Advanced Tools in Macroeconomics

Transcription

1 Advanced Tools in Macroeconomics Continuous time models (and methods) Pontus Rendahl August 21, 2017

2 Introduction In this lecture we will take a look at models in continuous, as opposed to discrete, time. There are some advantages and disadvantages Advantages: Can give closed form solutions even when they do not exist for the discrete time counterpart. Can be very fast to solve. Trendier than sourdough bread, fixed gear bicycles, and skinny jeans combined (so if you do continuous time you need neither). Disadvantages: Intuition is a bit tricky. Contraction mapping theorems / convergence results go out the window (but can be somewhat brought back). The latter can create issues for numerical computing. Difficult to deal with certain end-conditions (like finite lives etc.)

3 Plan for today Continuous time methods and models are not as well documented as the discrete time cases. Proceed through a series of examples 1. The Solow growth model (!) 2. The (stochastic) Ramsey growth model 3. A monetary economy 4. Search and matching How to solve (turns out to be pretty easy, and we can apply methods we know from earlier parts of the course)

4 Plan for tomorrow Speeding things up and making it more robust Little tips and tricks Heterogenous agent models in continuous time Focus on the Aiyagari model Useful list of papers to read 1. Finite?Difference Methods for Continuous?Time Dynamic Programming by Candler moll/hact.pdf 3. moll/wimm.pdf

5 The Solow growth model The Solow growth model is characterized by the following equations Y t = K α t (A t N t ) 1 α K t+1 = I t + (1 δ)k t S t = sy t I t = S t A t+1 = (1 + g)a t N t+1 = (1 + η)n t To solve this model we rewrite it in intensive form x t = X t A t N t, for X = {Y, K, S, I }

6 The Solow growth model Using this and substituting in gives K t+1 A t N t = sk α t + (1 δ)k t We can rewrite as K t+1 A t+1 N t+1 A t+1 N t+1 A t N t = sk α t + (1 δ)k t k t+1 A t+1 N t+1 A t N t = sk α t + (1 δ)k t k t+1 (1 + g)(1 + η) = sk α t + (1 δ)k t

7 The Solow growth model Ta-daa! k t+1 = Balanced growth: k t+1 = k t = k sk α t (1 + g)(1 + η) + (1 δ)k t (1 + g)(1 + η) ( g + η + gη + δ k = s ) 1 α 1 This is not textbook stuff. Why? Discrete time. More elegant solution in continuous time.

8 The Solow growth model Continuous time is not a state in itself, but is the effect of a limit. A derivative is a limit, an integral is a limit, the sum to infinity is a limit, and so on. Continuous time is the name we use for the behavior of an economy as intervals between time periods approaches zero.

9 The Solow growth model The right approach is therefore to derive this behavior as a limit (much like you probably derived derivatives from its limit definition in high school). Eventually you may get so well versed in the limit behavior that you can set it up directly (like you can say that the derivative of ln x is equal to 1/x, without calculating lim ε 0 (ln(x + ε) ln(x))/ε) I m not there yet. I have to do this the complicated way. People like Ben Moll at Princeton is. Take a look at his lecture notes on continuous time stuff. They are great.

10 The Solow growth model Back to the model. Suppose that before the length of each time period was one month. Now we want to rewrite the model on a biweekly frequency. It seems reasonable to assume that in two weeks we produce half as much as we do in one month: Y t = 0.5K α t (A t N t ) 1 α. It also seems reasonable that capital depreciates slower, i.e. 0.5δ.

11 The Solow growth model Notice that we still have N t worker and K t units of capital: Stocks are not affected by the length of time intervals (although the accumulation of them will). The propensity to save is the same, but with half of the income saving is halved too (and therefore investment) What happens to the exogenous processes for A t and N t?

12 The Solow growth model Before A t+1 = (1 + g)a t, N t+1 = (1 + η)n t Now A t+0.5 = ( g)A t, N t+0.5 = ( η)N t or A t+0.5 = e 0.5g A t, N t+0.5 = e 0.5η N t?

