@@ -35,7 +35,7 @@ tags: [hide-output]
35353636## Overview
373738-This notebook formulates and computes a plan that a **Stackelberg
38+This lecture formulates and computes a plan that a **Stackelberg
3939leader** uses to manipulate forward-looking decisions of a **Stackelberg
4040follower** that depend on continuation sequences of decisions made once
4141and for all by the Stackelberg leader at time $0$.
@@ -77,13 +77,13 @@ $a_0$ and $a_1$ are both positive.
7777$q_{10}, q_{20}$ are given numbers that serve as initial
7878conditions at time $0$.
797980-By incurring a cost of change
80+By incurring a cost equal to
81818282$$
83-\gamma v_{it}^2
83+\gamma v_{it}^2, \quad \gamma > 0 ,
8484$$
858586-where $\gamma > 0$, firm $i$ can change its output according
86+firm $i$ can change its output according
8787to
88888989$$
@@ -217,8 +217,7 @@ $$
217217where
218218$c_0 = \frac{\beta a_0}{2 \gamma}, c_1 = \frac{\beta a_1}{\gamma}, c_2 = \frac{\beta a_1}{2 \gamma}$.
219219220-This equation can in turn be rearranged to become the second-order
221-difference equation
220+This equation can in turn be rearranged to become
222221223222```{math}
224223:label: sstack1
@@ -272,7 +271,7 @@ $$
272271Operating on both sides of equation {eq}`sstack2` with
273272$\beta^{-1}$ times this inverse operator gives the follower's
274273decision rule for setting $q_{1t+1}$ in the
275-**feedback-feedforward** form.
274+**feedback-feedforward** form
276275277276```{math}
278277:label: sstack3
@@ -304,7 +303,7 @@ and formulate the following Lagrangian for the Stackelberg leader firm
304303305304subject to initial conditions for $q_{1t}, q_{2t}$ at $t=0$.
306305307-**Comments:** We have formulated the Stackelberg problem in a space of
306+**Remarks:** We have formulated the Stackelberg problem in a space of
308307sequences.
309308310309The max-min problem associated with Lagrangian
@@ -314,11 +313,11 @@ future of its choices of $\{q_{1t+j}\}_{j=0}^\infty$.
314313315314This renders a direct attack on the problem cumbersome.
316315317-Therefore, below, we will formulate the Stackelberg leader's problem
316+Therefore, below we will formulate the Stackelberg leader's problem
318317recursively.
319318320319We'll put our little duopoly model into a broader class of models with
321-the same conceptual structure.
320+the same structure.
322321323322## Stackelberg Problem
324323@@ -342,7 +341,7 @@ of the Stackelberg **follower**.
342341Let $u_t$ be a vector of decisions chosen by the Stackelberg leader
343342at $t$.
344343345-The $z_t$ vector is inherited physically from the past.
344+The $z_t$ vector is inherited from the past.
346345347346But $x_t$ is a decision made by the Stackelberg follower at time
348347$t$ that is the follower's best response to the choice of an
@@ -464,13 +463,13 @@ Subproblem 2 is solved by the **Stackelberg leader** at $t=0$.
464463465464The two subproblems are designed
466465467-- to respect the protocol in which the follower chooses
466+- to respect the timing protocol in which the follower chooses
468467 $\vec q_1$ after seeing $\vec q_2$ chosen by the leader
469468- to make the leader choose $\vec q_2$ while respecting that
470469 $\vec q_1$ will be the follower's best response to
471470 $\vec q_2$
472471- to represent the leader's problem recursively by artfully choosing
473- the state variables confronting and the control variables available
472+ the leader's state variables and the control variables available
474473 to the leader
475474476475**Subproblem 1**
@@ -1012,8 +1011,9 @@ plt.show()
1012101110131012We'll compute the present value earned by the Stackelberg leader.
101410131015-We'll compute it two ways (they give identical answers -- just a check
1016-on coding and thinking)
1014+We'll compute it two ways and get the same answer.
1015+1016+In addition to being a useful check on the accuracy of our coding, computing things in these two ways helps us think about the structure of the problem.
1017101710181018```{code-cell} python3
10191019v_leader_forward = np.sum(βs * π_leader)