q_{2t}) ) q_{2t} - \gamma (q_{2t+1} - q_{2t})^2 \} \cr
& + \sum_{t=0}^\infty \beta^t \theta_t \{ \delta_1 q_{1t} - c_0 \delta_2^{-1} \beta^{-1} \frac{1}{1 -\delta_2^{-1}} + c_2 \delta_2^{-1} \beta^{-1}
\sum_{j=0}^\infty \delta_2^{-j} q_{2t+j+1} - q_{1t+1} \} \end{aligned}
subject to initial conditions for q_{1t}, q_{2t} at t=0.
Comments: We have formulated the Stackelberg problem in a space of sequences.
The max-min problem associated with Lagrangian :eq:`sstack4` is unpleasant because the time t component of firm 1's payoff function depends on the entire future of its choices of \{q_{1t+j}\}_{j=0}^\infty.
This renders a direct attack on the problem cumbersome.
Therefore, below, we will formulate the Stackelberg leader's problem recursively.
We'll put our little duopoly model into a broader class of models with the same conceptual structure.
We formulate a class of linear-quadratic Stackelberg leader-follower problems of which our duopoly model is an instance.
We use the optimal linear regulator (a.k.a. the linear-quadratic dynamic programming problem described in LQ Dynamic Programming problems) to represent a Stackelberg leader's problem recursively.
Let z_t be an n_z \times 1 vector of natural state variables.
Let x_t be an n_x \times 1 vector of endogenous forward-looking variables that are physically free to jump at t.
In our duopoly example x_t = v_{1t}, the time t decision of the Stackelberg follower.
Let u_t be a vector of decisions chosen by the Stackelberg leader at t.
The z_t vector is inherited physically from the past.
But x_t is a decision made by the Stackelberg follower at time t that is the follower's best response to the choice of an entire sequence of decisions made by the Stackelberg leader at time t=0.
Let
y_t = \begin{bmatrix} z_t \\ x_t \end{bmatrix}
Represent the Stackelberg leader's one-period loss function as
r(y, u) = y' R y + u' Q u
Subject to an initial condition for z_0, but not for x_0, the Stackelberg leader wants to maximize
-\sum_{t=0}^\infty \beta^t r(y_t, u_t)
The Stackelberg leader faces the model
\begin{bmatrix} I & 0 \\ G_{21} & G_{22} \end{bmatrix}
\begin{bmatrix} z_{t+1} \\ x_{t+1} \end{bmatrix}
= \begin{bmatrix} \hat A_{11} & \hat A_{12} \\ \hat A_{21} & \hat A_{22} \end{bmatrix} \begin{bmatrix} z_t \\ x_t \end{bmatrix} + \hat B u_t
We assume that the matrix \begin{bmatrix} I & 0 \\ G_{21} & G_{22} \end{bmatrix} on the left side of equation :eq:`new2` is invertible, so that we can multiply both sides by its inverse to obtain
\begin{bmatrix} z_{t+1} \\ x_{t+1} \end{bmatrix}
= \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix}
\begin{bmatrix} z_t \\ x_t \end{bmatrix} + B u_t
or
y_{t+1} = A y_t + B u_t
The Stackelberg follower's best response mapping is summarized by the second block of equations of :eq:`new3`.
In particular, these equations are the first-order conditions of the Stackelberg follower's optimization problem (i.e., its Euler equations).
These Euler equations summarize the forward-looking aspect of the follower's behavior and express how its time t decision depends on the leader's actions at times s \geq t.
When combined with a stability condition to be imposed below, the Euler equations summarize the follower’s best response to the sequence of actions by the leader.
The Stackelberg leader maximizes :eq:`maxeq` by choosing sequences \{u_t, x_t, z_{t+1}\}_{t=0}^{\infty} subject to :eq:`constrainteq` and an initial condition for z_0.
Note that we have an initial condition for z_0 but not for x_0.
x_0 is among the variables to be chosen at time 0 by the Stackelberg leader.
The Stackelberg leader uses its understanding of the responses restricted by :eq:`constrainteq` to manipulate the follower's decisions.
For any vector a_t, define \vec a_t = [a_t, a_{t+1} \ldots ].