13 The Solow growth model It turns out that this choice does not matter much for our purpose Suppose that the time period is not one month but one month. And suppose that Rearrange A t+ = (1 + g)a t A t+ A t = ga t. And take limit 0 A t = ga t

14 The Solow growth model Suppose that the time period is not one month but one month. And suppose that Rearrange A t+ = e g A t Notice A t+ A t = (e g 1) A t. (e g 1) lim 0 = lim 0 (ge g ) 1 = g

15 The Solow growth model So and thus lim 0 (e g 1) A t = ga t A t = ga t Therefore, it doesn t matter if A t+ = (1 + g)a t or A t+ = e g A t. The limits are the same.

16 The Solow growth model Solow growth model in units of time Y t = Kt α (A t N t ) 1 α K t+ = I t + (1 δ)k t S t = sy t I t = S t A t+ = (1 + g)a t N t+ = (1 + η)n t

17 The Solow growth model Substitute and rearrange as before K t+ A t+ N t+ A t+ N t+ A t N t = s k α t + (1 δ)k t k t+ A t+ N t+ A t N t = s k α t + (1 δ)k t k t+ (1 + g)(1 + η) = s k α t + (1 δ)k t

18 The Solow growth model Simplify and rearrange k t+ (1+ g)(1 + η) = s k α t + (1 δ)k t k t+ k t = s kt α δk t (g + η + gη)k t+ k t+ k t = skt α δk t (g + η + gη)k t+ Take limits 0 k t = sk α t (g + η + δ)k t With steady state ( g + η + δ k = s ) 1 α 1

19 The Solow growth model: Solution How do you solve this model? The equation k t = sk α t (g + η + δ)k t is an ODE. Declare it as a function with respect to time, t, and capital, k, in Matlab as solow k) sk α (g + η + δ)k; The simulate it for, say 100 units of time, with initial condition k 0 as [time, capital] = ode45(solow, [0 100], k 0 );

20 Investment Consumption Output Capital The Solow growth model: Solution Solow growth model: Saving like China Time (years) Time (years)

21 The Solow growth model: Solution A few pointers Once you got the solution of a deterministic continuous time model, the solution will always be of the form x t = f (x t ), whether or not x t is a vector. The matlab function ode45 (or other versions) can then simulate a transition (such as an impulse response). You could also simulate on your own through the approximation x t x t+ x t and thus find your solution as x t+ = x t + f (x t ). For this to be accurate, must be small if there are a lot of nonlinearities. The ODE function in matlab uses so-called Runge Kutta methods to vary the step-size in an optimal way.

22 The Ramsey growth model Now consider the Ramsey growth model (without growth) In units of time v(k t ) = max c t,k t+1 {u(c t ) + (1 ρ)v(k t+1 )} s.t. c t + k t+1 = k α t + (1 δ)k t v(k t ) = max c t,k t+ { u(c t ) + (1 ρ)v(k t+ )} s.t. c t + k t+ = k α t + (1 δ)k t

23 The Ramsey growth model Notice that all flows change when the length of the time period on which they are defined changes. Stocks, k, are the same. I discount the future with 1 ρ instead of (1 ρ) (or with e ρ instead of e ρ, but these are, in the limit, equivalent). One funny thing: Consumption, c, is still monthly consumption, but it now only cost as much, and I only get a fraction of the utility! These assumptions are for technical reasons, and it will (hopefully) soon be clear why they are made.

24 The Ramsey growth model Bellman equation v(k t ) = max c t,k t+ { u(c t ) + (1 ρ)v(k t+ )} s.t. c t + k t+ = k α t + (1 δ)k t Subtract v(k t ) from both sides and insert the budget constraint into v(k t+ ) 0 = max c t { u(c t ) + v(k t + (k α t δk t c t )) v(k t ) ρv(k t + (k α t δk t c t ))}

25 The Ramsey growth model From before 0 = max c t { u(c t ) + v(k t + (k α t δk t c t )) v(k t ) Divide by ρv(k t + (k α t δk t c t ))} 0 = max c t {u(c t ) + v(k t + (k α t δk t c t )) v(k t ) ρv(k t + (k α t δk t c t ))}

26 The Ramsey growth model From before 0 = max c t {u(c t ) + v(k t + (k α t δk t c t )) v(k t ) ρv(k t + (k α t δk t c t ))} Take limit 0 and rearrange ρv(k t ) = max c t {u(c t ) + v (k t )(k α t δk t c t )} This is know as the Hamilton-Jacobi-Bellman (HJB) equation.

