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Computational EconomicsMunich Graduate Program

Burkhard Heer

University of Bolzano-Bozen, Italy

4. September 2006

Burkhard Heer (University of Bolzano-Bozen, Italy) Computational Economics 4. September 2006 1 / 393

Objectives of the Course

1 Formulate your own business cycle model2 Compute your own business cycle model with Matlab


References

References in order of importance for the course:

1 Heer, B., and A. Maussner, 2005, Dynamic General Equilibrium Modelling:Computational Methods and Applications, Springer.

2 Miranda, M.J., and P.L. Fackler, 2002, Applied Computational Economics andFinance, MIT Press.

3 Judd, K.L., 1998, Numerical Methods in Economics, MIT Press.4 Burden, R.L., and J.D. Faires, 1993, Numerical Analysis, 5th edition, PWS

Publishing Company.5 Marimon, R., and A. Scott, eds., 1999, Computational Methods for the Study of

Dynamic Economics, Oxford University Press.6 Burnside, C., 1999, Real Business Cycle Models: Linearization and GMM

estimation, 7th revision, notes, World Bank.7 Cooley, T.F., ed., 1995, Frontiers of Business Cycle Research, Princeton University

Press.8 Ljungqvist, L., and T.J. Sargent, 2005, Recursive Macroeconomic Theory, 2nd

edition, MIT Press.


Contents I1 The Deterministic Finite-Horizon Ramsey Model

The Ramsey ProblemThe Kuhn-Tucker ProblemNumerical solution

2 Numerical Methods: Non-Linear EquationsThe problemBisectionNewton’s Method

3 The Infinite-Horizon Ramsey ModelThe modelDynamic ProgrammingTransition DynamicsDifference Equations

4 Linear AlgebraReview: Basics from Linear AlgebraLinear Equations

5 Application: Reverse Shooting6 The Stochastic Ramsey Model

The modelStochastic Dynamic Programming


Contents II7 Real Business Cycle Model

Growth, Leisure and Business CyclesThe benchmark real business cycle modelCalibration of the benchmark modelComputation of the benchmark model 1: Value function iterationEvaluation of the Model

8 Loglinear ApproximationIllustrative ExampleThe General MethodApplication 1: The real business cycle modelApplication 2: A monetary business cycle model

9 More Numerical MethodsFunction ApproximationNumerical Integration

10 Projection MethodsA simple exampleThe general frameworkApplication: Deterministic Ramsey model

11 OutlookPerturbation methods


Contents IIIParameterized expectationsThe curse of dimensionalityHeterogeneous agents


The Deterministic Finite-Horizon Ramsey Model The Ramsey Problem










Ramsey Problem: Ramsey (1928)

The Deterministic Finite-Horizon Ramsey Model

representative agent

lives T + 1 periods

maximizes utility U(C0,C1, . . . ,CT )

produces output with capital Kt in period t :

Yt = F (Kt) (1)

Production Yt in period t can be used for either consumption Ct or next periodcapital stock Kt :

Yt = F (Kt) ≥ Kt+1 + Ct (2)

Initial capital stock K0 is given.

no uncertainty

no labor supply decision



Properties of the Production Function

No production without capital:

F (0) = 0

F is twice continuously differentiable

F is strictly increasing, F ′ > 0

F is strictly concave, F ′′ < 0



The Kuhn-Tucker ProblemThe Ramsey Problem

max(C0,...,CT )

U(C0, . . . ,CT )

s.t.

Kt+1 + Ct ≤ f (Kt),0 ≤ Ct ,0 ≤ Kt+1,

9=; t = 0, . . . ,T ,

K0 given.

(3)

optimization subject to constraints2 kinds of constraints:

1 equalities2 inequalities

⇒The Kuhn-Tucker theorem provides a set of necessary and sufficient conditions for asolution to exist.


The Deterministic Finite-Horizon Ramsey Model The Kuhn-Tucker Problem










The Kuhn-Tucker ProblemKuhn-Tucker Theorem

Theorem (Kuhn-Tucker)

Let f be a concave C1 function mapping U into R, where U ⊂ Rn is open and convex.For i = 1, . . . , l , let hi : U → R be concave C1 functions. Suppose there is some x ∈ Usuch that

hi(x) > 0, i = 1, . . . , l.

Then x∗ maximizes f over D = {x¯∈ U|hi(x) ≥ 0, i = 1, . . . l} if and only if there is

λ∗ ∈ Rl such that the Kuhn-Tucker first order conditions hold:

∂f (x∗)∂xj

+lX

i=1

λ∗i∂hi(x∗)∂xj

= 0, j = 1, . . . n,

λ∗i ≥ 0 i = 1, . . . , l,

λ∗i hi(x∗) = 0 i = 1, . . . , l.



The Kuhn-Tucker ProblemFirst-Order Conditions of the Ramsey Problem

The Ramsey problem fits the conditions of the Kuhn-Tucker theorem:

0 =∂U(C0, . . . ,CT )

∂Ct− λt + µt , t = 0, . . . ,T , (4a)

0 = −λt + λt+1f ′(Kt+1) + ωt+1, t = 0, . . .T − 1, (4b)

0 = −λT + ωT+1, (4c)

0 = λt (f (Kt)− Ct − Kt+1) , t = 0, . . .T , (4d)

0 = µtCt , t = 0, . . .T , (4e)

0 = ωt+1Kt+1, t = 0, . . .T , (4f)

λt — Lagrange multiplier associated with the resource constraint:

f (Kt)− Ct − Kt+1 ≥ 0

µt — multiplier for Ct ≥ 0ωt+1 — multiplier for Kt+1 ≥ 0.




Assumption: Marginal utility U(C0,C1, . . . ,CT ) goes to ∞ if consumption Ct → 0:

⇒ µt = 0 for all t = 0, . . . ,T (why?), the resource constraint always binds⇒

∂U(C0, . . . ,CT )

∂Ct= λt

⇒ Ct > 0 and f (0) = 0 imply Kt > 0 for t = 0, . . . ,T and KT+1 = 0.⇒ ωt = 0 for t = 1, . . . ,T .




Optimal solution:

Kt+1 = f (Kt)− Ct , (5a)

∂U(C0, . . .CT )/∂Ct

∂U(C0, . . .CT )/∂Ct+1= f ′(Kt+1). (5b)

Interpretation:

Resource constraint

Marginal rate of substitution between consumption in two adjacent periods =return on savings


The Deterministic Finite-Horizon Ramsey Model Numerical solution










Numerical Solution of the Finite-Horizon Ramsey Problem

Numerical solution:Calibration: matching empirical observations

1 choice of the functional types2 choice of the parameter values

Computation: solving the numerical problem




Example

Let U be given by a constant elasticity of substitution function

U(C0, . . . ,CT ) :=

(TX

t=0

C%t

)1/%

, % ∈ (−∞, 1],

and define f (Kt) := K αt , α ∈ (0, 1). Given this specification, equations (5) become

Kt+1 = K αt − Ct , t = 0, 1, . . . ,T ,»

Ct

Ct+1

–1−%

αK α−1t+1 = 1, t = 0, 1, . . . ,T − 1.




If we eliminate consumption in the second set of equations using the first T + 1equations we arrive at a set of T non-linear equations in the T unknowns(K1,K2, . . . ,KT ):

0 =

„K α

1 − K2

K α0 − K1

«1−%

− αK α−11 ,

0 =

„K α

2 − K3

K α1 − K2

«1−%

− αK α−12 ,

...

0 =

„K α

T

K αT−1 − KT

«1−%

− αK α−1T .

(6)



Reading Assignments and Problems

Reading Assignment

Heer, Maussner, Chapter 1.1.

Problems1 Review the concept of the CES utility function. How can we interpret ρ?2 Formulate the finite-horizon Ramsey with endogenous labor supply. Let utility be a

function of both consumption Ct and labor Nt ≥ 0. Utility in period t is given by:

C1−ηt (1− Nt)

θ(1−η)

1− η

Production is a function of both capital and labor, F (K ,N) = K αN1−α. Capitaldepreciates at rate δ. Compute the first-order conditions.


Numerical Methods: Non-Linear Equations The problem










Nonlinear equations

Problem:How do we find a numerical solution of a non-linear equation?

f (x) = 0

1 Bisection Method2 Newton’s Method


Numerical Methods: Non-Linear Equations Bisection










Bisection MethodThe idea

f (.) is continuous

f (a) · f (b) < 0

⇒ at least one root x in [a, b]

Idea of bisection: Half the interval that brackets the zero

Application: Find the root of f (x) = x3 − 2 on [1, 2]



Bisection MethodThe Matlab program

The matlab program bisect1.m:

a=1;b=2;tol=1e-10;

s=sign(f(a));x=(a+b)/2;d=(b-a)/2;

while d>tol;d=d/2;if s==sign(f(x))

x=x+d;else

x=x-d;end

end



Bisection MethodThe Matlab program, cont.