Define a feasible set of (\vec y_1, \vec u_0) sequences
\Omega(y_0) = \left\{ (\vec y_1, \vec u_0) : y_{t+1} = A y_t + B u_t, \forall t \geq 0 \right\}
Please remember that the follower's Euler equation is embedded in the system of dynamic equations y_{t+1} = A y_t + B u_t.
Note that in the definition of \Omega(y_0), y_0 is taken as given.
Although it is taken as given in \Omega(y_0), eventually, the x_0 component of y_0 will be chosen by the Stackelberg leader.
Once again we use backward induction.
We express the Stackelberg problem in terms of two subproblems.
Subproblem 1 is solved by a continuation Stackelberg leader at each date t \geq 0.
Subproblem 2 is solved by the Stackelberg leader at t=0.
The two subproblems are designed
- to respect the protocol in which the follower chooses \vec q_1 after seeing \vec q_2 chosen by the leader
- to make the leader choose \vec q_2 while respecting that \vec q_1 will be the follower's best response to \vec q_2
- to represent the leader's problem recursively by artfully choosing the state variables confronting and the control variables available to the leader
v(y_0) = \max_{(\vec y_1, \vec u_0) \in \Omega(y_0)} - \sum_{t=0}^\infty \beta^t r(y_t, u_t)
w(z_0) = \max_{x_0} v(y_0)
Subproblem 1 takes the vector of forward-looking variables x_0 as given.
Subproblem 2 optimizes over x_0.
The value function w(z_0) tells the value of the Stackelberg plan as a function of the vector of natural state variables at time 0, z_0.
We now describe Bellman equations for v(y) and w(z_0).
The value function v(y) in subproblem 1 satisfies the Bellman equation
v(y) = \max_{u, y^*} \left\{ - r(y,u) + \beta v(y^*) \right\}
where the maximization is subject to
y^* = A y + B u
and y^* denotes next period’s value.
Substituting v(y) = - y'P y into Bellman equation :eq:`bellman-stack` gives
-y' P y = {\rm max}_{ u, y^*} \left\{ - y' R y - u'Q u - \beta y^{* \prime} P y^* \right\}
which as in lecture linear regulator gives rise to the algebraic matrix Riccati equation
P = R + \beta A' P A - \beta^2 A' P B ( Q + \beta B' P B)^{-1} B' P A
and the optimal decision rule coefficient vector
F = \beta( Q + \beta B' P B)^{-1} B' P A
where the optimal decision rule is
u_t = - F y_t
We find an optimal x_0 by equating to zero the gradient of v(y_0) with respect to x_0:
-2 P_{21} z_0 - 2 P_{22} x_0 =0,
which implies that
x_0 = - P_{22}^{-1} P_{21} z_0
Now let's map our duopoly model into the above setup.
We will formulate a state space system
y_t = \begin{bmatrix} z_t \cr x_t \end{bmatrix}
where in this instance x_t = v_{1t}, the time t decision of the follower firm 1.
Now we'll proceed to cast our duopoly model within the framework of the more general linear-quadratic structure described above.
That will allow us to compute a Stackelberg plan simply by enlisting a Riccati equation to solve a linear-quadratic dynamic program.
As emphasized above, firm 1 acts as if firm 2's decisions \{q_{2t+1}, v_{2t}\}_{t=0}^\infty are given and beyond its control.
We again formulate firm 1's optimum problem in terms of the Lagrangian
L=\sum_{t=0}^{\infty}\beta^{t}\{a_{0}q_{1t}-a_{1}q_{1t}^{2}-a_{1}q_{1t}q_{2t}-\gamma v_{1t}^{2}+\lambda_{t}[q_{1t}+v_{1t}-q_{1t+1}]\}
Firm 1 seeks a maximum with respect to \{q_{1t+1}, v_{1t} \}_{t=0}^\infty and a minimum with respect to \{ \lambda_t\}_{t=0}^\infty.
