Solving Models with Heterogeneous Agents Xpa algorithm

Transcription

1 Solving Models with Heterogeneous Agents Xpa algorithm Wouter J. Den Haan London School of Economics c by Wouter J. Den Haan

2 Individual agent Subject to employment shocks ε i,t {0, 1} or ε i,t {u, e} Incomplete markets only way to save is through holding capital

3 Alternatives to inequality constraint Penalty function in the utility function Assumptions about P(k i,t ) u(c i,t, k i,t ) = ln(c i,t ) P(k i,t+1 ) standard inequality: k i,t+1 0 { if ki,t+1 < 0 P(k i,t+1 ) = 0 if k i,t+1 0 differentiable alternative p(k i,t+1 ) = P(k i,t+1) k i,t+1 0

4 Alternatives to inequality constraint Alternative to P ( ) in utility function Individual interest rate depends on amount invested r i,t = r t P(k i,t+1 ) with P(k i,t+1) k i,t+1 > 0 Advantage: nicer economic story Disadvantage: you have to take a stand on what happens with the resources (thrown away? intermediary?) both Euler and budget constraint are affected

5 Laws of motion aggregate productivity, z t, can take on two values employment status, ε t, can take on two values probability of being (un)employed depends on z t transition probabilities are such that unemployment rate only depends on current z t = u t = u(z t ) with u b = u(1 z ) > u g u(1 + z ).

6 Individual agent max E {c i,t,k i,t+1 } t=0 t=0 βt (ln(c i,t ) P(k i,t+1 )) s.t. c i,t + k i,t+1 = r t k i,t + (1 τ t )w t lε i,t + µw t (1 ε i,t ) + (1 δ)k i,t for given processes of r t and w t, this is a relatively simple problem

7 Individual agent: Euler equation 1 c i,t = p(k i,t+1 ) + E t [ ] β(rt δ) c i,t+1 cost of k i,t+1 = reduction in penalty + usual term

8 Firm problem ( r t = z t α K t l(1 u(zt )) ) α 1 ( K t w t = z t (1 α) l(1 u(zt )) ) α

9 Government τ t w t l(1 u(zt ) = µw t u(z t ) τ t = µu(z t ) l(1 u(zt ))

10 Aggregate variables agents care about r t and w t They only depend on aggregate capital stock and z t!!! This is not true in general for equilibrium prices Agents are interested in all information that forecasts K t In principle that is the complete cross-sectional distribution of employment status and capital levels

11 Equilibrium - first part Individual policy functions solving the agent s max problem A wage and a rental rate given by equations above.

12 Equilibrium - second part A transition law for the cross-sectional distribution of capital, consistent with the individual policy function. f t = beginning-of-period cross-sectional distribution of capital and the employment status after the employment status has been realized. f t+1 = Υ(z t+1, z t, f t ) z t+1 does not affect the period t cross-sectional distribution of capital z t+1 does affect the joint cross-sectional distribution of capital and employment status

13 Overview 1 What if individual policy rules are linear in levels? 2 What if individual policy rules are polynomials in levels 3 What if individual policy rules are not polynomials in the levels 4 Topics: What if there are non-differentiabilities Economy with bonds the price of a bond unlike rental rate is not a simple function of K t

14 Linear policy rule for individual problem Suppose policy rules are given and equal to: if ε = u : k u = Ψ u,0 (s) + Ψ u,1 (s)k if ε = e : k e = Ψ e,0 (s) + Ψ e,1 (s)k linear in k, completely general in all other dimensions

15 Linear policy rule for individual problem s is the set of aggregate state variables and consist for sure of z, K u, K e Xpa determines "endogenously" whether other elements should be added to this list

16 Linear policy rule for individual problem Policy rules: if ε = u : k u = Ψ u,0 (s) + Ψ u,1 (s)k if ε = e : k e = Ψ e,0 (s) + Ψ e,1 (s)k Can we calculate K from this? If the answer is yes, then we can calculate r (given z )

17 Notation End of this period K u : end-of-period aggregate capital stock of unemployed K e : end-of-period aggregate capital stock of employed Beginning of the next period K u and K e: corresponding beginning-of-period equivalents

18 Transition laws From end of this period to beginning of next period K u = K u and K e = K e because employment status changes Apply transition laws to go from K e, K u to K e, K u

19 More notation g εε zz : mass of agents with employment status ε now and ε next period for given values of z and z g uuzz + g euzz + g eezz + g uezz = 1

20 From end to beginning-of-period moments K u = g uuzz K u + g euzz K e g uuzz + g euzz K e = g uezz K u + g eezz K e g uezz + g eezz K = u(z )K u + (1 u(z ))K e Given the formulas for K u and K e we can calculate these.