27 The Ramsey growth model: Solving Dropping time notation we have ρv(k) = max{u(c) + v (k)(k α δk c)} c This is simple to solve and (can be) blazing fast!

28 The Ramsey growth model: Solving Dropping time notation we have ρv(k) = max{u(c) + v (k)(k α δk c)} c This is simple to solve and (can be) blazing fast! Why fast? Maximization is trivial: First order condition u (c) = v (k) So if we know v (k) we know optimal c without searching for it!

29 The Ramsey growth model: Solving How do we find v (k)? Suppose we have hypothetical values of v(k) on a uniformly spaced grid of k, K = {k 0, k 1,..., k N } with stepsize k. We can then approximate v (k) at gridpoint k i (i 1, N) as or v (k i ) = 0.5(v(k i+1 ) v(k i ))/ k + 0.5(v(k i ) v(k i 1 ))/ k v (k i ) = v(k i+1) v(k i 1 ) 2 k

30 The Ramsey growth model: Solving and for k 1 and k N and v (k 1 ) = (v(k 2 ) v(k 1 ))/ k v (k N ) = (v(k N ) v(k N 1 ))/ k There are many ways of doing this. If you have a vector of v(k) values call it V then dv=gradient(v)/dk.

31 The Ramsey growth model: Solving I prefer an alternative method. Construct the matrix D as 1/dk 1/dk /dk 0 0.5/dk D = 0 0.5/dk 0 0.5/dk /dk 1/dk Then v (k) D v(k)

32 The Ramsey growth model: Solving Algorithm 1. Construct a grid for k. 2. For each point on the grid, guess for a value of V Calculate the derivative as dv 0 =D*V Find V 1 from ρv 1 = u(c 0 ) + dv 0 (k α δk c 0 ), with u (c 0 ) = dv 0 5. Back to step 3 with V 1 replacing V 0. Repeat until convergence.

33 The Ramsey growth model: Solving Algorithm 1. Construct a grid for k. 2. For each point on the grid, guess for a value of V Calculate the derivative as dv 0 =D*V Find V 1 from ρv 1 = u(c 0 ) + dv 0 (k α δk c 0 ), with u (c 0 ) = dv 0 5. Back to step 3 with V 1 replacing V 0. Repeat until convergence. Beware: The contraction mapping theorem does not work, so convergence is an issue. Solution: update slowly. That is, V 1 = γv 1 + (1 γ)v 0, for a low value of γ.

34 The Ramsey growth model: Solving Alternative algorithm 1. Construct a grid for k. 2. For each point on the grid, guess for a value of V Calculate the derivative as dv 0 =D*V Find V 1 from V 1 = Γ(u(c 0 ) + dv 0 (k α δk c 0 ) ρv 0 ) + V 0, with u (c 0 ) = dv 0 5. Back to step 3 with V 1 replacing V 0. Repeat until convergence. We will take a look at an alternative way of doing things tomorrow.

35 The Ramsey growth model: Solving Alternative vs. standard algorithm But to me they sort of look the same I.e. V 1 = γ 1 ρ (u(c 0) + dv 0 (k α δk c 0 )) + (1 γ)v 0 = γ ρ (u(c 0) + dv 0 (k α δk c 0 ) ρv 0 ) + V 0 So as long as γ = Γ they should be identical. ρ

36 The Ramsey growth model: Euler equation Let s go back to the HJB equation. ρv(k) = u(c) + v (k)(k α δk c) with u (c) = v (k) Thus ρv (k) = v (k)(k α δk c) + v (k)(αk α 1 δ) And v (k) = u (c)c (k)

37 The Ramsey growth model: Euler equation Using ρv (k) = v (k)(k α δk c) + v (k)(αk α 1 δ) Together with v (k) = u (c) and v (k) = u (c)c (k) gives or ρu (c) = u (c)c (k)(k α δk c) + u (c)(αk α 1 δ) u (c)c (k)(k α δk c) = u (c)(αk α 1 δ ρ) Suppose CRRA utility, such that u (c)c u (c) = γ

38 The Ramsey growth model: Euler equation Then the last equation u (c)c (k)(k α δk c) = u (c)(αk α 1 δ ρ) is equal to γ c (k) (k α δk c) = (αk α 1 δ ρ) c This is the Euler equation in continuous time.