First-order conditions for this problem are
\begin{aligned} \frac{\partial L}{\partial q_{1t}} & = a_0 - 2 a_1 q_{1t} - a_1 q_{2t} + \lambda_t - \beta^{-1}
\lambda_{t-1} = 0 , \quad t \geq 1 \cr
\frac{\partial L}{\partial v_{1t}} & = -2 \gamma v_{1t} + \lambda_t = 0 , \quad t \geq 0 \cr \end{aligned}
These first-order order conditions and the constraint q_{1t+1} = q_{1t} + v_{1t} can be rearranged to take the form
\begin{aligned} v_{1t} & = \beta v_{1t+1} + \frac{\beta a_0}{2 \gamma} - \frac{\beta a_1}{\gamma} q_{1t+1} -
\frac{\beta a_1}{2 \gamma} q_{2t+1} \cr
q_{t+1} & = q_{1t} + v_{1t} \end{aligned}
We use these two equations as components of the following linear system that confronts a Stackelberg continuation leader at time t
\begin{bmatrix} 1 & 0 & 0 & 0 \cr
0 & 1 & 0 & 0 \cr
0 & 0 & 1 & 0 \cr
\frac{\beta a_0}{2 \gamma} & - \frac{\beta a_1}{2 \gamma} & -\frac{\beta a_1}{\gamma} & \beta \end{bmatrix}
\begin{bmatrix} 1 \cr q_{2t+1} \cr q_{1t+1} \cr v_{1t+1} \end{bmatrix}
= \begin{bmatrix} 1 & 0 & 0 & 0 \cr
0 & 1 & 0 & 0 \cr
0 & 0 & 1 & 1 \cr
0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \cr q_{2t} \cr q_{1t} \cr v_{1t} \end{bmatrix}
+ \begin{bmatrix} 0 \cr 1 \cr 0 \cr 0 \end{bmatrix} v_{2t}
Time t revenues of firm 2 are \pi_{2t} = a_0 q_{2t} - a_1 q_{2t}^2 - a_1 q_{1t} q_{2t} which evidently equal
z_t' R_1 z_t \equiv \begin{bmatrix} 1 \cr q_{2t} \cr q_{1t} \end{bmatrix}'
\begin{bmatrix} 0 & \frac{a_0}{2}& 0 \cr
\frac{a_0}{2} & -a_1 & -\frac{a_1}{2}\cr
0 & -\frac{a_1}{2} & 0 \end{bmatrix}
\begin{bmatrix} 1 \cr q_{2t} \cr q_{1t} \end{bmatrix}
If we set Q = \gamma, then firm 2's period t profits can then be written
y_t' R y_t - Q v_{2t}^2
where
y_t = \begin{bmatrix} z_t \cr x_t \end{bmatrix}
with x_t = v_{1t} and
R =
\begin{bmatrix} R_1 & 0 \cr 0 & 0 \end{bmatrix}
We'll report results of implementing this code soon.
But first, we want to represent the Stackelberg leader's optimal choices recursively.
It is important to do this for several reasons:
- properly to interpret a representation of the Stackelberg leader's choice as a sequence of history-dependent functions
- to formulate a recursive version of the follower's choice problem
First, let's get a recursive representation of the Stackelberg leader's choice of \vec q_2 for our duopoly model.
In order to attain an appropriate representation of the Stackelberg leader's history-dependent plan, we will employ what amounts to a version of the Big K, little k device often used in macroeconomics by distinguishing z_t, which depends partly on decisions x_t of the followers, from another vector \check z_t, which does not.
We will use \check z_t and its history \check z^t = [\check z_t, \check z_{t-1}, \ldots, \check z_0] to describe the sequence of the Stackelberg leader's decisions that the Stackelberg follower takes as given.
Thus, we let \check y_t' = \begin{bmatrix}\check z_t' & \check x_t'\end{bmatrix} with initial condition \check z_0 = z_0 given.
That we distinguish \check z_t from z_t is part and parcel of the Big K, little k device in this instance.
We have demonstrated that a Stackelberg plan for \{u_t\}_{t=0}^\infty has a recursive representation
\begin{aligned} \check x_0 & = - P_{22}^{-1} P_{21} z_0 \cr
u_t & = - F \check y_t, \quad t \geq 0 \cr
\check y_{t+1} & = (A - BF) \check y_t, \quad t \geq 0 \end{aligned}
From this representation, we can deduce the sequence of functions \sigma = \{\sigma_t(\check z^t)\}_{t=0}^\infty that comprise a Stackelberg plan.