The Matlab program f.m:

function f1 = f(x)f1 = x^3-2;

Problem with Bisection:1 Find two values that bracket the root2 Slowness


Numerical Methods: Non-Linear Equations Newton’s Method










Netwon-Rhapson MethodThe idea

Idea: Successive linearization

Linear approximation of of f : [a, b] → R at x0:

g0(x) := f (x0) + f ′(x0)(x − x0)

Root of g0(x):

0 = f (x0) + f ′(x0)(x1 − x0) ⇒ x1 = x0 −f (x0)

f ′(x0).

Iteration on the sequence:

xk+1 = xk −f (xk )

f ′(xk ).



Netwon-Rhapson Method

Rates of Convergence

Suppose limk→∞ xk = x∗ with f (x∗) = 0.We say that xk converges at rate q to x∗ if

limk→∞

‖xk+1 − x∗‖‖xk − x∗‖q <∞. (7)

q = 2: xk converges quadratically

limk→∞

‖xk+1 − x∗‖‖xk − x∗‖ ≤ β ≤ 1,

→ xk converges linearly at rate β.



Netwon-Rhapson MethodRate of convergence

Theorem

Suppose that f (x) is C2 and that f (x∗) = 0. If x0 is sufficiently close to x∗, f ′(x∗) 6= 0,and |f ′′(x∗)/f ′(x∗)| <∞, the Newton sequence xk converges to x∗, and it isquadratically convergent:

limk→∞

|xk+1 − x∗||xk − x∗|2 <∞.



Netwon-Rhapson MethodRate of convergence

Proof:

Taylor’s theorem:

f (xk ) = f (x∗) + f ′(x∗)(xk − x∗) +12

f ′′(η(x))(xk − x∗)2

for some function η(x) ∈ [xk , x∗] ∪ [x∗, xk ].⇒

˛xk −

f (xk )

f ′(xk )− x∗

˛= |xk+1 − x∗| = 1

2

˛f ′′(η(x))(xk − x∗)2

f ′(x∗)

˛For any β ∈ (0, 1), if |xk − x∗| is sufficiently small:

|xk+1 − x∗| < β |xk − x∗|

⇒ xk → x∗




Stopping criteria

When shall we stop the algorithm?

Two Problems:

1 "Have we solved the problem?": f (xk ) close to zero2 "Have we ground to a halt?": xk+1 and xk are close together

1 Stop if |f (xk )| < ε.I |f (x)| < 10−5 for all x ∈ [x0, x∗]: ε = 10−3 too largeI f (x) differ greatly over x : ε = 10−3 may be overly restrictive

2 Stop if

|xsi − xs+1

i |1 + |xs

i |≤ ε ∀ i = 1, 2, . . . , n, ε ∈ R++. (8)



Netwon-Rhapson MethodAlgorithm

Algorithm: Newton’s Method

Purpose: Solve f (x) = 0, where x ∈ R.

Step 1: Initialize: choose x0 ∈ [x , x ].

Step 2: Compute f (x0), f ′(x0) and iterate on the sequence:

xk+1 = xk −f (xk )

f ′(xk ).

Step 3: Check for convergence: if ‖f (xk )‖∞ < ε and/or |xk+1 − xk |/(1 + |xk |) ≤ ε for agiven tolerance ε ∈ R++ stop, else return to step 2.




Problems:

1 xk+1 may not be defined2 it may be impossible to derive f ′



Numerical Methods: Non-Linear EquationsQuasi Newton



Numerical Methods: Non-Linear Equations

Solutions to the problems:

1 Backtrack xs+1 along the direction of f ′(xs) to a point x ′s+1 at which f can beevaluated: See problem 1.

2 Use the slope of the secant in order to compute the derivative:

xs+2 = xs+1 −xs+1 − xs

f (xs+1)− f (xs)

→ Modified or Quasi-Newton method



Numerical Methods: Non-Linear EquationsSecant Method



Numerical Methods: Non-Linear EquationsNumerical Differentiation

First Difference Formulas

Taylor’s Theorem:

f (x + h) = f (x) + f (1)(x)h + f (2)(ξ)h2

2. (9)

Approximation of the first derivative: forward difference formula

DFD f (x , h) :=f (x + h)− f (x)

h. (10)

The approximation error is proportional to h:˛DFD f (x , h)− f (1)(x)

˛=

˛f (2)(ξ)/2

˛h.

Backward difference formula: −h in place of h.




Taylor’s theorem for n = 2, h, and −h:

f (x + h) = f (x) + f (1)(x)h + f (2)(x)h2

2+ f (3)(ξ1)

h3

6, (11a)

f (x − h) = f (x)− f (1)(x)h + f (2)(x)h2

2− f (3)(ξ2)

h3

6, (11b)

Subtract the second line from the first:

f (x + h)− f (x − h) = 2f (1)(x)h +“

f (3)(ξ1) + f (3)(ξ2)” h3

6

⇒ Central difference formula

DCD f (x , h) :=f (x + h)− f (x − h)

2h(12)

C denoteS the maximum of (f (3)(ξ1) + f (3)(ξ2))/6 in [x , x + h]Approximation error is proportional to Ch2 and, thus, of second order.




Central difference formula for the second derivative:

D2CD f (x , h) :=

f (x + h) + f (x − h)− 2f (x)

h2 , (13)

Approximation error is bound by Ch and, thus, of first order.

Choice of h

As small as possible? Finite precision of computer arithmetic. Only accurate forthe first 10 digits to the right of the floating point (depending on PC, program).

depending on f (3) (see Heer/Maussner)

rule-of-thumb for two-sided rules: h = max (|x |, 1)ε1/3, (ε is the machine precision)

rule-of-thumb for one-sided rules: h = max (|x |, 1)√ε



Netwon-Rhapson MethodExtension to Rn

Assume we want to solve the system of n equations

0=f 1(x1, x2, . . . , xn),

0=f 2(x1, x2, . . . , xn),...=

...,0=f n(x1, x2, . . . , xn),

9>>>=>>>; ⇐⇒ 0 = f(x) (14)

The equivalent to f ′(x) in the multi-variable case is the Jacobian matrix J(x) of partialderivatives of f = [f1, f2, . . . , fn]′ with respect to xi , i = 1, 2, . . . n. We use the notation

f ij :=

∂f i(x)

∂xj



Netwon-Rhapson MethodExtension to Rn, Definitions

Jacobian:

J(x) :=

26664f 11 f 1

2 . . . f 1n

f 21 f 2

2 . . . f 2n

......

. . ....

f n1 f n

2 . . . f nn

37775 . (15)

Open ball with center x0 and radius r :

N (x0, r) := {x ∈ Rn : ‖x− x0‖ < r}.

J is said to be Lipschitz on N (x0) of x0 with constant γ if for x1, x2 ∈ N (x0) thefollowing condition holds

‖J(x1)− J(x2)‖ ≤ γ‖x1 − x2‖.




TheoremLet f : Rn → Rn be continuously differentiable in an open convex set D ⊂ Rn. Assumethat there exists x∗ ∈ Rn and r , β > 0, such that N (x∗, r) ⊂ D, f(x∗) = 0, J(x∗)−1

exists with ‖J(x∗)−1‖ ≤ β, and J Lipschitz with constant γ on N (x∗, r). Then thereexists ε > 0 such that for all x0 ∈ N (x∗, r) the sequence x1, x2, . . . generated by

xk+1 = xk − J(xk)−1f(xk), k = 0, 1, . . . ,

is well defined, converges to x∗, and obeys

‖xk+1 − x∗‖ ≤ βγ‖xk − x∗‖2, k = 0, 1, . . . .

Proof: Dennis and Schnabel (1983)




Algorithm: Modified Newton-Raphson

Purpose: Solve 0 = f(x), where f = [f1, f2, .., fn]′ and x ∈ Rn.

Step 1: Initialize: choose x0 ∈ [x, x].

Step 2: Compute J(x0) the Jacobian of f at x0 and solve J(x0)dx = −f(x0). Ifx1 = x0 + dx /∈ [x, x] choose λ ∈ (0, 1) such that x2 = x0 + λdx ∈ [x, x] andset x1 = x2.

Step 3: Check for convergence: if ‖f(x1)‖∞ < ε and/or |x1i − x0

i |/(1 + |x0i |) ≤ ε ∀i for a

given tolerance ε ∈ R++ stop, else set x0 = x1 and return to step 2.