39 The Ramsey growth model: Euler equation Before we attempt to solve the Euler equation, recall that we had k t+ + c t = k α t + (1 δ)k t rearrange k t+ k t = (k α t δk t c t ) Divide with and take limit 0 to get k t = k α t δk t c t Or dropping time notation k = k α δk c

40 The Ramsey growth model: Euler equation Our Euler equation is or now c (k) (k α δk c) = 1 c γ (αk α 1 δ ρ) γ c (k) c k = (αk α 1 δ ρ) What is c (k) k? Recall chain rule Thus ċ = c t t = c t k k t = c (k) k ċ c = 1 γ (αk α 1 δ ρ)

41 The Ramsey growth model: Exploring Two equations ċ = c γ (αk α 1 δ ρ) k = k α δk c Nullclines 0 = αk α 1 δ ρ 0 = k α δk c

42 Consumption, c The Ramsey growth model: Exploring 3.2 " k=0 " c= Capital,k

43 Consumption, c The Ramsey growth model: Exploring 3.2 " k=0 " c= Capital,k

44 Consumption, c The Ramsey growth model: Exploring " k=0 " c=0 Explosive paths Capital,k

45 Consumption, c The Ramsey growth model: Exploring " k=0 " c=0 Explosive paths Saddle path Capital,k

46 The Ramsey growth model: Exploring How did I do that? I created a grid for k and c, and found ċ and k through ċ = c γ (αk α 1 δ ρ) k = k α δk c Then I used Matlab s command quiver(k,c, k,ċ) This creates the swarm of arrows I then used Matlab s command streamline(k,c, k,ċ) at various starting values to get the explosive paths. Lastly I solved for the saddle path and plotted it.

47 The Ramsey growth model: Euler equation solution Back to the recursive Euler Solve for c c (k) (k α δk c) = 1 c γ (αk α 1 δ ρ) c (k)(k α δk) c = 1 (αk γ α 1 δ ρ) + c (k)

48 The Ramsey growth model: Euler equation solution Algorithm 1. Construct a grid for k. 2. For each point on the grid, guess for a value of c Calculate the derivative as dc 0 =D*c Find c 1 from dc 0 (k α δk) c 1 = 1 (αk γ α 1 δ ρ) + dc 0 5. Back to step 3 with c 1 replacing c 0. Repeat until convergence.

49 The Ramsey growth model: Euler equation solution Algorithm 1. Construct a grid for k. 2. For each point on the grid, guess for a value of c Calculate the derivative as dc 0 =D*c Find c 1 from dc 0 (k α δk) c 1 = 1 (αk γ α 1 δ ρ) + dc 0 5. Back to step 3 with c 1 replacing c 0. Repeat until convergence. Beware: No guaranteed convergence. Update slowly. Fewer gridpoints appears to provide some stability.

50 _ The Ramsey growth model: Solution k t Value function iteration Euler equation iteration k t

51 Capital The Ramsey growth model: Solution 6.65 Comparison Euler equation iteration Value function iteration Time (quarters)

52 More on continuous time Euler equations We derived the Euler equation in a slightly roundabout way 1. Discrete time Bellman equation 2. To continuous time HJB equation 3. To continuous time Euler equation using the envelope condition. This can be done more directly from the discrete time Euler equation.