For convenience, let \check A \equiv A - BF and partition \check A conformably to the partition y_t = \begin{bmatrix}\check z_t \cr \check x_t \end{bmatrix} as
\begin{bmatrix}\check A_{11} & \check A_{12} \cr \check A_{21} & \check A_{22} \end{bmatrix}
Let H^0_0 \equiv - P_{22}^{-1} P_{21} so that \check x_0 = H^0_0 \check z_0.
Then iterations on \check y_{t+1} = \check A \check y_t starting from initial condition \check y_0 = \begin{bmatrix}\check z_0 \cr H^0_0 \check z_0\end{bmatrix} imply that for t \geq 1
x_t = \sum_{j=1}^t H_j^t \check z_{t-j}
where
\begin{aligned} H^t_1 & = \check A_{21} \cr
H^t_2 & = \check A_{22} \check A_{21} \cr
\ \ \vdots \ \ & \ \ \quad \vdots \cr
H^t_{t-1} & = \check A_{22}^{t-2} \check A_{21} \cr
H^t_t & = \check A_{22}^{t-1}(\check A_{21} + \check A_{22} H^0_0 ) \end{aligned}
An optimal decision rule for the Stackelberg's choice of u_t is
u_t = - F \check y_t \equiv - \begin{bmatrix} F_z & F_x \cr \end{bmatrix}
\begin{bmatrix}\check z_t \cr x_t \cr \end{bmatrix}
or
u_t = - F_z \check z_t - F_x \sum_{j=1}^t H^t_j z_{t-j} = \sigma_t(\check z^t)
Representation :eq:`finalrule` confirms that whenever F_x \neq 0, the typical situation, the time t component \sigma_t of a Stackelberg plan is history-dependent, meaning that the Stackelberg leader's choice u_t depends not just on \check z_t but on components of \check z^{t-1}.
After all, at the end of the day, it will turn out that because we set \check z_0 = z_0, it will be true that z_t = \check z_t for all t \geq 0.
Then why did we distinguish \check z_t from z_t?
The answer is that if we want to present to the Stackelberg follower a history-dependent representation of the Stackelberg leader's sequence \vec q_2, we must use representation :eq:`finalrule` cast in terms of the history \check z^t and not a corresponding representation cast in terms of z^t.
Given the sequence \vec q_2 chosen by the Stackelberg leader in our duopoly model, it turns out that the Stackelberg follower's problem is recursive in the natural state variables that confront a follower at any time t \geq 0.
This means that the follower's plan is time consistent.
To verify these claims, we'll formulate a recursive version of a follower's problem that builds on our recursive representation of the Stackelberg leader's plan and our use of the Big K, little k idea.
We now use what amounts to another “Big K, little k” trick (see rational expectations equilibrium) to formulate a recursive version of a follower’s problem cast in terms of an ordinary Bellman equation.
Firm 1, the follower, faces \{q_{2t}\}_{t=0}^\infty as a given quantity sequence chosen by the leader and believes that its output price at t satisfies
p_t = a_0 - a_1 ( q_{1t} + q_{2t}) , \quad t \geq 0
Our challenge is to represent \{q_{2t}\}_{t=0}^\infty as a given sequence.
To do so, recall that under the Stackelberg plan, firm 2 sets output according to the q_{2t} component of
y_{t+1} = \begin{bmatrix} 1 \cr q_{2t} \cr q_{1t} \cr x_t \end{bmatrix}
which is governed by
y_{t+1} = (A - BF) y_t
To obtain a recursive representation of a \{q_{2t}\} sequence that is exogenous to firm 1, we define a state \tilde y_t
\tilde y_t = \begin{bmatrix} 1 \cr q_{2t} \cr \tilde q_{1t} \cr \tilde x_t \end{bmatrix}
that evolves according to
\tilde y_{t+1} = (A - BF) \tilde y_t
subject to the initial condition \tilde q_{10} = q_{10} and \tilde x_0 = x_0 where x_0 = - P_{22}^{-1} P_{21} as stated above.
Firm 1's state vector is
X_t = \begin{bmatrix} \tilde y_t \cr q_{1t} \end{bmatrix}
It follows that the follower firm 1 faces law of motion
\begin{bmatrix} \tilde y_{t+1} \\
q_{1t+1} \end{bmatrix} = \begin{bmatrix} A - BF & 0 \\
0 & 1 \end{bmatrix} \begin{bmatrix} \tilde y_{t} \\
q_{1t} \end{bmatrix} + \begin{bmatrix} 0 \cr 1 \end{bmatrix} x_t
This specification assures that from the point of the view of a firm 1, q_{2t} is an exogenous process.
