Economics 2010c: Lecture 1 Introduction to Dynamic Programming

Economics 2010c: Lecture 1 Introduction to Dynamic Programming David Laibson 9/02/2014 Outline of my half-semester course: 1. Discrete time method...
Author: Arlene Wilcox
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Economics 2010c: Lecture 1 Introduction to Dynamic Programming David Laibson

9/02/2014

Outline of my half-semester course:

1. Discrete time methods (Bellman Equation, Contraction Mapping Theorem, and Blackwell’s Sufficient Conditions, Numerical methods) • Applications to growth, search, consumption, asset pricing 2. Continuous time methods (Bellman Equation, Brownian Motion, Ito Process, and Ito’s Lemma) • Applications to search, consumption, price-setting, investment, industrial organization, asset-pricing

Outline of today’s lecture:

1. Introduction to dynamic programming

2. The Bellman Equation

3. Three ways to solve the Bellman Equation

4. Application: Search and stopping problem

1

Introduction to dynamic programming. • Course emphasizes methodological techniques and illustrates them through applications.

• We start with discrete-time dynamic optimization. • Is optimization a ridiculous model of human behavior? Why or why not? • Today we’ll start with an ∞-horizon stationary problem: The Sequence Problem (cf. Stokey and Lucas)

Notation:  is the state vector at date   ( +1) is the flow payoff at date  ( is ‘stationary’)   is the exponential discount function  is referred to as the exponential discount factor The discount rate is the rate of decline of the discount function, so  ≡ − ln  = −

h i    

Note that exp(−) =  and exp(−) =  





Definition of Sequence Problem: Find () such that (0) =

sup

∞ X

{+1 }∞ =0 =0

  ( +1)

subject to +1 ∈ Γ() with 0 given. Remark 1.1 When I omit time subscripts, this implies that an equation holds for all relevant values of . In the statement above, +1 ∈ Γ() implies, +1 ∈ Γ() for all  = 0 1 2 

Example 1.1 Optimal growth with log utility and Cobb-Douglas technology: sup

∞ X

{}∞ =0 =0

  ln()

subject to the constraints,   ≥ 0  + +1 =  and 0 given. Translate this problem into Sequence Problem notation by (1) eliminating redundant variables and (2) introducing constraint correspondence Γ Example 1.2 Optimal growth translated into Sequence Problem notation: (0) =

sup

∞ X

{+1}∞ =0 =0

  ln( − +1)

such that +1 ∈ [0 ] ≡ Γ() and 0 given.

2

Bellman Equation

Compare Sequence Problem and Bellman Equation. Definition: Bellman Equation expresses the value function as a combination of a flow payoff and a discounted continuation payoff: () =

sup +1∈Γ()

{ ( +1) + (+1)}

∀

• Flow payoff is  ( +1) • Current value function is () Continuation value function is (+1) • Equation holds for all (feasible) values of 

• We call (·) the solution to the Bellman Equation. • Note that any old function won’t solve the Bellman Equation. • We haven’t yet demonstrated that there exists even one function (·) that will satisfy the Bellman equation.

• We will show that the (unique) value function defined by the Sequence Problem is also the unique solution to the Bellman Equation.

A solution to the Sequence Problem is also a solution to the Bellman Equation. (0) = =

∞ X

sup

+1∈Γ() =0 ⎧ ⎨

sup

+1∈Γ()

= = =

sup

 (0 1) +

⎩ ⎧ ⎨

1∈Γ(0)

sup

1∈Γ(0)

∞ X

=1 ∞ X

 (0 1) + 



⎫ ⎬

  ( +1)

 (0 1) + 

⎩ +1∈Γ() ⎧ ⎨

sup

  ( +1)



⎫ ⎬

 −1 ( +1)

=1

sup

∞ X

+1 ∈Γ() =0

{ (0 1) + (1)}



⎫ ⎬

  (+1 +2)



A solution to the Bellman Equation is also a solution to the Sequence Problem. (0) = =

sup 1∈Γ(0)

{ (0 1) + (1)}

sup

+1∈Γ()

..