Two Problems:

1 Solution of the system of linear equations:

(x1 − x0)J(x0) ≡ dxJ(x0) = f(x0)

The issue will be taken up in chapter 5 of this course.2 Computation of the Jacobian J(x0). Two solutions:

1 Central differences2 Broyden’s algorithm




Central differences:

Approximate the element f ij of the Jacobian matrix at the point x by

f ij :=

∂f i(x)

∂xj' f i(x + ejh)− fi(x− ejh)

2h, (16)

where ej is the unit (row) vector with one in the j–th position and zeros elsewhere.

⇒ n2 computations.




Broyden’s Algorithm:

Initialize the Jacobian with a guess J(x0) = A0.

Given the points xk and xk−1,

f(xk)− f(xk−1) = Ak (xk − xk−1)

→ does not provide enough conditions for unique matrix Ak .→ no information on vectors orthogonal to (xk − xk−1)

We require

Ak z = J(x0)z, whenever (xk − xk−1)z.

→ details of the computation of Ak : see Miranda/Fackler, p.38→ less accurate, but fewer (n) computations.→ implemented in broyden.m



Computation of the Example 1TURNPIKE Property

0 10 20 30 40 50 600.16

0.165

0.17

0.175

0.18

0.185

0.19

0.195

0.2

0.205



Computation of the Example 1

Program problem1.m

function y = problem1(x)rho=0.5;:alpha=0.35;T=59;K0=0.10;y=zeros(T,1);x1=zeros(T+2,1);x1(1)=K0;x1(2:T+1)=x;x1(T+2)=0;for i=2:T+1

y(i-1)=(x1(i)^alpha-x1(i+1))/(x1(i-1)^alpha-x1(i));y(i-1)=y(i-1)^(1-rho)-alpha*x1(i)^(alpha-1);

endend




Script example1.m:

clear;T=59;alpha=0.35;K=ones(T,1)*alpha^(1/(1-alpha));K1=zeros(T,1);K1=broyden(’problem1’,K);t1=1:1:T;plot(t1,K1);




Assignment: Program Example 1

1 Create Directory ’/compecon/matlab/nonlinear/’.2 Write the two programs example1.m and problem1.m and store them in the new

directory.3 Change the current directory to ’/compecon/matlab/nonlinear/’.4 Download the toolbox from Fackler’s homepage and store it in the directory

’/compecon/matlab/nonlinear/’.5 Load the program broyden.m in the editor and add the path (’File ⇒ Set Path ⇒

Add Folder’).6 Type ’Example1’ in order to start the program.



Netwon-Rhapson MethodProblems

Problems with Newton’s Method

1 Starting points needs to be close to the true solution.

Possible remedies:1 Grid Search. The algorithm may have to evaluate the function at a point that results in a

breakdown (e.g. 1/0 or (−2.5)1/2)2 Educative guess3 Genetic Search4 Backstepping

→ diligent discussion in Heer/Maussner



Reading Assignments and ProblemsReading Assignment

Heer, Maussner, Chapter 8.3. and 8.5.Miranda, Fackler, Chapter 3.1.-3.6.

Problems1 Newton-Rhapson

1 Solve the equation

f (x) = x3 − 2 = 0.

Write a MATLAB program that solves the non-linear equation with the help of theNewton-Rhapson algorithm. For this reason, specify the two functions f (x) = x3 − 2.0and f ′(x) = 3x2. Use x0 = 1.0 as a starting value.

2 Solve the equation

f (x) = ln x = 0.

What happens if you use the Newton-Rhapson algorithm with x0 = 5? Write a MATLABprogram that solves the equation with the modified Newton-Rhapson algorithm. Specifyboundaries [a, b] for xk . In order to compute the step size reduce the step size by 1/2whenever xk is outside [a, b].

3 Assume that we are unable to compute the derivative f ′() analytically. Write a programthat computes the derivative f ′() numerically and compute the root of

f (x) = x3 − 2 = 0.


The Infinite-Horizon Ramsey Model The model










Infinite-Horizon Ramsey Model

The Ramsey Model

maxC0,C1,...

U0 =∞Xt=0

βtu(Ct)

s.t.

Kt+1 + Ct ≤ f (Kt),0 ≤ Ct ,0 ≤ Kt+1,

9=; t = 0, 1, . . . ,

K0 given.

(17)




maximizing of the Lagrange function with respect to C0,C1, . . . ,K1,K2, . . . :

L = βt∞Xt=0

hu(Ct) + λt (f (Kt)− Ct − Kt+1)

+ µtCt + ωt+1Kt+1

i.

First order conditions for maximizing L are given by:

u′(Ct) = λt − µt , (18a)

λt = βλt+1f ′(Kt+1) + ωt+1, (18b)

0 = λt(f (Kt)− Ct − Kt+1), (18c)

0 = µtCt , (18d)

0 = ωt+1Kt+1. (18e)




Transversality condition:

limt→∞

βtλtKt+1 = 0, (19)

limC→0 u′(C) = ∞

⇒ Euler Equation:

u′(f (Kt)− Kt+1)

u′(f (Kt+1)− Kt+2)− βf ′(Kt+1) = 0. (20)


The Infinite-Horizon Ramsey Model Dynamic Programming











Dynamic Programming

Bellman equation:

v(K ) = max0≤K ′≤f (K )

u(f (K )− K ′) + βv(K ′). (21)

v(.) — value function, time-invariant

The first order condition for this problem is

u′(f (K )− K ∗) = βv ′(K ∗). (22)

v(K ) implies the policy function:

K ∗ = g(K ).



Infinite-Horizon Ramsey ModelDynamic Programming

If u() and f () are strictly increasing, strictly concave and twice continuouslydifferentiable functions, the following results hold:

1 The function v exists, is differentiable, strictly increasing, and strictly concave.2 The policy function g is increasing and differentiable.3 The function v is the limit of the following sequence of steps s = 0, 1, . . . :

v s+1(K ) = max0≤K ′≤f (K )

u(f (K )− K ′) + βv s(K ′),

with v0 = 0.

Proof: Harris (1987)




Example

Let the one-period utility function u and the production function f be given by

u(C) := ln C

f (K ) := K α, α ∈ (0, 1),

respectively.The policy function K ∗ = g(K ) that solves the Ramsey problem (17) is given by

K ∗ = αβK α.

Furthermore, the value function is linear in ln K and given by

v(K ) = a + b ln K ,

a :=1

1− β

»ln(1− αβ) +

αβ

1− αβlnαβ

–, b :=

α

1− αβ.




Analytical Solution of the Example

Let v0 = 0, we solve

v1 = max ln(K α − K ′)

yielding K ′ = 0 and v1 = α ln K . In the next step we seek K ′ that solves

v2 = max ln(K α − K ′) + βα ln K ′.

From the first order condition

1K α − K ′ =

αβ

K ′

we get

K ′ =αβ

1 + αβK α,

v2 = α(1 + αβ) ln K + A1,

A1 := ln(1/(1 + αβ)) + αβ ln(αβ/(1 + αβ)).




The value function in step s = 3 is given by

v3 = max ln(K α − K ′) + βα(1 + αβ) ln K ′ + βA1

yielding

K ′ =αβ + (αβ)2

1 + αβ + (αβ)2 K α,

v3 = α(1 + αβ + (αβ)2) ln K + A2,

A2 = ln»

11 + αβ + (αβ)2

–+

“αβ + (αβ)2

”ln

»αβ + (αβ)2

1 + αβ + (αβ)2

–+ βA1.

Continuing in this fashion we find the policy function in step s given by

K ′ =

Ps−1i=1 (αβ)sPs−1i=0 (αβ)s

K α

with limit s →∞ equal to

K ′ = αβK α.




⇒ the value function is a linear function of ln K : v = a + b ln K .

Solution for a, b: Method of undetermined coefficients:

maxK ′

ln(K α − K ′) + β(a + b ln K ′)

⇒

K ′ =βb

1 + βbK α.

Therefore

v = α(1 + βb)| {z }b

ln K + βa + ln»

11 + βb

–+ βb ln

»βb

1 + βb

–| {z }

a

.




Equating the constant on the right hand side of this equation to a and the slopeparameter to b, we get:

b = α(1 + βb) ⇒ b =α

1− αβ,

a = βa + ln»

11 + βb

–+ βb ln

»βb

1 + βb

–,

⇒ a =1

1− β

»ln(1− αβ) +

αβ

1− αβlnαβ

–.




Dynamic Programming versus Euler Equation

Let

K ∗ = g(K ) := argmax0≤K ′≤f (K )

u(f (K )− K ′) + βv(K ′),

and consider the identity

v(K ) = u(f (K )− g(K )) + βv(g(K )).

Differentiation with respect to K yields

v ′(K ) = u′(C)`f ′(K )− g′(K )

´+ βv ′(K ∗)g′(K ) = 0.