21 Simpler case (to learn method) Suppose individual policy function are as follows: if ε = u : k u = Ψ u,0 + Ψ u,k k + Ψ u,z z + Ψ u,ku K u + Ψ u,ke K e if ε = e : k e = Ψ e,0 + Ψ e,k k + Ψ e,z z + Ψ e,ku K u + Ψ e,ke K e This immediately gives K u = M u (1) = k u( )df u (k) = (Ψ u,0 + Ψ u,z z) + (Ψ u,k + Ψ u,ku ) K u + Ψ u,ke K e, K e = M e (1) = k e( )df e (k) = (Ψ e,0 + Ψ e,z z) + (Ψ e,k + Ψ e,ke ) K e + Ψ e,ku K u, Applying transition laws gives K u, K e, and K

22 What about the state variables? Above, we assumed that s consists of k, ε, z, K u, K e and possibly other characteristics of the distribution. But with linear policy rules, no other moments are needed. Intuition?

23 Simple way to solve for coeffi cients 1 Assume linear law of motion for K u and K e 2 Find linear approximation for k u and k e 3 Get new linear law of motion for K u and K e by explicit aggregation (and transition laws) 4 Iterate until convergence

24 More general linear policy rules Policy rules: This immediately gives if ε = u : k u = Ψ u,0 (s) + Ψ u,1 (s)k if ε = e : k e = Ψ e,0 (s) + Ψ e,1 (s)k K u = M u (1) = k u(k, )df u (k) = Ψ u,0 (s) + Ψ u,1 (s)m u (1), K e = M e (1) = k e(k, )df e (k) = Ψ e,0 (s) + Ψ e,1 (s)m e (1), This law of motion could be non-linear in K u and K e!!!! Applying transition laws gives K u, K e, and K

25 Progress so far Given linear individual policy rule: we can get law of motion for aggregate variables determine what the set of aggregate state variables are Haven t said anything yet on how to find individual policy rules We ll answer that for somewhat more general case with quadratic individual policy rules

26 Second-order policy rule Policy rules: k u = Ψ u,0 (s) + k e = Ψ e,0 (s) + 2 i=1 2 i=1 Ψ u,i (s)k i and Ψ e,i (s)k i, quadratic in k, completely general in all other dimensions

27 Second-order policy rule s is the set of state variables and consist for sure of z, K u, K e Xpa determines "endogenously" whether other elements should be added to this list

28 Aggregate second-order policy Aggregation gives K u = M u (1) = Ψ u,0 (s) + 2 i=1 Ψ u,i(s)m u (i), K e = M e (1) = Ψ e,0 (s) + 2 i=1 Ψ e,i(s)m e (i), What are the state variables? Not only M u (1) and M e (1), but also M u (2) and M e (2)

29 Aggregate second-order policy = we need laws of motion for M u (2) and M e (2); these together with transition laws would give laws of motion for M u(2) and M e(2)

30 Laws of motion for second-order terms Moments are (k M ε (2) = ε (k, ) ) 2 dfε (k) Do I have a law of motion for (k ε) 2? Yes, of course, namely ( k ε ) 2 = Ψ 2 ε,0 (s) + 2Ψ ε,0(s)ψ ε,1 (s)k+ (2Ψ ε,0 (s)ψ ε,2 + (Ψ ε,1 (s)) 2 )k 2 What is the problem? +2Ψ ε,1 (s)ψ ε,2 (s)k 3 + (Ψ ε,2 (s)) 2 k 4.