=

sup +1∈Γ()

=

sup

{ (0 1) +  [ (1 2) + (2)]} n

o −1   (0 1) + · · · +   (−1 ) +  ()

∞ X

+1∈Γ() =0

  ( +1)

Sufficient condition: lim→∞  () = 0 ∀ feasible  sequences (Stokey and Lucas Thm. 4.3). In summary, a solution to the Bellman Equation will also be a solution to the Sequence Problem and vice versa.

Example 2.1 Optimal growth in Sequence Problem notation: (0) =

sup

∞ X

{+1}∞ =0 =0

  ln( − +1)

such that +1 ∈ [0 ] ≡ Γ() and 0 given. Optimal growth in Bellman Equation notation: () =

sup +1∈Γ()

{ln( − +1) + (+1)}

∀

3

Solving the Bellman Equation • Three methods 1. guess a solution (that’s no typo) 2. iterate functional operator analytically (what’s a functional operator?) 3. iterate functional operator numerically

• Method 1 today. • Guess a function (), and then check to see that this function satisfies the Bellman Equation at all possible values of 

• For our growth example, guess that the solution of the growth problem takes the form: () =  +  ln() where  and  are constants for which we need to find solutions.

• Here value function inherits functional form of utility function (ln). • To solve for constants rewrite Bellman Equation: () =

sup +1∈Γ()

 +  ln() =

sup +1∈Γ()

{ln( − +1) + (+1)}

∀

{ln( − +1) +  [ +  ln(+1)]}

∀

First order condition (FOC) on the right-hand-side of the Bellman Equation:  ( +1) +  0(+1) = 0 +1 Envelope Theorem:  ( +1) 0  () =  

Heuristic Proof of Envelope Theorem:  ( +1)  ( +1) +1  + +  0(+1) +1  +1   " #  ( +1) +1  ( +1) + 0(+1) = +  +1   ( +1) =  

 0() =

Problem Set 1 asks you to use the FOC and the Envelope Theorem to solve for  and . You will also confirm that () =  +  ln() is a solution to the Bellman Equation.

4

Search and optimal stopping

Example 4.1 An agent draws an offer,  from a uniform distribution with support in the unit interval. The agent can either accept the offer and realize net present value  (ending the game), or the agent can reject the offer and draw again a period later. All draws are independent. Rejections are costly because the agent discounts the future exponentially with discount factor . This game continues until the agent receives an offer that she is willing to accept.

• The Bellman equation for this problem is (relatively) easy to write: () = max{  [(+1)]}

(1)

Our problem is to find the value function (·) that solves equation (1). We’ll also want to find the associated policy rule. Definition: A policy is a function that maps  to the action space. Definition: An optimal policy achieves payoff () for all feasible .

Proposition: In the search and optimal stopping problem, the threshold policy with cutoff ∗ is a best response to any continuation value function, b if and only if (iff) ∗ =  [b(+1)] 

Proof: Optimization generates the following policy: ACCEPT iff   ∗ =  [b(+1)] REJECT iff   ∗ =  [b(+1)]

If  = ∗ =  [b(+1)]  then ACCEPT and REJECT generate the same payoff. ¥

• Find threshold ∗ so that the associated value function, () =

(

 if  ≥ ∗ ∗ if  ≤ ∗

)



(2)

satisfies the Bellman Equation. • In other words, find the value of ∗ so that () (defined in equation 2) solves the Bellman Equation (equation 1).

If  = ∗ you should be indifferent between stopping and continuing. (∗) = ∗ = (+1) = 

Z =∗

∗ ()  + 

=0 1 1 ∗ 2 =  ( ) + 

2

So the final result is

Z =1

=∗

  () 

2

i h ∗ 2 ∗ ( ) + 1  =

2

which has solution

µ

∗ =  −1 1 −

q



1 − 2 

Always think about comparative statics and sensibility of the answer.

Optimal threshold in stopping problem 1 0.9 converges to 1 as discount rate goes to 0

0.8

optimal threshold

0.7 0.6 0.5 0.4 0.3 converges to 0 as discount rate goes to ∞

0.2 0.1 0

0.1

0.2

0.3

0.4 0.5 0.6 0.7 discount rate = -ln(delta)

0.8

0.9

1

Outline of today’s lecture:

1. Introduction to dynamic programming

2. The Bellman Equation

3. Three ways to solve the Bellman Equation

4. Application: Search and stopping problem