Here
- \tilde q_{1t}, \tilde x_t play the role of Big K
- q_{1t}, x_t play the role of little k
The time t component of firm 1's objective is
\tilde X_t' \tilde R x_t - x_t^2 \tilde Q = \begin{bmatrix} 1 \cr q_{2t} \cr \tilde q_{1t} \cr \tilde x_t \cr q_{1t} \end{bmatrix}'
\begin{bmatrix} 0 & 0 & 0 & 0 & \frac{a_0}{2} \cr
0 & 0 & 0 & 0 & - \frac{a_1}{2} \cr
0 & 0 & 0 & 0 & 0 \cr
0 & 0 & 0 & 0 & 0 \cr
\frac{a_0}{2} & -\frac{a_1}{2} & 0 & 0 & - a_1 \end{bmatrix}
\begin{bmatrix} 1 \cr q_{2t} \cr \tilde q_{1t} \cr \tilde x_t \cr q_{1t} \end{bmatrix} - \gamma
x_t^2
Firm 1's optimal decision rule is
x_t = - \tilde F X_t
and it's state evolves according to
\tilde X_{t+1} = (\tilde A - \tilde B \tilde F) X_t
under its optimal decision rule.
Later we shall compute \tilde F and verify that when we set
X_0 = \begin{bmatrix} 1 \cr q_{20} \cr q_{10} \cr x_0 \cr q_{10} \end{bmatrix}
we recover
x_0 = - \tilde F \tilde X_0
which will verify that we have properly set up a recursive representation of the follower's problem facing the Stackelberg leader's \vec q_2.
Since the follower can solve its problem using dynamic programming its problem is recursive in what for it are the natural state variables, namely
\begin{bmatrix} 1 \cr q_{2t} \cr \tilde q_{10} \cr \tilde x_0 \end{bmatrix}
It follows that the follower's plan is time consistent.
Here is our code to compute a Stackelberg plan via a linear-quadratic dynamic program as outlined above
# Parameters
a0 = 10
a1 = 2
β = 0.96
γ = 120
n = 300
tol0 = 1e-8
tol1 = 1e-16
tol2 = 1e-2
βs = np.ones(n)
βs[1:] = β
βs = βs.cumprod()# In LQ form
Alhs = np.eye(4)
# Euler equation coefficients
Alhs[3, :] = β * a0 / (2 * γ), -β * a1 / (2 * γ), -β * a1 / γ, β
Arhs = np.eye(4)
Arhs[2, 3] = 1
Alhsinv = la.inv(Alhs)
A = Alhsinv @ Arhs
B = Alhsinv @ np.array([[0, 1, 0, 0]]).T
R = np.array([[0, -a0 / 2, 0, 0],
[-a0 / 2, a1, a1 / 2, 0],
[0, a1 / 2, 0, 0],
[0, 0, 0, 0]])
Q = np.array([[γ]])
# Solve using QE's LQ class
# LQ solves minimization problems which is why the sign of R and Q was changed
lq = LQ(Q, R, A, B, beta=β)
P, F, d = lq.stationary_values(method='doubling')
P22 = P[3:, 3:]
P21 = P[3:, :3]
P22inv = la.inv(P22)
H_0_0 = -P22inv @ P21
# Simulate forward
π_leader = np.zeros(n)
z0 = np.array([[1, 1, 1]]).T
x0 = H_0_0 @ z0
y0 = np.vstack((z0, x0))
yt, ut = lq.compute_sequence(y0, ts_length=n)[:2]
π_matrix = (R + F. T @ Q @ F)
for t in range(n):
π_leader[t] = -(yt[:, t].T @ π_matrix @ yt[:, t])
# Display policies
print("Computed policy for Stackelberg leader\n")
print(f"F = {F}")The following code plots the price and quantities
q_leader = yt[1, :-1]
q_follower = yt[2, :-1]
q = q_leader + q_follower # Total output, Stackelberg
p = a0 - a1 * q # Price, Stackelberg
fig, ax = plt.subplots(figsize=(9, 5.8))
ax.plot(range(n), q_leader, 'b-', lw=2, label='leader output')
ax.plot(range(n), q_follower, 'r-', lw=2, label='follower output')
ax.plot(range(n), p, 'g-', lw=2, label='price')
ax.set_title('Output and prices, Stackelberg duopoly')
ax.legend(frameon=False)
ax.set_xlabel('t')
plt.show()We'll compute the present value earned by the Stackelberg leader.
