Decentralised economies I

Transcription

1 Decentralised economies I Martin Ellison 1 Motivation In the first two lectures on dynamic programming and the stochastic growth model, we solved the maximisation problem of the representative agent. More generally, these were examples of a social planner s problem, in which welfare is maximised. At first, it appears that this is a very restricted class of problems, since normally we tend to have a mixture of firms maximising profits, consumers maximising utility and workers making optimal labour supply decisions. Fortunately, the social planner s problem is more general than at first appears. The second fundamental welfare theorem states that any equilibrium which is Pareto optimal ( as the social planner s solution is by definition) can be achieved as a decentralised equilibrium of a competitive economy, provided there is a suitable reallocation of initial endowments. Our solutions therefore do correspond to equilibria in decentralised economies. However, often we are interested in equilibria which are not Pareto optimal, such as those that prevail under sticky prices. In such cases, it is not possible to solve for the decentralised equilibrium by solving the social planner s problem. Instead, we need to work directly with the first order conditions of optimising agents. In this lecture, we will take a log-linear approximation of the first order conditions, and proceed to solve the model using the 1

2 eigenvalue-eigenvector decomposition first proposed by Olivier Blanchard and Charles Kahn. In the nest lecture, we will obtain an exact solution by using the method of parameterised expectations developed by Albert Marcet and Wouter den Haan. 2 Key reading The best explanation of log-linearisation and eigenvalue-eigenvector decomposition in a macroeconomic context is the unpublished manuscript Production, Growth and Business Cycles by King, Plosser and Rebelo, This is the seminal paper in the area. Our example will be a simplified version of their basic neoclassical model. 3 Other reading The original paper on applying eigenvalue-eigenvector decompositions to linear rational expectations models is The solution of linear difference models under rational expectations by Blanchard and Kahn, Econometrica, Despite being in Econometrica, it is very accessible (and very short). Other papers based on different eigenvalue-eigenvector decompositions are Solving linear rational expectations models by Chris Sims, 2000 and Solution and estimation of RE models with optimal policy by Paul Söderlind, European Economic Review, The latter provides Gauss codes at 4 Approximation By far the most common approach to solving decentralised economies is to take log-linear approximations around the steady state and then solve 2

3 the resulting linear expressions to arrive at AR processes for the various endogenous variables (see King, Plosser and Rebelo (1999)). This approach therefore has four main steps: 1. Calculate the steady state. 2. Derive analytical expressions for the approximation around the steady state. 3. Feed in the model parameter values. 4. Solve for the decision rules linking endogenous variables with predetermined and exogenous variables. The main reason why this approach is so common is it relative cheapness - the approximation leads to linear expressions for which there is a plentiful supply of cheap solution techniques available. The main cost comes in deriving analytical expressions for the approximation, whereas the actual computing time is reasonably trivial, which is a major gain compared to all other solution techniques. Naturally, this computation cheapness comes at a cost. Firstly, the model takes an approximation around the steady state. If the underlying model is fairly log-linear then this approximation will be a good one. However, the more nonlog-linear the model the worse the approximation and the more misleading the resulting simulations will be. For many ofthesimplemodelsthatacademicsexamine(suchasthestochasticgrowth model with only one source of uncertainty) this is unlikely to be a problem. However, as the size of the model increases and as risk aversion and volatility become more important these log-linear approximations become increasingly unreliable. Secondly, this approach only works if it is possible to solve for the steady state. For some models, a unique steady state may not exist. In spite of these drawbacks, it would be fair to say that this approach is most prevalent in the literature. 3