31 Laws of motion for second-order terms Aggregating gives ( k ε ) 2 = Ψ 2 ε,0 (s) + 2Ψ ε,0(s)ψ ε,1 (s)k+ (2Ψ ε,0 (s)ψ ε,2 + (Ψ ε,1 (s)) 2 )k 2 +2Ψ ε,1 (s)ψ ε,2 (s)k 3 + (Ψ ε,2 (s)) 2 k 4. M ε (2) = (Ψ ε,0 (s)) 2 + 2Ψ ε,0 (s)ψ ε,1 (s) M ε (1) +(2Ψ ε,0 (s)ψ ε,2 + (Ψ ε,1 (s)) 2 ) M ε (2) +2Ψ ε,1 (s)ψ ε,2 (s) M ε (3) + (Ψ ε,2 (s)) 2 M ε (4).

32 Avoiding infinite-regress problem Define y ε = ( k ε ) 2 Come up with a separate 2 nd -order approximation for y ε: y ε = ( k ε ) 2 Ψε,(k ) 2,0 (s) + Ψ ε,(k ) 2,1 (s)k + Ψ ε,(k ) 2,2 (s)k2 Ψ ε,j coeffs have no direct relationship to Ψ ε,(k ) 2,j coeffs

33 Aggregation in second-order case and k ε = Ψ ε,0 (s) + 2 i=1 K ε = M ε (1) = Ψ ε,0 (s) + Ψ ε,i (s)k i gives 2 i=1 Ψ ε,i (s)m ε (i), ( ) 2 k 2 ε = Ψ ε,(k ) 2,0 (s) + Ψ ε,(k ) 2,i (s)ki gives i=1 M ε (2) = Ψ ε,(k ) 2,0 (s) + 2 i=1 Ψ ε,(k ) 2,i (s)m ε(i),

34 Basic formulation Policy rules: k u = Ψ u,0 (s) + Aggregation gives I i=1 Ψ u,i (s)k i and k e = Ψ e,0 (s) + I i=1 K u = M u (1) = Ψ u,0 (s) + I i=1 Ψ u,i(s)m u (i), K e = M e (1) = Ψ e,0 (s) + I i=1 Ψ e,i(s)m e (i), Ψ e,i (s)k i, Get additional separate policy rules for each polynomial term in policy function

35 Full program How to find the Ψ coeffi cients? Iterative perturbation procedures general projection method

36 Iterative perturbation solutions 1 Guess a law of motion for aggregate law of motion 2 Conditional on this solve for individual law of motion 3 Explicitly aggregate and update aggregate law of motions

37 Perturbation solution with discrete support Write law of motion for z t and ε t as [ ] z t = z + ρ z z t 1 + e z,t with E e 2 z,t = σ 2 z [ ] ε i,t = ε + ρ ε ε t 1 + e ε,t with E e 2 ε,t = σ 2 ε ρ z, ρ ε, σ 2 z, σ 2 ε are such that autocovariances and variances correspond to original process

38 General projection procedure Suppose a second-order solution is used: s i,t = [ε i,t, k i,t, z t, M u,t (1), M e,t (1), M u,t (2), M e,t (2)] Basic idea: set up a grid define error term at each grid point define loss function use minimization routine to find Ψ coeffi cients

39 General projection procedure Solve for ψ k (s; Ψ) by making model equations "fit" on grid Notation: we have included ε i,t in s and ψ k (s; Ψ) describes behavior of both employed and unemployed agent

40 Individual policy rules & projection methods {s κ } χ κ=1 the set of state variables with χ nodes s κ = {z κ, M u,κ (1), M e,κ (1), M u,κ (2), M e,κ (2)} κ indicates a grid point not a period

41 First-order condition (r(z κ, K κ ) + 1 δ)k κ +(1 τ(z κ ))w(z κ, K κ )lε κ + µw(z κ, K κ )(1 ε κ ) ψ k (s κ ; Ψ) β(r(z, K ) + (1 δ)) (r(z, K ) + 1 δ)ψ k (s κ ; Ψ) = p(ψ k (s κ ; Ψ)) + E +(1 τ(z ))w(z, K )lε +µw(z, K )(1 ε ) ψ k (s ; Ψ) ν ν