We'll compute it two ways (they give identical answers -- just a check on coding and thinking)
v_leader_forward = np.sum(βs * π_leader)
v_leader_direct = -yt[:, 0].T @ P @ yt[:, 0]
# Display values
print("Computed values for the Stackelberg leader at t=0:\n")
print(f"v_leader_forward(forward sim) = {v_leader_forward:.4f}")
print(f"v_leader_direct (direct) = {v_leader_direct:.4f}")# Manually checks whether P is approximately a fixed point
P_next = (R + F.T @ Q @ F + β * (A - B @ F).T @ P @ (A - B @ F))
(P - P_next < tol0).all()# Manually checks whether two different ways of computing the
# value function give approximately the same answer
v_expanded = -((y0.T @ R @ y0 + ut[:, 0].T @ Q @ ut[:, 0] +
β * (y0.T @ (A - B @ F).T @ P @ (A - B @ F) @ y0)))
(v_leader_direct - v_expanded < tol0)[0, 0]In the code below we compare two values
- the continuation value - y_t P y_t earned by a continuation Stackelberg leader who inherits state y_t at t
- the value of a reborn Stackelberg leader who inherits state z_t at t and sets x_t = - P_{22}^{-1} P_{21}
The difference between these two values is a tell-tale sign of the time inconsistency of the Stackelberg plan
# Compute value function over time with a reset at time t
vt_leader = np.zeros(n)
vt_reset_leader = np.empty_like(vt_leader)
yt_reset = yt.copy()
yt_reset[-1, :] = (H_0_0 @ yt[:3, :])
for t in range(n):
vt_leader[t] = -yt[:, t].T @ P @ yt[:, t]
vt_reset_leader[t] = -yt_reset[:, t].T @ P @ yt_reset[:, t]fig, axes = plt.subplots(3, 1, figsize=(10, 7))
axes[0].plot(range(n+1), (- F @ yt).flatten(), 'bo',
label='Stackelberg leader', ms=2)
axes[0].plot(range(n+1), (- F @ yt_reset).flatten(), 'ro',
label='continuation leader at t', ms=2)
axes[0].set(title=r'Leader control variable $u_{t}$', xlabel='t')
axes[0].legend()
axes[1].plot(range(n+1), yt[3, :], 'bo', ms=2)
axes[1].plot(range(n+1), yt_reset[3, :], 'ro', ms=2)
axes[1].set(title=r'Follower control variable $x_{t}$', xlabel='t')
axes[2].plot(range(n), vt_leader, 'bo', ms=2)
axes[2].plot(range(n), vt_reset_leader, 'ro', ms=2)
axes[2].set(title=r'Leader value function $v(y_{t})$', xlabel='t')
plt.tight_layout()
plt.show()We now formulate and compute the recursive version of the follower's problem.
We check that the recursive Big K , little k formulation of the follower's problem produces the same output path \vec q_1 that we computed when we solved the Stackelberg problem
A_tilde = np.eye(5)
A_tilde[:4, :4] = A - B @ F
R_tilde = np.array([[0, 0, 0, 0, -a0 / 2],
[0, 0, 0, 0, a1 / 2],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0],
[-a0 / 2, a1 / 2, 0, 0, a1]])
Q_tilde = Q
B_tilde = np.array([[0, 0, 0, 0, 1]]).T
lq_tilde = LQ(Q_tilde, R_tilde, A_tilde, B_tilde, beta=β)
P_tilde, F_tilde, d_tilde = lq_tilde.stationary_values(method='doubling')
y0_tilde = np.vstack((y0, y0[2]))
yt_tilde = lq_tilde.compute_sequence(y0_tilde, ts_length=n)[0]# Checks that the recursive formulation of the follower's problem gives
# the same solution as the original Stackelberg problem
fig, ax = plt.subplots()
ax.plot(yt_tilde[4], 'r', label="q_tilde")
ax.plot(yt_tilde[2], 'b', label="q")
ax.legend()
plt.show()Note: Variables with _tilde are obtained from solving the follower's
problem -- those without are from the Stackelberg problem
# Maximum absolute difference in quantities over time between
# the first and second solution methods
np.max(np.abs(yt_tilde[4] - yt_tilde[2]))# x0 == x0_tilde
yt[:, 0][-1] - (yt_tilde[:, 1] - yt_tilde[:, 0])[-1] < tol0If we inspect the coefficients in the decision rule - \tilde F, we can spot the reason that the follower chooses to set x_t = \tilde x_t when it sets x_t = - \tilde F X_t in the recursive formulation of the follower problem.