4 5 The stochastic growth model revisited To illustrate the technique of log-linearisation and eigenvalue-eigenvector decomposition, we return to the simple stochastic growth model studied in the second lecture. The equilibrium in this economy is Pareto optimal so, by the second fundamental welfare theorem, the social planner s solution and the decentralised equilibrium coincide. On the one hand, this means that for this model it is not necessary to use the approximation method. On the other hand, it means we can stick with a familiar model and at the end compare the exact solution from lecture 2 to the approximation solution derived in this lecture. In the standard stochastic growth model, the representative agent solves the following maximisation problem. µ max P 1 { } 1 =0 + = 1 +(1 ) 1 ln +1 = ln + The problem is identical to that studied in lecture 2, except now the log of the stochastic term follows an AR(1) process with persistence parameter. is the discount factor, is the depreciation rate and is the coefficient of relative risk aversion. We want to solve this model, by which we mean we wish to calculate sequences for consumption, output and capital which represent the equilibrium of the economy as it unfolds over time. The first order condition for this model is = [ +1 ( )] 4

5 6 Steady-state calculation In steady state, consumption, output and capital are all constant. The logarithm of the technology term is zero so itself is unity. In terms of steady-state values and, the budget constraint and first order conditions are + = +(1 ) 1 = [( 1 +1 )] Solving for and, and adding from the production function: = µ 1 (1 ) = = Log-linearisation The budget constraint and first order conditions are both non-linear so we proceed with a log-linear approximation. The basic idea is to rewrite the equations in terms of variables that define how much a variable is deviating from its steady-state value. To aid exposition, we introduce the hat notation. ˆ = ln ln In this case, rather than saying is 12 and is 10, we refer to ˆ as 0.2, meaning that is 20% above its steady-state value. To transpose the first order condition into hat notation, we first take logs. 5

6 ln =ln + [ ln +1 +ln( )] (1) Notice that already at this stage we have performed a trick by taking the expectations operator outside the logarithmic operator. In other words, we replace ln () with ln + ln. This is of course not strictly correct but is a necessary part of the approximation process. Thelefthandsideandfirst two terms of the right hand side of the first order condition (1) will be easy to deal with. More problematic is the third term of the right hand side, which is a complex function of two variables, and. To deal with this, we take a first order Taylor approximation of ln ( ) arround ln ( ). ln ( ) ln ( )+ ( ) ( ) ( )+ ( ) ( ) ( ) Applying this to the third term in the right hand side of (1), we obtain ln( ) ln( ) ( +1 ) 2 ( 1) ( ) The expression can be simplified by recognising that in steady state, 1 +1 = 1, 1 = 1 (1 ) and =1. ln( ) ln +(1 (1 )) ˆ +1 +( 1)(1 (1 ))ˆ Notice again that this only holds to a first order approximation. There will inevitably be some loss of accuracy compared to the exact solution. Returning to condition (1), we write 6

7 ln = [ ln +1 +(1 (1 )) ˆ +1 +( 1)(1 (1 ))ˆ ] Adding ln to each side and writing ln ln =ˆ gives the final form. ˆ = [ ˆ +1 +(1 (1 )) ˆ +1 +( 1)(1 (1 ))ˆ ] A similar process can be used to log-linearise the budget constraint and the law of motion for technology. ˆ + ˆ = ( +(1 ) )ˆ 1 + ˆ ˆ +1 = ˆ + 8 State space form It is convenient to express the three equations of the model (first order condition, budget constraint and law of motion for technology) in matrix form. 1 (1 ) ( 1)(1 (1 )) ˆ = +(1 ) ˆ ˆ ˆ +1 ˆ ˆ +1 More succinctly, +1 =. Wewillusestatespaceformsintherest of the lecture. What is required is to find the solution of +1 =. Assuming is invertible, we can premultiply each side of the equation by 1 to obtain 7

8 +1 = where = 1. Our technique does require that the matrix is invertible. However, other equally simple techniques such as Sims (2000) and Söderlind (1999) exist for models where cannot be inverted. 9 Eigenvalue-eigenvector decomposition The technique suggested by Blanchard and Kahn solves the system +1 = by decomposing the matrix into its eigenvalues and eigenvectors. Other techniques exist which do the job equally well, most notably the method of undetermined coefficients, which is the basis of Harald Uhlig s toolkit for analysing nonlinear economic dynamic models easily (see The Blanchard-Kahn algorithm begins by partitioning the variables in into predetermined and exogenous variables and controls. In our model, ( ˆ ˆ 1 ) 0 since technology is exogenous and the capital stock is predetermined. The control variable is consumption so ˆ. With the variables partitioned, we have = (2) The heart of the Blanchard-Kahn approach is the Jordan decomposition of the matrix. Under quite general conditions, is diagonalisable and we can write = 1 Λ In this Jordan canonical form, Λ is a diagonal matrix with the eigenvalues of along its leading diagonal and zeros in the off-diagonal elements. is a matrix of the corresponding eigenvectors. In order to continue, we need the 8