53 More on continuous time Euler equations The discrete time Euler equation is given by In units of time u (c t ) = (1 ρ)(1 + αk α 1 t+1 δ)u (c t+1 ) u (c t ) = (1 ρ)(1 + (αk α 1 t+ δ))u (c t+ ) Use the approximation x t+ x t + x t to get u (c t ) = (1 ρ)(1 + (αk α 1 t+ δ))u (c t + c t )

54 More on continuous time Euler equations u (c t ) = (1 ρ)(1 + (αk α 1 t+ δ))u (c t + c t ) Move the u (c t + c t ) term to the left-hand side and expand u (c t ) u (c t + c t ) = [αk α 1 α 1 t+ δ ρ ρ (αkt+ δ)]u (c t + c t ) Divide by and take limits 0 u (c t ) c t = [αk α 1 t δ ρ]u (c t )

55 More on continuous time Euler equations u (c t ) c t = [αkt α 1 δ ρ]u (c t ) Lastly, use the CRRA property to get c t = 1 α 1 [αkt δ ρ] c t γ

56 More on continuous time Euler equations Now consider a stochastic model with a good, g, and a bad, b, state and u (c g t ) = (1 ρ)[(1 p)(1+z g t+1 α(k g t+1 )α 1 δ)u (c g t+1 ) + p(1 + z b t+1α(k b t+1) α 1 δ)u (c b t+1)] u (c b t ) = (1 ρ)[(1 q)(1+z g t+1 α(k g t+1 )α 1 δ)u (c g t+1 ) + q(1 + z b t+1α(k b t+1) α 1 δ)u (c b t+1)] We will focus on the good state (the treatment of the bad state is symmetric)

57 More on continuous time Euler equations Good state Euler equation in units of time u (c g t ) = (1 ρ)[(1 p)(1+ (z g t+ α(k g t+ )α 1 δ)) u (c g t+ ) + p(1 + (z b t+ α(k b t+ ) α 1 δ))u (c b t+ )] Use u (c g t+ ) u (c g t + c g t ) again, move to the left-hand side, divide by and take limits Or u (c g t ) c g t = (z g t α(k g t ) α 1 δ ρ))u (c g t ) + p(u (c b t ) u (c g t )) c g t c g t = 1 γ (z g t α(k g t ) α 1 δ ρ)) + p (u (c b t ) u (c g t )) u (c g t )

58 More on continuous time Euler equations For the bad state c b t c b t = 1 γ (z b t α(k b t ) α 1 δ ρ)) + q (u (c b t ) u (c g t )) u (c b t ) These can be solved using the previous methods. The only difference is that we now iterate on two equations instead of one. But the procedure is the same.

59 More on continuous time Euler equations As a last step, I just want to give you a hint on how these ideas can be applied in different settings. For instance, the Euler equation for a standard deterministic monetary model is given by In units of time u (c t ) = (1 ρ)(1 + i t+1 ) p t p t+1 u (c t+1 ) u (c t ) = (1 ρ)(1 + i t+1 ) p t p t+ u (c t+ )

60 More on continuous time Euler equations Use the approximations u (c t+ ) u (c t + c t ), and p t p t+ p t and rewrite u (c t ) = (1 ρ)(1 + i t+ ) p t+ p t u (c t + c t ) Expand p t+ u (c t ) = (1 ρ + i t+ 2 i t+ ρ)(1 Thus p t )u (c t + c t ) p t+ u (c t ) u (c t + c t ) = ( ρ + i t+ 2 i t+ ρ) (1 p t )u (c t + c t ) p t u (c t + c t ) p t+ p t+

61 More on continuous time Euler equations Previous equation u (c t ) u (c t + c t ) = ( ρ + i t+ 2 i t+ ρ) (1 p t )u (c t + c t ) p t u (c t + c t ) p t+ p t+ Divide by u (c t ) u (c t + c t ) (1 p t )u (c t + p t+ And take limit 0 = ( ρ + i t+ i t+ ρ) c t ) c t = 1 c t γ (i t p t ρ) p t p t u (c t + c t ) p t+

62 The Mortensen-Pissarides model Continuous time is frequently used in the theoretical labor literature. In the remainder of this lecture I will go through the workhorse model developed by Christopher Pissarides and Dale Mortensen. In today s exercise you will be asked to solve this model.