Using the first order condition (22), we find

v ′(K ) = u′(C)f ′(K ). (23)

Finally, letting C′ = F (N,K ′)− K ′′ denote next period’s consumption, (22) and (23)may be combined to yield

1 = βu′(f (K ′)− K ′′)

u′(f (K )− K ′)f ′(K ′),

which is identical to the Euler equation (20).Burkhard Heer (University of Bolzano-Bozen, Italy) Computational Economics 4. September 2006 82 / 393



Numerical Solution of the Example

1 Value function iteration2 Backward iteration



Infinite-Horizon Ramsey Model IDynamic Programming

Algorithm: Value Function Iteration 1

Purpose: Find an Approximate Solution of the Policy Function for the Ramsey model(17)

Step 1: Choose a grid of n equally spaced points

G = [K1,K2, . . . ,Kn], Ki < Kj∀i < j = 1, 2, . . . n.

Step 2: Initialize the value function: ∀i = 1, . . . , n set

v0i =

u(f (K ∗)− K ∗)

1− β,

where K ∗ denotes the stationary solution to the continuous valued Ramseyproblem.



Infinite-Horizon Ramsey Model IIDynamic Programming

Step 3: Compute a new value function and associated policy function, v1 and g1,respectively: Put j∗0 ≡ 1. For i = 1, 2, . . . , n, and j∗i−1 find the index j∗i thatmaximizes

u(f (Ki)− Kj) + βv0j

in the set of indices j∗i−1, j∗i−1 + 1, . . . n. Set g1

i = j∗i andv1

i = u(f (Ki)− Kj∗i) + βv0

j∗i.

Step 4: Check for convergence: If ‖v0 − v1‖∞ < ε(1− β), ε ∈ R++ (or if the policyfunction has remained unchanged for a number of consecutive iterations) stop,else set v0 = v1 and g0 = g1 and return to step 3.



Value Function Iteration

Algorithm: See value1.m, value2.m

Computation of the infinite-horizon Ramsey model with depreciation

Parameter choice: quarterly periods

β = 0.99

σ = 2.0

δ = 0.02

grid: K ∗ ± 0.1 ∗ K ∗

1000 grid points over K



Value Function IterationConsumption

5 10 15 20 25 30 35 40 45 501

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

capital stock

cons

umpt

ion



Value Function IterationSavings

5 10 15 20 25 30 35 40 45 50−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

capital stock

savi

ngs



Value Function IterationValue function

5 10 15 20 25 30 35 40 45 5044

46

48

50

52

54

56

58

capital stock

v an

d tv

tvv



Value Function Iteration IAccuracy

Accuracy check

We stop the algorithm if the maximum of the absolute difference between the oldand the new value function is less than ε = 0.001.

For this accuracy, the policy function k ′(k) does not change any more.

Question: Is ε small enough?

Accuracy measure: Violation of the Euler equation

For δ = 0, we know the exact solution of k ′(k) and c(k) and can compute thepercentage error.

Given the solution k ′(k), c(k), and c′(k ′) we can compute the residual of the Eulerequation for all k ∈ K:

r = 1− βu′(c′(k ′))u′(c(k))

h1− δ + αk ′α−1

iNotice that we transformed the equation to get a measure-unit free variable.



Value Function Iteration IIAccuracy

In addition, we may compute the change in the policy function c(k), ∆, that isnecessary in order to fulfill the Euler equation:

f (∆) = 1− βu′(c′(k ′))

u′(c(k) · (1 + ∆))

h1− δ + αk ′α−1

i= 0

→ good measure for accuracy of the policy functions!





Remarks:

1 A good initial guess speeds up the computation significantly. Try v = 0 instead.2 Make use of the characteristics of the value and policy functions:

1 policy function monotone: for Ki > Ki−1, K ′(Ki ) ≥ K ′(Ki−1)2 value function strictly concave: stop iteration over K ′ if for K ′ = Ki the right-hand side

of the Bellman equation is smaller than for K ′ = Ki−1




Binary Search

Purpose: Find the maximum of a strictly concave function f (x) defined over a grid of npoints G = [x1, ..., xn]

Step 1: Initialize: Put imin = 1 and imax = n.

Step 2: Select two points: il = floor(imin + imax)/2 and iu = il + 1, where floor(i)denotes the largest integer smaller or equal to i ∈ R.

Step 3: If f (xiu ) > f (xil ) set imin = il . Otherwise put imax = iu .

Step 4: If imax − imin = 2, stop and choose the largest element among f (ximin ), f (ximin+1),and f (ximax ). Otherwise return to Step 2.





Purpose: Find an Approximate Solution of the Policy Function for the Ramsey model(17)

Step 1: Choose a grid of n equally spaced points

G = [K1,K2, . . . ,Kn], Ki < Kj∀i < j = 1, 2, . . . n.

Step 2: Initialize the value function: ∀i = 1, . . . , n set

v0i =

u(f (K ∗)− K ∗)

1− β,

where K ∗ denotes the stationary solution to the continuous valued Ramseyproblem.



Infinite-Horizon Ramsey Model IIDynamic Programming

Step 3: Compute a new value function and associated policy function, v1 and g1,respectively: Put j∗0 ≡ 1. For i = 1, 2, . . . , n, and j∗i−1 use Algorithm BinarySearch to find the index j∗i that maximizes

u(f (Ki)− Kj) + βv0j

in the set of indices j∗i−1, j∗i−1 + 1, . . . n. Set g1

i = j∗i andv1

i = u(f (Ki)− Kj∗i) + βv0

j∗i.

Step 4: Check for convergence: If ‖v0 − v1‖∞ < ε(1− β), ε ∈ R++ (or if the policyfunction has remained unchanged for a number of consecutive iterations) stop,else set v0 = v1 and g0 = g1 and return to step 3.


The Infinite-Horizon Ramsey Model Transition Dynamics










Infinite-Horizon Ramsey ModelSaddle Path

Dynamics of the Ramsey Model

Kt+1 = f (Kt)− Ct , (24a)

1 = βu′(Ct+1)

u′(Ct)f ′(Kt+1). (24b)

and the transversality condition.

Steady State:

1β

= f ′(K ∗).


The Infinite-Horizon Ramsey Model Difference Equations










Difference Equations

Linear Difference Equations

Most economic data are compiled at a yearly, quarterly, or monthly frequency:

discrete variables x only at equally spaced points in time: x(t), x(t + 1), x(t + 2), . . . .

The first difference of xt , ∆xt , is defined as

∆xt := xt − xt−1,

and further differences are computed according to

∆2xt := ∆xt −∆xt−1 = xt − 2xt−1 + xt−2,

∆3xt := ∆2xt −∆2xt−1 = xt − 3xt−1 + 3xt−2 − xt−3,

...,

∆nxt := ∆n−1xt −∆n−1xt−1.

A difference equation of order n relates the function x to its n differences.



Difference EquationsLinear Difference Equations

First-Order Linear Difference Equation with Constant Coefficient:

∆xt = xt − xt−1 = axt−1, a ∈ R. (25)

Assume we know the time t = 0 value x0. We can then determine all future (or past)values of xt by iterating forward (or backward) on (25):

x1 = λx0, λ := 1 + a,

x2 = λx1 = λ2x0,

x3 = λx2 = λ3x0,

...,

xt = λxt−1 = λtx0.

Asymptotic stability if |λ| < 1:

limt→∞

xt = 0




Generalization to n variables x := [x1, x2, . . . , xn]′ ∈ Rn:

xt = Axt−1. (26)

Suppose the matrix possesses n distinct real eigenvalues and associated eigenvectorsv1, v2, . . . , vn.

Let

P = [v1, v2, . . . , vn]

then

Λ =

26664λ1 0 . . . 00 λ2 . . . 0...

.... . .

...0 0 . . . λn

37775 = P−1AP

where Λ is the diagonal matrix with the eigenvalues λ1, λ2, . . . , λn on its main diagonal.




Transformation of the original system xt = Axt−1 to new coordinates y = P−1x:

yt = P−1xt = P−1Axt−1 = P−1APyt−1 = Λyt−1,

with solution

yt = Λty0.

Since

Λt =

26664λt

1 0 . . . 00 λt

2 . . . 0...

.... . .

...0 0 . . . λt

n

37775 ,a necessary and sufficient condition for asymptotic stability of yt and xt = Pyt is that alleigenvalues λi have absolute value of less than unity.




Now assume (without loss of generality) that the first n1 eigenvalues are less than onein absolute value while the remaining n2 = n − n1 eigenvalues exceed one in absolutevalue.

