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Taylor Collins 1 RECURSIVE MACROECONOMIC THEORY, LJUNGQVIST AND SARGENT, 3 RD EDITION, CHAPTER 19 DYNAMIC STACKELBER G PROBLEMS.

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Presentation on theme: "Taylor Collins 1 RECURSIVE MACROECONOMIC THEORY, LJUNGQVIST AND SARGENT, 3 RD EDITION, CHAPTER 19 DYNAMIC STACKELBER G PROBLEMS."— Presentation transcript:

1 Taylor Collins 1 RECURSIVE MACROECONOMIC THEORY, LJUNGQVIST AND SARGENT, 3 RD EDITION, CHAPTER 19 DYNAMIC STACKELBER G PROBLEMS

2 BACKGROUND INFORMATION A new type of problem Optimal decision rules are no longer functions of the natural state variables A large agent and a competitive market A rational expectations equilibrium Recall Stackelberg problem from Game Theory The cost of confirming past expectations Taylor Collins2

3 THE STACKELBERG PROBLEM Solving the problem – general idea Defining the Stackelberg leader and follower Defining the variables: Z t is a vector of natural state variables X t is a vector of endogenous variables U t is a vector of government instruments Y t is a stacked vector of Z t and X t Taylor Collins3

4 THE STACKELBERG PROBLEM The government’s one period loss function is Government wants to maximize subject to an initial condition for Z 0, but not X 0 Government makes policy in light of the model The government maximizes (1) by choosing subject to (2) Taylor Collins4 (1) (2)

5 PROBLEM S “The Stackelberg Problem is to maximize (2) by choosing an X 0 and a sequence of decision rules, the time t component of which maps the time t history of the state Z t into the time t decision of the Stackelberg leader.” The Stackelberg leader commits to a sequence of decisions The optimal decision rule is history dependent Two sources of history dependence Government’s ability to commit at time 0 Forward looking ability of the private sector Dynamics of Lagrange Multipliers The multipliers measure the cost today of honoring past government promises Set multipliers equal to zero at time zero Multipliers take nonzero values thereafter Taylor Collins5

6 SOLVING THE STACKELBERG PROBLEM 4 Step Algorithm Solve an optimal linear regulator Use stabilizing properties of shadow prices Convert Implementation multipliers into state variables Solve for X 0 and μ x0 Taylor Collins6

7 STEP 1: SOLVE AN O.L.R. Assume X 0 is given This will be corrected for in step 3 With this assumption, the problem has the form of an optimal linear regulator The optimal value function has the form where P solves the Riccati Equation The linear regulator is subject to an initial Y 0 and the law of motion from (2) Then, the Bellman Equation is Taylor Collins7 (3)

8 STEP 1: SOLVE AN O.L.R. Taking the first order condition of the Bellman equation and solving gives us Plugging this back into the Bellman equation gives us such that ū is optimal, as described by (4) Rearranging gives us the matrix Riccati Equation Denote the solution to this equation as P * Taylor Collins8 (4)

9 STEP 2: USE THE SHADOW PRICE Decode the information in P * Adapt a method from 5.5 that solves a problem of the form (1),(2) Attach a sequence of Lagrange multipliersto the sequence of constraints (2) and form the following Lagrangian Partition μ t conformably with our partition of Y Taylor Collins9

10 STEP 2: USE THE SHADOW PRICE Want to maximize L w.r.t. U t and Y t+1 Solving for U t and plugging into (2) gives us Combining this with (5), we can write the system as Taylor Collins10 (5 ) (6 )

11 STEP 2: USE THE SHADOW PRICE We now want to find a stabilizing solution to (6) ie, a solution that satisfies In section 5.5, it is shown that a stabilizing solution satisfies Then, the solution replicates itself over time in the sense that Taylor Collins11 (7)

12 STEP 3: CONVERT IMPLEMENTATION MULTIPLIERS We now confront the inconsistency of our assumption on Y 0 Forces multiplier to be a jump variable Focus on partitions of Y and μ Convert multipliers into state variables Write the last n x equations of (7) as Pay attention to partition of P Solving this for X t gives us Taylor Collins12 (8)

13 STEP 3: CONVERT IMPLEMENTATION MULTIPLIERS Using these modifications and (4) gives us We now have a complete description of the Stackelberg problem Taylor Collins13 (9) (9’’) (9’)

14 STEP 4: SOLVE FOR X 0 AND The value function satisfies Now, choose X 0 by equating to zero the gradient of V(Y 0 ), w.r.t. X 0 Then, recall (8) Finally, the Stackelberg problem is solved by plugging in these initial conditions to (9), (9’), and (9’’) and iterating the process to get Taylor Collins14 μ x0

15 CONCLUSION Brief Review Setup and Goal of problem 4 step Algorithm Questions, Comments, or Feedback Taylor Collins15


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