63 The Mortensen-Pissarides model Continuous time is frequently used in the theoretical labor literature. In the remainder of this lecture I will go through the workhorse model developed by Christopher Pissarides and Dale Mortensen. In today s exercise you will be asked to solve this model.

64 The Mortensen-Pissarides model: Workers In the simplest case workers are risk-neutral and value a job according to V t = w t + (1 ρ)[(1 δ)v t+1 + δu t+1 ] The value of being unemployed is given by U t = b + (1 ρ)[(1 f t+1 )U t+1 + f t+1 V t+1 ] The variables w t and f t+1 will be endogenously determined, but we will, for the moment, treat them as exogenous.

65 The Mortensen-Pissarides model: Workers Let s rewrite these equations in units of time V t = w t + (1 ρ)[(1 δ)v t+ + δu t+ ] U t = b + (1 ρ)[(1 f t+ )U t+1 + f t+ V t+1 ] Or V t V t+ = w t [δ + ρ δρ]v t+ + (1 ρ) δu t+ U t U t+ = b [f t+ + ρ f t+ ρ]u t+ + (1 ρ) f t+ V t+

66 The Mortensen-Pissarides model: Workers Dividing through by gives V t V t+ U t U t+ = w t [δ + ρ δρ]v t+ + (1 ρ)δu t+ = b [f t+ + ρ f t+ ρ]u t+ + (1 ρ)f t+ V t+ And taking limits 0 yields V t = w t (δ + ρ)v t + δu t U t = b (f t + ρ)u t + f t V t

67 The Mortensen-Pissarides model: Workers These equations are commonly written as ρv t = w t + V t + δ(u t V t ) ρu t = b + U t + f t (V t U t ) Define the surplus of having a job as S t = V t U t, that is (ρ + δ + f t )S t = w t b + S t

68 The Mortensen-Pissarides model: Firms The value to a firm of having an employed worker is J t = z t w t + (1 ρ)[(1 δ)j t+1 + δw t+1 ] We will assume free entry, such that W t = 0 for all t. Following the same procedure as before we find that in continuous time (ρ + δ)j t = z t w t + J t

69 The Mortensen-Pissarides model: Wages Collecting equations ρs t = w t b + S t + (δ + f t )S t ρj t = z t w t + J t δj t Wages are set according to Nash bargaining, which are renegotiated period-by-period First order condition w t = argmax{j η t S 1 η t } ηs t = (1 η)j t

70 The Mortensen-Pissarides model: Wages Expanding η(w t b + S t + (δ + f t )S t ) = (1 η)(z t w t + J t δj t ) and using the fact that gives S t = 1 η J t, and S η t = 1 η η J t w t = ηb + (1 η)z t + f t (1 η)j t Inserting into the firm s value function gives ρj t = η(z t b) + J t (f t (1 η) + δ)j t

71 The Mortensen-Pissarides model: Matching Suppose that there are v t vacancies posted and u t unemployed individuals. Then the measure of matches in a given period is given as M t = ψv ω t u 1 ω t The probability that an unemployed individual finds a job, f t, is then given as f t = M t u t = ψθ ω t, with θ = v t u t

72 The Mortensen-Pissarides model: Matching The probability that a vacant position is filled, h t, is then given as h t = M t v t = ψθ ω 1 t Suppose the cost of posting a vacancy is given by κ. Free entry then ensures that κ = h t J t To see this more clearly, a firm that is considering posting a vacancy faces the optimization problem max v t,i { κ v t,i + v t,i h t J t }

73 The Mortensen-Pissarides model: Matching Lastly, employment, n t = 1 u t, satisfies the law of motion n t+ = (1 n t ) f t + (1 δ)n t which can be rearranged to taking limits n t+ n t = (1 n t )f t δn t n t = (1 n t )f t δn t

74 The Mortensen-Pissarides model: The standard Mortensen-Pissarides model is therefore characterized by the three equations ρj t = η(z t b) + J t (f t (1 η) + δ)j t κ = h t J t n t = (1 n t )f t δn t in the three unknowns J t, θ t, n t.