Can you spot what features of \tilde F imply this?
Hint: remember the components of X_t
# Policy function in the follower's problem
F_tilde.round(4)# Value function in the Stackelberg problem
P# Value function in the follower's problem
P_tilde# Manually check that P is an approximate fixed point
(P - ((R + F.T @ Q @ F) + β * (A - B @ F).T @ P @ (A - B @ F)) < tol0).all()# Compute `P_guess` using `F_tilde_star`
F_tilde_star = -np.array([[0, 0, 0, 1, 0]])
P_guess = np.zeros((5, 5))
for i in range(1000):
P_guess = ((R_tilde + F_tilde_star.T @ Q @ F_tilde_star) +
β * (A_tilde - B_tilde @ F_tilde_star).T @ P_guess
@ (A_tilde - B_tilde @ F_tilde_star))# Value function in the follower's problem
-(y0_tilde.T @ P_tilde @ y0_tilde)[0, 0]# Value function with `P_guess`
-(y0_tilde.T @ P_guess @ y0_tilde)[0, 0]# Compute policy using policy iteration algorithm
F_iter = (β * la.inv(Q + β * B_tilde.T @ P_guess @ B_tilde)
@ B_tilde.T @ P_guess @ A_tilde)
for i in range(100):
# Compute P_iter
P_iter = np.zeros((5, 5))
for j in range(1000):
P_iter = ((R_tilde + F_iter.T @ Q @ F_iter) + β
* (A_tilde - B_tilde @ F_iter).T @ P_iter
@ (A_tilde - B_tilde @ F_iter))
# Update F_iter
F_iter = (β * la.inv(Q + β * B_tilde.T @ P_iter @ B_tilde)
@ B_tilde.T @ P_iter @ A_tilde)
dist_vec = (P_iter - ((R_tilde + F_iter.T @ Q @ F_iter)
+ β * (A_tilde - B_tilde @ F_iter).T @ P_iter
@ (A_tilde - B_tilde @ F_iter)))
if np.max(np.abs(dist_vec)) < 1e-8:
dist_vec2 = (F_iter - (β * la.inv(Q + β * B_tilde.T @ P_iter @ B_tilde)
@ B_tilde.T @ P_iter @ A_tilde))
if np.max(np.abs(dist_vec2)) < 1e-8:
F_iter
else:
print("The policy didn't converge: try increasing the number of \
outer loop iterations")
else:
print("`P_iter` didn't converge: try increasing the number of inner \
loop iterations")# Simulate the system using `F_tilde_star` and check that it gives the
# same result as the original solution
yt_tilde_star = np.zeros((n, 5))
yt_tilde_star[0, :] = y0_tilde.flatten()
for t in range(n-1):
yt_tilde_star[t+1, :] = (A_tilde - B_tilde @ F_tilde_star) \
@ yt_tilde_star[t, :]
fig, ax = plt.subplots()
ax.plot(yt_tilde_star[:, 4], 'r', label="q_tilde")
ax.plot(yt_tilde[2], 'b', label="q")
ax.legend()
plt.show()# Maximum absolute difference
np.max(np.abs(yt_tilde_star[:, 4] - yt_tilde[2, :-1]))The state vector is
z_t = \begin{bmatrix} 1 \cr q_{2t} \cr q_{1t} \end{bmatrix}
and the state transition dynamics are
z_{t+1} = A z_t + B_1 v_{1t} + B_2 v_{2t}
where A is a 3 \times 3 identity matrix and
B_1 = \begin{bmatrix} 0 \cr 0 \cr 1 \end{bmatrix} ,
\quad B_2 = \begin{bmatrix} 0 \cr 1 \cr 0 \end{bmatrix}
The Markov perfect decision rules are