9 number of unstable eigenvalues (i.e. of modulus less than one) to be exactly equal to the number of controls. This is known as the Blanchard-Kahn condition. In a two dimensional model (such as the Ramsey growth model) with one predetermined variable and one control, it is equivalent to requiring one stable root and one unstable root to guarantee saddle path stability in the phase diagram. If there are two many unstable roots than the system is explosive and we run into problems with the transversality conditions. If there are too few unstable roots then the system is super stable, which means there will be indeterminacy. Techniques do exist for handling models with indeterminacy, see The Macroeconomics of Self-Fulfilling Prophecies by Roger Farmer, MIT Press, 1993, but we restrict our attention here to models that satisfy the Blanchard-Kahn conditions. We progress by partitioning the matrix of eigenvalues Λ. Λ 1 contains the stable eigenvalues (of number equal to the number of predetermined and exogenous variables) and Λ 2 the unstable eigenvalues (of number equal to the number of controls). The matrix is similarly partitioned. Λ Λ = = 0 Λ Using this partition and premultiplying each side by, equation (2) becomes Λ Λ = This is a cumbersome expression to work with so we prefer to solve a transformed problem, with =

10 so that the equation to solve becomes Λ1 0 0 Λ = We will solve this equation for and +1 ( is either exogenous or predetermined) and then work backwards to recover and +1. In the two dimensional case, the transformation rotates the phase diagram so that the stable eigenvector lies on the x-axis and the unstable eigenvector lies on the y-axis. The beauty of working with the transformed problem is that the two equations are now decoupled. In other words, we can write each time +1 variable as solely a function of predetermined variables, exogenous variables and controls at time. Λ 1 +1 = Λ 2 +1 = The second equation shows the evolution of the controls. Solving forward to time + gives + =(Λ 1 2 ) Since Λ 2 contains the unstable eigenvalues (of modulus less than one), this is an explosive process. In this case, the only solution which satisfies the transversality conditions is, inwhichcase + =0. The condition =0 translates back into the original programme as =0 0= We can therefore write the decision rule for the controls as =

11 The reaction function defines the controls as a linear function of the predetermined and exogenous variables. The linearity of the decision rule is a general feature of solution by log-linearisation. It has a direct analogy to the linear decision rules we derived for optimal linear-quadratic control in the last lecture. To derive the evolution of the predetermined variables, we return to the first equation of the transformed problem. + =(Λ 1 1 ) In this case Λ 1 contains the stable eigenvalues (of modulus greater than one) and the system is stable. It already shows the expected evolution of the vector. To return to the original problem, we recognise that Hence, = = ( ) +1 =( ) 1 Λ ) and the evolution of the predetermined and exogenous variables is also linear. 10 A numerical example To demonstrate the technique of eigenvalue-eigenvector decomposition in practice, we present a Matlab code to solve the stochastic growth model. To maintain comparability with lecture 2, we use the calibration =09, =04, =1and =03. The persistence parameter in the law of 11

12 motion for technology is calibrated at 0.95, implying very high persistence. We begin by clearing the screen and defining the calibrated parameters. CLEAR; beta=0.9; alpha=0.75; sigma=1; delta=0.3; rho=0.95; Next, solve for steady state using = µ 1 (1 ) = = 1 1 kbar=((1-(1-delta)*beta)/(alpha*beta))^(1/(alpha-1)); cbar=kbar^alpha-delta*kbar; ybar=kbar^alpha; The numerical values in our calibrated model are =1108 =275 and =607. To write the model in state-space form, we define the matrices and in +1 =. A(1,1)=1-(1-delta)*beta; A(1,2)=(1-(1-delta)*beta)*(alpha-1); A(1,3)=-sigma; A(2,2)=kbar; A(3,1)=1; B(1,3)=-sigma; 12