If we are free to choose the initial point y0, we can pick

y0 := [y1, y2, . . . , yn1 , 0, 0, . . . , 0| {z }n2 elements

]′,

for arbitrary yi ∈ R, i = 1, 2, . . . , n1.

⇒ y0 converges to zero.

→ this is also true for all x0 = Py0 of the original system.

The set of all such starting points constitutes the stable eigenspace of the differenceequation (26). In the two-dimensional case, we refer to this space sometimes as thesaddle path.




Complex eigenvalues

λ := α+ βi

In this case one of the columns of P is a complex vector and the respective newcoordinate, say y = y1 + iy2, is a complex number, too.

For a given initial coordinate y0 the solution is given by

(y1t + iy2t) = (α+ iβ)t(y10 + iy20). (27)

which may equivalently be stated as:»y1t

y2t

–= At

»y10

y20

–= r t

»cos(tθ) − sin(tθ)sin(tθ) cos(tθ)

– »y10

y20

–. (28)

with r =pα2 + β2.




Extension to the case of mixed real and complex eigenvalues of multiplicity k ≥ 2:

TheoremLet A be a real square matrix. Then

limt→∞

At = 0n×n

if and only if every eigenvalue of A is less than unity in modulus.

Proof: See Murata (1977), p. 85.



Difference EquationsNon–Linear Difference Equations

Non–Linear Difference Equations

Let f i : Rn → R, i = 1, 2, . . . , n denote arbitrary differentiable functions.

A system of n non–linear first–order difference equations is defined by the following setof equations:26664

x1t

x2t...

xnt

37775 =

26664f 1(x1 t−1, x2 t−1, . . . , xn t−1)

f 2(x1 t−1, x2 t−1, . . . , xn t−1)...

f n(x1 t−1, x2 t−1, . . . , xn t−1)

37775 , (29)

or

xt = f(xt−1).



Difference EquationsNon–Linear Difference Equations

Assume there is a rest point x that satisfies

x = f(x).

The behavior of the dynamical system (29) near its rest point can be inferred from theproperties of the linear system

xt = J(x)xt−1.

J denotes the Jacobian matrix of f at x:

J(x) :=

0BBB@f 11 (x) f 1

2 (x) . . . f 1n (x)

f 21 (x) f 2

2 (x) . . . f 2n (x)

......

. . ....

f n1 (x) f n

2 (x) . . . f nn (x)

1CCCA , (30)

where

f ij (x) :=

∂f i(x1, x2, . . . , xn)

∂xj.



Reading Assignments and Problems I

Reading Assignment

Heer, Maussner, Chapter 1.2. and Chapter 9.1

Problems1 Compute the accuracy of the program value1.m with the help of the percentage

error in the Euler equation. For δ = 0, compare the solution with the analyticalsolution and report the percentage deviation. How does the choice of the periodlength (quarter, year) influence the accuracy of the policy function?

2 Use the Matlab functions tic and toc to compute the time needed to compute theinfinite-horizon Ramsey model with inelastic labor with the help of value functionwhere the value function is initialized i) by zeros, ii) by the steady state value, andiii) by a flat consumption profile.

3 Introduce binary search in the program value2.m in order to speed up thecomputation. Compare with value1.m.



Reading Assignments and Problems II

4 Compute the infinite-horizon Ramsey model with elastic labor supply using valuefunction iteration. Use the following utility function:

u(c, n) =1

1− η

hc1−η(1− n)θ(1−η) − 1

iUse the production function y = kαn1−α. Consider quarters and use theparameter values α = 0.4, β = 0.99, η = 1.5.

1 Calibrate θ so that the steady-state supply of labor n is equal to 0.3.2 Compute the policy functions k ′(k), n(k), and c(k) with the help of value function

iteration. Plot the policy functions.3 Report the maximum absolute Euler equation residual and plot ∆ as a function of k .


Linear Algebra Review: Basics from Linear Algebra










Review: Linear Algebra

VectorsA real (complex) vector of dimension n is a n-tuple of numbers xi ∈ R (xi ∈ C)i = 1, 2, . . . n, denoted by

x =

26664x1

x2...

xn

37775 .The space of all n-tuples is Rn (Cn).

Vector addition and scalar multiplication are defined by

y = a + bx¯

=

26664a + bx1

a + bx2...

a + bxn

37775 .




Norms

Norm: measure of distance

Def.: Norm on Rn is a real valued function ‖x‖ that obeys:

‖x‖ ≥ 0 for every x ∈ Rn, and ‖x‖ = 0 if and only if x = 0 ∈ Rn,

‖ax‖ = |a| · ‖x‖ for every x ∈ Rn and a ∈ R,‖x + y‖ ≤ ‖x‖+ ‖y‖ for every x, y ∈ Rn.

(31)

the `∞ or sup norm: ‖x‖∞ := max1≤i≤n

|xi|, where |xi | denotes the absolute value of xi .

the `2 or Euclidean norm: ‖x‖2 :=`Pn

i=1 x2i´1/2

.



Review: Linear AlgebraMatrices

A = (aij) :=

26664a11 a12 . . . a1m

a21 a22 . . . a2m...

.... . .

...an1 an2 . . . anm

37775 .If n = m, A is called a square matrix. Other special matrices:2664

a11 0 . . . 00 a22 . . . 0...

.... . .

...0 0 . . . ann

3775diagonal matrix

,

2664a11 a12 . . . a1n0 a22 . . . a2n...

.... . .

...0 0 . . . ann

3775upper triangular matrix

,

26641 0 . . . 00 1 . . . 0...

.... . .

...0 0 . . . 1

3775identity matrix

.

the `2 norm of A is

‖A‖ =

0@ nXj=1

nXi=1

a2ij

1A1/2

.



Review: Linear AlgebraMatrices

Matrix addition and scalar multiplication:

C = A + dB =

26664a11 + db11 a12 + db12 . . . a1m + db1m

a21 + db21 a22 + db22 . . . a2m + db2m...

.... . .

...an1 + dbn1 an2 + dbn2 . . . anm + dbnm

37775 (32)

The product of two matrices A ∈ Rn×m and B ∈ Rm×n, is the n × n matrix C = (cij),defined by

cij =mX

k=1

aik bkj . (33)

For suitable matrices A, B, C, and D matrix multiplication satisfies the following rules

AB 6= BA, (34a)

A(B + C) = AB + AC, (34b)

A(BC) = (AB)C, (34c)

A(B + C)D = ABD + ACD. (34d)




Kronecker product ⊗ of two matrices A, B:

A⊗ B :=

26664a11B a12B . . . a1mBa21B a22B . . . a2mB

......

. . ....

an1B an2B . . . anmB

37775 .




Trace of a square matrix A: sum of the elements of its main diagonal

trA =nX

i=1

aii . (35)

Determinant of a 2× 2 matrix A: |A| or det(A)

|A| = a11a22 − a12a21. (36)

Recursive formula for computation:

|A| =nX

j=1

aij(−1)i+j |Aij | =nX

i=1

aij(−1)i+j |Aij |, (37)

where Aij is the matrix obtained from A by deleting the i-th row and j-th column.



Review: Linear AlgebraThe transpose of A, denoted by A′ or AT :

A′ = (a′ij) = (aji) =

26664a11 a21 . . . an1

a12 a22 . . . an2...

.... . .

...a1m a2m . . . anm

37775 .

Inverse of a square matrix A, denoted A−1 = (aij) solves the problem AA−1 = I. If itexists, the inverse is unique and given by

aij =aij(−1)i+j |Aji |

|A| . (38)

Rules for transpose:

(A′)′ = A, (39a)

(A + B)′ = A′ + B′, (39b)

(AB)′ = B′A′, (39c)

(A−1)′ = (A′)−1. (39d)



Review: Linear AlgebraA right eigenvector of A is a vector v that solves

Av = λv ⇔ (A− λI)v = 0. (40)

Similarly, the solution of

v¯′A = λv

¯′

is named a left eigenvector of A.

The condition |A− λI| = 0 results in a polynomial of degree n in λ.These n roots are the eigenvalues of the matrix A.

Eigenvectors are unique up to a scalar multiple and, thus, may be normalized to haveunit length.

Properties of the eigenvalues λi :nX

i=1

λi =nX

i=1

aii , (41a)

nYi=1

λi = |A|. (41b)




eigenspace

(A− λI)mvm = 0

If there are non-trivial solutions vm for m ≥ 2 but not for m − 1, the vector vm is called ageneralized right eigenvector of rank m for the square matrix A.

The space spanned by the (generalized) eigenvectors of A is called the eigenspace ofA.

The eigenspace can be partitioned in three subspaces formed by generalizedeigenvectors that belong to the eigenvalues with

modulus less than one (stable eigenspace, Es),

modulus equal to one (center eigenspace, Ec),

modulus greater than one (unstable eigenspace, Eu).