42 Individual problem errors (usual part): u κ = +p(ψ k (s κ ; Ψ)) + z (r(z κ, K κ ) + 1 δ)k κ +(1 τ(z κ ))w(z κ, K κ )lε κ + µw(z κ, K κ )(1 ε κ ) ψ k (s κ ; Ψ) ε β(r(z, K ) + (1 δ)) (r(z, K ) + 1 δ)ψ k (s κ ; Ψ) +(1 τ(z ))w(z, K )lε +µw(z, K )(1 ε ) ψ k (s ; Ψ) π(z, ε z κ, ε κ ) ν ν

43 Individual problem errors (new part): Error for y = (k ) 2 : u κ = ψ (k ) 2(s κ; Ψ) + (ψ k (s κ ; Ψ)) 2

44 Errors only depend only on known things and Ψ r(z κ, K κ ) = αz κ (K κ /( l(1 u(z κ )))) α 1 w(z κ, K κ ) = (1 α)z α κ(k κ /( l(1 u(z κ )))) α r(z, K ) = αz (K /( l(1 u(z κ))) α 1 w(z, K ) = (1 α)z (K /L(z )) α µu(z) τ(z) = and τ(z l(1 ) = µu(z ) u(z)) l(1 u(z )) s = { k, ε, z, M u(1), M e(1), M u(2), M e(2) } = { ψ k (s κ ; Ψ), ε, z, M u(1),, M e(1), M u(2), M e(2) }

45 Errors only depend only on known things and Ψ How to get K, M u(1), M e(1), M u(2), M e(2) in terms of current state variables? 1 K = u(z )M u(1) + (1 u(z ))M e(1) 2 Express M u(j) and M e(j) in terms of z, z, and M u (j) and M e (j) 3 Explicitly aggregate to get expressions for M u (j) and M e (j)

46 Topics non-polynomial basis functions procedure bias correction Bond economy

47 If individual policy rules are not polynomials in levels Suppose b(k u) = Ψ u,0 (s) + Ψ u,1 (s) b(k u ) if ε = u and b(k e) = Ψ e,0 (s) + Ψ e,1 (s) b(k e ) if ε = e, where b(k) is some function, e.g. ln(k)

48 If individual policy rules are not polynomials in levels Since you need to know K you need a law of motion for k. Thus, in addition to policy rule for b(k ) also obtain policy rules for k. How to get policy rule for k? Use linear approximation to b(k ), or Solve two individual policy rules one to solve for aggregate law of motion and one to describe individual behavior

49 linear approximation for linear spline Taking linear approximation when using linear spline is trivial: Get aggregate law of motion for K u simply by evaluating policy rule of unemployed at k u = K u Get aggregate law of motion for K u simply by evaluating policy rule of unemployed at k e = K u

50 Bias correction If policy rules are not polynomials = inconsistency between individual policy rules and aggregate law of motion Estimate of the mean bias can be found from model without aggregate uncertainty

51 Bias correction K ε and K ε very accurate solution for beginning and end-of-period capital stocks in model without aggregate undertainty Xpa aggregate law of motion without bias correction: Bias correction ζ ε : K ε = Ψ ε,0 (M) + Ψ ε,1 (M)K ε. ζ ε = K ε Ψ ε,0 ( M) Ψ ε,1 ( M) K ε.

52 No bias correction Figure: Simulated values of K u and K e without bias correction

53 With bias correction Figure: Simulated values of K u and K e with bias correction

54 Would the R2 pick this up? No, the R 2 > in both cases

55 Imposing equilibrium KS economy: at r t = z(k t /L t ) α 1 equilibrium is ensured for any law of motion for K t For other types of assets this is not that easy But exactly imposing equilibrium is important errors are unlikely to be exactly average on zero = errors accumulate and at some point simulation is meaningless

56 Bond economy endowment economy borrowing and lending in one and two-period riskless bonds penalty functions instead of inequality constraints

57 Bond economy - equations q 1 t c i,t = βe t 1 c i,t+1 + p(b 1 i,t+1 ) q 2 t c i,t = βe t q 1 t+1 c i,t+1 + p(b 2 i,t+1 ) c i,t + q 1 t b 1 i,t+1 + q2 t b 2 i,t+1 = y i,t + b 1 t + q 1 t b 2 t,

58 Bond economy - (too) simple approach guess law of motion for q 1 t and q2 t solve individual problem problem with simulation: equilibrium is not imposed problem with Xpa: equilibrium not imposed off grid points and during simulation

59 Bond economy - imposing equilibrium Instead of solving for b j (s i,t, m t ) solve for b j (q j t, s i,t, m t ) Solve for q j t from b j (q j t, s i,t, m t )di = 0 How do I get these b j (q j t, s i,t, m t )?