13 B(2,1)=ybar; B(2,2)=alpha*ybar+(1-delta)*kbar; B(2,3)=-cbar; B(3,1)=rho; The numeric state-space form is {z 0 } ˆ +1 ˆ ˆ = {z 0 } ˆ ˆ 1 ˆ Inverting and defining = 1,wehave C=inv(B)*A; {z 1 } ˆ +1 ˆ ˆ +1 = ˆ ˆ 1 With the model in state-space form, we can perform the Jordan decomposition of into eigenvalues and eigenvectors. The eigenvalues are stored in the matrix, with corresponding normalised eigenvectors in the matrix. [ve,mu] = eig(c); P=inv(ve); The matrix of eigenvalues has the following numerical values. ˆ 13

14 Λ = In this case, we have one unstable eigenvalue (the 0.81) and two stable eigenvalues (the 1.11 and 1.05) so the Blanchard-Kahn condition is satisfied and we have saddle path stability. In general, before partitioning the and matrices, we would need to sort the eigenvalues so that the two stable eigenvalues are in the first two rows (corresponding to exogenous and predetermined variables) and the unstable eigenvalue is in the last row (corresponding to the control). Although in our case the eigenvalues are already sorted and we have saddle path stability, we present a more general algorithm which includes a sorting procedure and an eigenvalue stability test. IF (MU(1,1)1&MU(2,2)1) (MU(1,1)1 &MU(3,3)1) (MU(2,2)1 &MU(3,3)1) (MU(1,1)1 &MU(2,2)1 & MU(3,3)1) display( No saddle path stability. ); ELSE val= diag(mu); t=flipud(sortrows([val ve ])); MU=diag(t1(:,1)); ve=t1(:,2:4); P=inv(ve ); Partitioning and, MU1=MU(1:2,1:2); MU2=MU(3,3); P11=P(1:2,1:2); 14

15 P12=P(1:2,3); P21=P(3,1:2); P22=P(3,3); The model is now in standard Blanchard-Kahn form, i.e. Λ Λ = Λ 1 = Λ 2 = = = 0 h i 21 = = 083 The decision rule is obtained from the formula = and the expected evolution of the predetermined variables from +1 =( ) 1 Λ 1 1 ( ). -inv(p22)*p21 inv(p11-p12*inv(p22)*p21)*inv(mu1)*(p11-p12*inv(p22)*p21) END The full solution is therefore 15

16 ˆ = 057 ˆ +085ˆ 1 ( ˆ +1 ) = 095 ˆ (ˆ ) = 041 ˆ +090ˆ 1 The decision rule for consumption shows that consumption is increasing in both the predetermined capital stock and technology, a parallel result to that obtained in the stochastic growth model in lecture 2. Consumption increases less than one-for-one with technology and the capital stock because of consumption smoothing. 11 Comparison with value function iteration approach. In lecture 2, we solved a deterministic version of the stochastic growth model through value function iterations. Optimal policy defined the evolution of capital as a function of lagged capital 1. If we ignore the technology term in the model in this lecture by setting ˆ + =0 then the models are exactly equivalent. It is therefore interesting to compare the result obtained under each method. 1 by value function iteration by log-linear approximation 09 = = = = =

17 The results show that the log-linear approximation is very close to the policy with the value function iterations, especially if the capital stock is close to its steady-state value. As capital deviates further from the steady-state value, the quality of the approximation deteriorates, although hardly significantly. This is not surprising, given the assumption of logarithmic utility and a log-linear production function. The only source of approximation error in the model is the nonlinear budget constraint. In this case, an approximate method is almost as good as the exact method. More complex models with less benign nonlinearities may not be so well served by approximate methods. 17