Review: Linear AlgebraMATLAB application

In MATLAB, enter the matrix A, the vector b, and a vector b = 0 as follows:

A=[-3 2 3; -3 2 1; 3 0 0]b=[10; 8; -3]x=zeros(3,1)

Eigenvalues of A:

>> eig(A)ans =-4.00001.5000 + 0.8660i1.5000 - 0.8660i

Determinant of A:

>> det(A)

ans =-12



Review: Linear AlgebraMATLAB application

Inverse of a matrix:

>> inv(A)

ans =

0 0 0.3333-0.2500 0.7500 0.50000.5000 -0.5000 0

Trace: sum of the diagonal:

sum(diag(A))


Linear Algebra Linear Equations










Linear Equations

Problem: How shall we solve Ax = b?

LU and Cholesky Factorization

Consider a system of linear equations

a11x1 + a12x2 + · · ·+ a1nxn = b1,

a21x1 + a22x2 + · · ·+ a2nxn = b2,

... =...,

an1x1 + an2x2 + · · ·+ annxn = bn

9>>>>>=>>>>>;⇔ Ax = b. (42)

Assumption: Square matrix A has full rank



Linear Equations

In this case it is possible to factorize A as follows

LU = A,

L =

26666641 0 0 . . . 0

l21 1 0 0 . . . 0l31 l32 1 0 . . . 0...

......

.... . .

...ln1 ln2 ln3 ln4 . . . 1

3777775 , U =

2666664u11 u12 u13 . . . u1n0 u22 u23 . . . u2n0 0 u33 . . . u3n...

.... . .

......

0 0 0 . . . unn

3777775(43)

If A is symmetric and positive definite, its Cholesky factor is the lower triangular matrixL that solves

LL′ = A (44)



Linear EquationsBoth the LU and the Cholesky factorization can be used to solve the linear system.

Let x := Ux. Then it is easy to solve the system

Lx = x

by forward substitution:

x1 = b1,

x2 = b2 − l21x1,

x3 = b3 − l31x1 − l32x2,

... =...

Given the solution for x, one gets the desired solution for x via backward substitutionfrom Ux = ~x:

xn =xn

unn,

xn−1 =1

un−1 n−1(xn−1 − un−1 nxn) ,

xn−2 =1

un−2 n−2(xn−2 − un−2 n−1xn−1 − un−2 nxn) ,

... =...



LU Decomposition

Computation of the LU decomposition: Idea of Gaussian elimination

Solution of Ax = b:

We pre-multiply the equation Ax = b by a series of matrices M(k−1), k = 1, . . . , n thattransforms the matrix A into an upper-triangular one.

For k = 1: A(1) = A, M(0) = I, b(1) = b.

For k = 2: A(2)x = M(1)A(1)x = M(1)b(1) = b(2), with

M(1) =

2666641 0 . . . 0

− a21a11

1 . . . 0...

.... . .

...− an1

a110 . . . 1

377775 .

Question: How does M(k) look like?



LU Decomposition

If akk = 0, we need to pre-multiply the system by a permutation matrix that changes theorder of the rows. In this case, Mk is replaced by a permutation matrix P.

Let A have the dimension 3× 3. Then PA changes the third and second row of A:

P =

241 0 00 0 10 1 0

35 .The Gaussian elimination results in:

M(n−1) ·M(n−2) . . .M(2) ·M(1)Ax = Ux = M(n−1) ·M(n−2) . . .M(2) ·M(1)b = b(n).



LU Decomposition

In order to transform b(n) back into b and reverse these transformations, we need tomultiply the equations by the inverse of the matrix M(k) starting with k = n − 1.

Questions: How do these simple matrices L(k) =“

M(k)”−1

look like?

The final result of this Gaussian elimination: LU-Decompostion

LU = L(1)L(2)L(3) . . . L(n−1) ·M(n−1)M(n−2) . . .M(2)M(1)A = A.



Linear EquationsMATLAB application

LU-Factorization of A:

>> [L,U]=lu(A)

L =

1 0 01 0 1-1 1 0

U =

-3 2 30 2 30 0 -2

Type LU in MATLAB help index to find out more



Linear EquationsMATLAB application

Solution of Ax = b in Matlab: based on LU-decomposition

>> x=A\b

x =

-121



LU DecompositionPivoting Strategies

0.003000x1 + 59.14x2 = 59.17 (45)

5.291x1 − 6.130x2 = 46.78 (46)

has the solution x = (10.0, 1.0)′:

>> A=[0.0030 59.14; 5.291 -6.130]; b=[59.17; 46.78]; x=[10.0; 1.0];>> A*x

ans =

59.170046.7800

>> x=A\b

x =

101



LU DecompositionPivoting Strategies

Assume: Accuracy is only 4 digits

First pivot element: a11 = 0.003000,

Associated multiplier, a21a11

= 5.2910.003000 = 1763.66, rounds to 1764.

After Gaussian elimination, the system is transformed to:

0.003000x1 + 59.14x2 ≈ 59.17

−104300x2 ≈ −104400,

instead of

0.003000x1 + 59.14x2 = 59.17

−104309.376x2 ≈ −104309.376,

The solutions are:

x2 ≈ 1.001

x1 ≈ 59.17− (59.14)(1.001)

0.003000= −10.00



Problem: Small error in x2 (0.001) is multiplied by a large number

59.140.003000

⇒ pivot element aii is too small

Good pivot strategy: change rows such that the diagonal elements are not small,reduces round-off error!



LU DecompositionProblems

Ill conditioning

Assumption:b is perturbated by rounding errors rno errors in A

Let x be the solution of Ax = b.

Define e ≡ x− x = A−1r

Condition number:‖e‖‖x‖/

‖r‖‖b‖ (47)

Ae = r ⇒

‖A‖ ‖e‖ ≥ ‖r‖

e = A−1r⇒

‖A−1‖ ‖r‖ ≥ ‖e‖



LU DecompositionProblems

Together with Ax = b:

‖r‖‖A‖ ‖A−1‖ ‖b‖ ≤

‖e‖‖x‖ ≤

‖A−1‖ ‖A‖ ‖r‖‖b‖

Condition number of A:

cond(A) = ‖A‖ ‖A−1‖ (48)

Upper and lower bounds:

1cond(A)

‖r‖‖b‖ ≤

‖e‖‖x‖ ≤

‖r‖‖b‖ cond(A)

Rough rule of thumb: As the condition number increase by a factor of 10, you lose onesignificant digit

Any condition over 10ε−5 is considered large if ε is the number of the machine accuracyBurkhard Heer (University of Bolzano-Bozen, Italy) Computational Economics 4. September 2006 144 / 393


Schur Decomposition

Schur Factorization

The Schur factorization of a square matrix A is given by

A = TST−1. (49)

The complex matrix S is upper triangular with the eigenvalues of A on the maindiagonal. It is possible to choose T such that the eigenvalues appear in any desiredorder along the diagonal of S.

The transformation matrix T has the following properties:

its complex conjugate transpose T ′ equals the inverse of T ,

therefore: TT ′ = TT−1 = I, i.e., T is an unitary matrix,

all eigenvalues of T have absolute value equal to 1.




Reading Assignment

Heer, Maussner, Chapter 8.1.

Miranda, Fackler, Chapter 2

Burden, Faires, Chapter 6

Problems1 Solve the problem Ax = b with

A =

»M−1 1

1 1

–, P =

»12

–1 Type help lu and help norm in Matlab.2 Write a short program by increasing M from 100 to 500. Report the solution vector and

the [P, L, U] decomposition.3 Verify the solution.

2 Use the Matlab function randn to generate a 100-by-100 matrix A and a random100-vector b. Then use the Matlab functions tic and toc to compute the timeneeded to solve the linear equation Ax = b for the algorithms i) LU decompositionand ii) x = A−1b.



Reading Assignments and Problems II3 Consider the Hilbert matrix

Hn =

266641 1

213 . . . 1

n12

13

14 . . . 1

n+1...

......

......

1n

1n+1

1n+2 . . . 1

2n

377751 Type help cond in Matlab.2 Increase N from 3 to 10 and report the condition number.

4 Consider the Vandermonde matrix

V =

266641 x1 . . . xN−1

11 x2 . . . xN−1

2...

......

...1 xN . . . xN−1

N

377751 Type help randn, help plot and help print in Matlab.2 Write a Matlob code that generates N random numbers xi from the normal distribution

N(10, σ2) with σ = 5.3 Increase N from 2 to 5 and report the condition number.4 Compare your results to the Hilbert matrix.