60 Bond economy - imposing equilibrium Use model equations: q 1 t c i,t = βe t 1 c i,t+1 + p(b 1 i,t+1 ) q 2 t c i,t = βe t q 1 t+1 c i,t+1 + p(b 2 i,t+1 ) c i,t + q 1 t b 1 i,t+1 + q2 t b 2 i,t+1 = y i,t + b 1 t + q 1 t b 2 t and add the following two equations that define d 1 t+1 and d2 t+1 b 1 i,t+1 + q1 t = d1 t+1 and b 2 i,t+1 + q2 t = d2 t+1

61 Bond economy - imposing equilibrium This gives the following solutions q(m t ) b j (s i,t, m t ) d j (s i,t, m t ) Take none of these literally. Except use b j t+1 = bj (q j t, s i,t, m t ) = d j (s i,t, m t ) q j t

62 Bond economy - imposing equilibrium Imposing equilibrium b j (q t, s i,t, m t )di = 0 gives q j t = d j (s i,t, m t )di

63 Bond economy - imposing equilibrium Alternative definitions for d j ( ) are possible!!!! But one does need that b j (q t, s i,t, m t ) q t < 0 that is, you have a demand equation.

64 Something practical The following slides work out a particular case The individual policy rule is of a simpler form than used in discussing the general case above Given this simpler form the slides go through the steps of XPA

65 Something practical Suppose that z t = z + ρ z z t 1 + e z,t with E [e z,t ] = σ 2 z ε i,t = ε(1 ρ ε ) + ρ ε ε t 1 + e ε,t with E [e ε,t ] = σ 2 ε Note that E [ε i,t ] = ε

66 Something practical Individual policy rule is higher-order k = Ψ0 + Ψ kk + Ψ ε ε + Ψ kε kε + Ψ u,z z + Ψ u,k K This can be written as k = Ψ 0 + Ψ k k + Ψ ε ε + Ψ kε ρkε 1 + Ψ kε e ε + Ψ u,z z + Ψ u,k K

67 Something practical Aggregation of k = Ψ 0 + Ψ k k + Ψ ε ε + Ψ kε ρkε 1 + Ψ kε e ε + Ψ z z + Ψ K K gives K = (Ψ 0 + Ψ ε ε) + (Ψ k + Ψ u,k ) K + Ψ kε ρm kε + Ψ u,z z where M kε = kε 1 df(k, ε 1 )

68 Something practical Aggregate state variables: z, K, M kε 1 Why not K u and K e separately? Why don t we have to use transition laws? How we get law of motion for M kε? Answer: define y = k ε Ψ 0 + Ψ k k + Ψ ε ε + Ψ kε ρkε 1 + Ψ kε e ε + Ψ z z + Ψ K K

69 Something practical Aggregation of y = Ψ 0 + Ψ k k + Ψ ε ε + Ψ kε ρkε 1 + Ψ kε e ε + Ψ z z + Ψ K K gives ) M kε ( Ψ = 0 + Ψ ε ε + ( Ψ k + Ψ u,k ) K + Ψ kε ρm kε + Ψ u,z z

70 References Algan, Y., O. Allais, W.J. Den Haan, P. Rendahl, 2010, Solving and simulating models with heterogeneous agents and aggregate uncertainty survey article available online, also describes link between Xpa and the perturbation method of Preston & Roca Den Haan, W. J., and P. Rendahl, 2010, Solving the incomplete markets model with aggregate uncertainty using explicit aggregation, Journal of Economic Dynamics and Control article that develops Xpa