v_{1t} = - F_1 z_t , \quad v_{2t} = - F_2 z_t
and in the Markov perfect equilibrium, the state evolves according to
z_{t+1} = (A - B_1 F_1 - B_2 F_2) z_t
# In LQ form
A = np.eye(3)
B1 = np.array([[0], [0], [1]])
B2 = np.array([[0], [1], [0]])
R1 = np.array([[0, 0, -a0 / 2],
[0, 0, a1 / 2],
[-a0 / 2, a1 / 2, a1]])
R2 = np.array([[0, -a0 / 2, 0],
[-a0 / 2, a1, a1 / 2],
[0, a1 / 2, 0]])
Q1 = Q2 = γ
S1 = S2 = W1 = W2 = M1 = M2 = 0.0
# Solve using QE's nnash function
F1, F2, P1, P2 = qe.nnash(A, B1, B2, R1, R2, Q1,
Q2, S1, S2, W1, W2, M1,
M2, beta=β, tol=tol1)
# Simulate forward
AF = A - B1 @ F1 - B2 @ F2
z = np.empty((3, n))
z[:, 0] = 1, 1, 1
for t in range(n-1):
z[:, t+1] = AF @ z[:, t]
# Display policies
print("Computed policies for firm 1 and firm 2:\n")
print(f"F1 = {F1}")
print(f"F2 = {F2}")q1 = z[1, :]
q2 = z[2, :]
q = q1 + q2 # Total output, MPE
p = a0 - a1 * q # Price, MPE
fig, ax = plt.subplots(figsize=(9, 5.8))
ax.plot(range(n), q, 'b-', lw=2, label='total output')
ax.plot(range(n), p, 'g-', lw=2, label='price')
ax.set_title('Output and prices, duopoly MPE')
ax.legend(frameon=False)
ax.set_xlabel('t')
plt.show()# Computes the maximum difference between the two quantities of the two firms
np.max(np.abs(q1 - q2))# Compute values
u1 = (- F1 @ z).flatten()
u2 = (- F2 @ z).flatten()
π_1 = p * q1 - γ * (u1) ** 2
π_2 = p * q2 - γ * (u2) ** 2
v1_forward = np.sum(βs * π_1)
v2_forward = np.sum(βs * π_2)
v1_direct = (- z[:, 0].T @ P1 @ z[:, 0])
v2_direct = (- z[:, 0].T @ P2 @ z[:, 0])
# Display values
print("Computed values for firm 1 and firm 2:\n")
print(f"v1(forward sim) = {v1_forward:.4f}; v1 (direct) = {v1_direct:.4f}")
print(f"v2 (forward sim) = {v2_forward:.4f}; v2 (direct) = {v2_direct:.4f}")# Sanity check
Λ1 = A - B2 @ F2
lq1 = qe.LQ(Q1, R1, Λ1, B1, beta=β)
P1_ih, F1_ih, d = lq1.stationary_values()
v2_direct_alt = - z[:, 0].T @ lq1.P @ z[:, 0] + lq1.d
(np.abs(v2_direct - v2_direct_alt) < tol2).all()vt_MPE = np.zeros(n)
vt_follower = np.zeros(n)
for t in range(n):
vt_MPE[t] = -z[:, t].T @ P1 @ z[:, t]
vt_follower[t] = -yt_tilde[:, t].T @ P_tilde @ yt_tilde[:, t]
fig, ax = plt.subplots()
ax.plot(vt_MPE, 'b', label='MPE')
ax.plot(vt_leader, 'r', label='Stackelberg leader')
ax.plot(vt_follower, 'g', label='Stackelberg follower')
ax.set_title(r'MPE vs. Stackelberg Value Function')
ax.set_xlabel('t')
ax.legend(loc=(1.05, 0))
plt.show()# Display values
print("Computed values:\n")
print(f"vt_leader(y0) = {vt_leader[0]:.4f}")
print(f"vt_follower(y0) = {vt_follower[0]:.4f}")
print(f"vt_MPE(y0) = {vt_MPE[0]:.4f}")# Compute the difference in total value between the Stackelberg and the MPE
vt_leader[0] + vt_follower[0] - 2 * vt_MPE[0]