Application: Reverse Shooting


Backward Iteration: Reverse Shooting

Computation of the transition dynamics in the infinite-horizon Ramsey model with fulldepreciation of the capital stock:

Kt+1 = K αt − Ct , (50a)

Ct+1 = αβCt(K αt − Ct)

α−1. (50b)

Its stationary solution is

K ∗ = (αβ)1/(1−α) and C∗ =

„1− αβ

αβ

«(αβ)1/(1−α),

and the Jacobian matrix at this solution is

J =

»1/β −1

1−αβαβ

α−1β

1− 1−αβαβ

(α− 1)

–.

It can be shown analytically that J has one eigenvalue smaller than one and oneeigenvalue exceeding one. See Heer/Maussner, Section 2.3.1.



Infinite-Horizon Ramsey ModelComputation of the Saddle Path

Computation with Matlab: shooting.m

1 Compute the steady state (K ∗,C∗).2 Compute the eigenvalue and the eigenvector of the Jacobian.3 Start from a point close to the steady state where (1, v)′ denotes the eigenvector

associated with the stable eigenvalue: (K ∗ + dK ,C∗ + vdK ), dK small.4 Iterate backwards, i.e., compute (Kt ,Ct) given (Kt+1,Ct+1) from (50). Therefore,

you have to solve a system of non-linear equations with the Broyden algorithm.

Reason, why we start in K ∗ + dk rather than K0:

The terminal state is excessively sensitive to the initial guess.

The initial state corresponding to the any terminal state is relatively insensitive tothe terminal state.




0 2 4 6 8 10 12 140.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4ca

pita

l sto

ck a

nd c

onsu

mpt

ion

period

kc




0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20.22

0.24

0.26

0.28

0.3

0.32

0.34

0.36

0.38co

nsum

ptio

n

capital stock




Reading Assignment

Heer, Maussner, Chapter 1.2.5

Optional: Judd, Chapter 10.7 (requires advanced knowledge of numericalmathematics)

Problems1 Compute the transition dynamics in the above Example for given K0 = 0.1. Apply

the method of reverse shooting.1 For this reason, choose a time horizon T = 20 for the number of transition periods.

Choose an initial value of dK 0 = 10−6 as a starting value. Compute the sequence{K 0

t }Tt=0 backwards. Solve the non-linear equation f (K0) = K 0

0 − K0 by adjusting dK .Store the solution K0, C0.

2 Control your solution. Iterate forward your solution (K0, C0). Graph the transition.


The Stochastic Ramsey Model The model










The stochastic Ramsey model


maxC0

E0

"∞Xt=0

βtu(Ct)

#s.t.

Kt+1 + Ct ≤ Zt f (Kt) + (1− δ)Kt ,0 ≤ Ct ,0 ≤ Kt+1.

9=; t = 0, 1, . . .

K0, Z0 given.

(51)

Zt — stochastic productivity, source of uncertainty

Zt follows Markov process ⇒ recursive structure: policy function Kt+1 = g(Zt ,Kt).




Markov processes

Markov process: probability distribution of Zt+1 only depends on the realization of Zt

AR(1) Process:

Zt = (1− ρ)Z + ρZt−1 + εt , εt ∼ N(0, σ2).

Markov chains are discrete valued Markov processes. They are characterized bythree objects:

1 The vector z = [z1, z2, . . . , zm]′ summarizes the m different realizations of Zt .

2 The probability distribution of the initial date t = 0 is represented by the vectorπ0 = [π01, π02, . . . , π0n]

′, where π0i denotes the probability of the event Z0 = zi .

3 The dynamics of the process are represented by a transition matrix P = (pij), wherepij denotes the probability of the event Zt+1 = zj |Zt = zi , i.e., the probability that nextperiod’s state is zj given that this period’s state is zi . Therefore, pij ≥ 0 andPm

j=1 pij = 1.




Thus, given Zt = zi the conditional expectation of Zt+1 is given by:

E(Zt+1|Zt = zi) = Piz,

where Pi denotes the i–th row of P

Conditional variance: var(Zt+1|Zt = zi) =P

j Pij(zj − Piz)2.

The probability distribution of Zt evolves according to

π′t+1 = π′t P

Ergodic distribution:

π′ = limt→∞

π′t ,

defined by

π′ = π′P ⇔ (P′ − I)π = 0.




Does the ergodic distribution π exist?

If for some integer k , all elements of Pk = P × P × . . .× P are strictly positive,pk

ij > 0.

P =

„0.0 1.00.9 0.1

«, P2 =

„0.9 0.1

0.09 0.91

«.

computation of the ergodic distribution

π′

26664p11 − 1 p12 . . . p1,n−1 1

p21 p22 − 1 . . . p2,n−1 1...

......

......

pn1 pn2 . . . pn,n−1 1

37775 = (0, . . . , 0, 1)



The stochastic Ramsey modelMatlab application

>> x=[0,1];>> y=[-1.0 1; 0.9 1];>> x*inv(y)

ans =

0.4737 0.5263

>> p=x*inv(y)

p =

0.4737 0.5263

>> p*[0 1; 0.9 0.1]

ans =

0.4737 0.5263



The stochastic Ramsey modelMarkov Chain Approximations of AR(1) Processes

Consider the process

Zt+1 = %Zt + εt , εt ∼ N(0, σ2ε).

The unconditional mean and variance of this process are 0 and σ2Z = σ2

ε/(1− %2).

Tauchen (1986):

choose a grid Z = [z1, z2, . . . , zm] of equidistant points z1 < z2, . . . , < zm, whoseupper end point is a multiple, say λ, of the unconditional standard of theautoregressive process, zm = λσZ and whose lower end point is z1 = −zm.For a given realization zi ∈ Z the variable z := %zi + ε is normally distributed withmean %zi and variance σ2

ε .Let dz denote half of the distance between two consecutive grid points. Theprobability that z is in the interval [zj − dz, zj + dz] is given by

prob(zj − dz ≤ z ≤ zj + dz) = π(zj + dz)− π(zj − dz)

where π(·) denotes the cumulative distribution function of the normal distributionwith mean %zi and variance σ2.




Algorithm: Markov chain approximation

Purpose: Finite state Markov chain approximation of first order autoregressive process

Step 1: Compute the discrete approximation of the realizations: Let % and σε denotethe autoregressive parameter and the standard deviation of innovations,respectively. Select the size of the grid by choosing λ ∈ R++ so thatz1 = −λσε/

p1− %2. Choose the number of grid points m. Put

step = −2z1/(m − 1) and for i = 1, 2, ...,m compute zi = z1 + (i − 1)step.

Step 2: Compute the transition matrix P = (pij): Let π(·) denote the cumulativedistribution function of the standard normal distribution. For i = 1, 2, . . . ,m put

pi1 = π“

z1−%ziσε

+ step2

”,

pij = π“

zj−%ziσε

+ step2

”− π

“zj−%zi

σε− step

2

”,

j = 2, 3, . . . ,m − 1,pim = 1−

Pm−1j=1 pij .

Tauchen: m = 9, λ = 3




The stochastic Euler equations

L = E0

∞Xt=0

βthu(Ct) + µtCt + ωt+1Kt+1

+ λt (Zt f (Kt) + (1− δ)Kt − Ct − Kt+1)iff.

Differentiation with respect to C0 and K1:

∂L∂C0

= E0{u′(C0)− λ0 + µ0} = 0,

∂L∂K1

= βE0{−λ0 + ω1 + βλ1(1− δ + Z1f ′(K1))} = 0,

0 = λ0(Z0f (K0) + (1− δ)K0 − C0 − K1),

0 = µ0C0,

0 = ω1K1.



The stochastic Ramsey modelThe stochastic Euler equations

Since C0, K1, and hence the multipliers λ0, µ0, and ω1 are non-stochastic, we canreplace the first condition with

u′(C0) = λ0 − µ0

and the second with

λ0 = βE0λ1{1− δ + Z1f ′(K1)}+ ω1.

Now, consider the problem from t = 1 onwards, when Z1 is known and K1 given. TheLagrangean for this problem is

L = E1

∞Xt=1

βt−1hu(Ct) + µtCt + ωt+1Kt+1

+ λt (Zt f (Kt) + (1− δ)Kt − Ct − Kt+1)iff.

Proceeding as before, we find

u′(C1) = λ1 − µ1,

λ1 = βE1λ2{1− δ + Z2f ′(K2)}+ ω2.




Continuing in this way, we find, since Kt must be optimal at t , that the plan for choosingC0,C1, . . . and K1,K2, . . . must solve the system:

u′(Ct) = λt − µt , (52a)

λt = βEtλt+1ˆ1− δ + Zt+1f ′(Kt+1)

˜+ ωt+1, (52b)

0 = λt(Zt f (Kt) + (1− δ)Kt − Ct − Kt+1), (52c)

0 = µtCt , (52d)

0 = ωt+1Kt+1. (52e)

Thus, an interior solution with strictly positive consumption and capital at all dates t(i.e., ∀t : µt = ωt+1 = 0) must satisfy:

1 =βEtu′(Zt+1f (Kt+1) + (1− δ)Kt+1 − Kt+2)

u′(Zt f (Kt) + (1− δ)Kt − Kt+1)

× (1− δ + Zt+1f ′(Kt+1)).

(53)

In addition to the stochastic Euler equation (53) there is also the stochastic analog ofthe transversality condition:

limt→∞

βtEtλtKt+1 = 0 (54)




Example

Let the one-period utility function u and the production function f be given by

u(C) := ln C,

f (K ) := K α, α ∈ (0, 1),

K ′ is directly proportional to K α in the deterministic counterpart of the model. So let ustry

Kt+1 = g(Kt ,Zt) := AZtK αt

with the unknown parameter A as policy function. If this function solves the problem, itmust satisfy the stochastic Euler equation (53). To prove this assertion, we replace Kt+1

in equation (53) by the right hand side of the previous equation:

1 = βEt

»(1− A)ZtK α

t

(1− A)Zt+1[AZtK αt ]α

αZt+1[AZtK αt ]α−1

–=αβ

A.

If we put A = αβ the function g(Zt ,Kt) = αβZtK αt satisfies the Euler equation, and,

thus is the policy function we look for.


The Stochastic Ramsey Model Stochastic Dynamic Programming










The stochastic Ramsey modelStochastic Dynamic Programming

Value function:

v(K ,Z ) = max0≤K ′≤Zf (K )+(1−δ)K

u(Zf (K ) + (1− δ)K − K ′)

+ βEˆv(K ′,Z ′)|Z

˜,

In the case of a Markov chain with realizations [z1, z2, . . . , zn] and transition matrixP = (pij):

Eˆv(K ′,Z ′)|zi

˜=

nXj=1

pijv(K ′, zj)

In the case of the continuous valued Markov process with conditional probabilitydensity function π(z,Z ′) over the interval [a, b]:

Eˆv(K ′,Z ′)|z

˜=

Z b

av(K ′,Z ′)π(z,Z ′)dZ ′.

Existence and properties of the value function: Stokey, Lucas (1989)Burkhard Heer (University of Bolzano-Bozen, Italy) Computational Economics 4. September 2006 170 / 393



The procedure is built on the result that the sequence of value functions obtained under

v s+1(K , zi) =maxK ′

u(zi f (K ) + (1− δ)K − K ′)

+ β

mXj=1

pijv s(K ′, zj),

converges linearly at rate β.

Problems:1 Choice of a grid for K : minimum and maximum values2 Choice of the number of grid points3 Initialization of the value function



The stochastic Ramsey model IStochastic Dynamic Programming


Purpose: Compute an Approximation to the Policy Function of the Stochastic RamseyModel with m–state Markov Chain for the productivity shock.

Step 1: Choose a grid of n equally spaced points over [K1 = K ,Kn = K ]:

G = [K1,K2, . . . ,Kn], Ki < Ki+1 ∀i = 1, 2, . . . n − 1.

Step 2: Initialize the value function as follows: Let U = (uij). For all j = 1, 2, . . . , ncompute

uij := u(zj f (K ∗j ) + (1− δ)K ∗

j − K ∗j ),

where K ∗j solves 1 = β(1− δ + zj f ′(K ∗

j )) and i = 1, 2, . . . , n. Put

v0 = U(I − βP′)−1.



The stochastic Ramsey model IIStochastic Dynamic Programming

Step 3: Compute a new value function v1 and associated policy function g1: For eachj = 1, 2, . . . ,m repeat these steps: Put k∗0 = 1 and for i = 1, 2, . . . , n find theindex k∗i that maximizes

wk = u(zj f (Ki) + (1− δ)Ki − Kk ) + β

mXl=1

pjlv0kl

in the set of indices k = k∗i−1, k∗i−1 + 1, . . . , n. Set v1

ij = wk∗iand g1

ij = k∗i .

Step 4: Check for convergence: if

maxi=1,...nj=1,...m

|v1ij − v0

ij | ≤ ε(1− β), ε ∈ R++

(or if the policy function has remained unchanged for a number of consecutiveiterations) stop, else set v0 = v1 and g0 = g1 and return to Step 3.




Matlab program: svalue1.m and svalue2.m

Sensitivity to the grid [kmin, kmax ]1 upper and lower value: try kmin ∈ {0.3k∗, 0.5k∗, 0.8k∗} and

kmax ∈ {1.7k∗, 1.5k∗, 1.2k∗}.2 Number of grid points: Try n ∈ {500, 1000, 5000}.

Sensitivity: Initialization of v : try v ≡ 0Simulate time series:

1 stability? solution ok?2 does the simulated time series stay within the grid for k?

Reasonable choice of the asset grid? Check if the policy function k ′(z, k) hit thelower/upper grid point.



Infinite-Horizon Stochastic Ramsey ModelMatlab output, N=500 grid points over K

18 20 22 24 26 28 3018

20

22

24

26

28

30solution of example 1.3.4 in heer/maussner

capital stock

next

per

iod

capi

tal s

tock




18 20 22 24 26 28 301.9

1.95

2

2.05

2.1

2.15

2.2

2.25

2.3

2.35solution of example 1.3.4 in heer/maussner

capital stock

cons

umpt

ion




18 20 22 24 26 28 3029

30

31

32

33

34

35

36solution of example 1.3.4 in heer/maussner

capital stock

valu

e fu

nctio

n




18 20 22 24 26 28 301.9

1.95

2

2.05

2.1

2.15

2.2

2.25

2.3

2.35solution of example 1.3.4 in heer/maussner

capital stock

cons

umpt

ion



Infinite-Horizon Stochastic Ramsey ModelMatlab output, svalue2.m, N=2000 grid points over K

100 150 200 250 300 350 400 450 50021.8

22

22.2

22.4

22.6

22.8

23

time

capi

tal s

tock




Reading Assignment

Heer, Maussner, Chapter 1.3, 9.2

Problems1 Compute the ergodic distribution of

P =

»0.5 0.50.1 0.9

–.

2 Discretize the AR-(1) process Zt = ρZt−1 + εt with ρ = 0.95 and σ = 0.0072. UseTauchen’s algorithm with m = 9 and λ = 3. Report your results. Draw a sequenceof 10000 random numbers from the normal distribution with σ = 0.0072 andsimulate a time series for Zt given Z0 = 0. Drop the first 100 observations andcompute the mean and the variance of the series. Compare it to the mean and thevariance of the AR(1) process. Repeat the exercise for m ∈ {7, 15} andλ ∈ {2, 4}. Report your results in a table.


Real Business Cycle Model Growth, Leisure and Business Cycles










Real Business Cycle Model

The Real Business Cycle Model

⇒ The stochastic growth model with labor/leisure choice, U(C, 1− N)

Effect of positive productivity shock:1 Income effect: higher consumption, more leisure2 Substitution effect: more labor3 Intertemporal substitution: more labor today, less labor tomorrow, higher savings




Growth and Restrictions on Technology and Preferences

Balanced growth path = equivalent to steady state

1 Output per working hour grows at constant rate2 Share of net savings in output is constant

⇒ only in accordance with labor-augmenting technical progress:

Yt = ZtF (AtNt ,Kt)

Zt — disembodied technological changeAt — embodied technological change, quality of labor




Restrictions on preferences

max{Ct ,Nt}∞t=0

∞Xt=0

βtu(Ct , 1− Nt)

s.t.

Kt+1 + Ct ≤ F (AtNt ,Kt) + (1− δ)Kt ,0 ≤ Ct ,1 ≥ Nt ≥ 0,0 ≤ Kt+1.

9>>=>>; t = 0, 1, . . . ,

K0 given.

(55)

→ restriction on utility (see Heer/Maussner, Appendix 2, Ch.1 ):

u(C, 1− N) =

C1−ηv(1− N) if η 6= 1,

ln C + v(1− N) if η = 1.(56)

Of course, u(.) must be concave.



Real Business Cycle ModelLagrangean of Example (55):

L =∞Xt=0

βthu(Ct , 1− Nt) + Λt

`F (AtNt ,Kt)

+ (1− δ)Kt − Ct − Kt+1´i.

Differentiating this expression with respect to Ct , Nt , and Kt+1 provides the following setof first order conditions:

0 = u1(Ct , 1− Nt)− Λt , (57a)

0 = −u2(Ct , 1− Nt) + ΛtF1(AtNt ,Kt)At , (57b)

0 = −Λt + βΛt+1(1− δ + F2(At+1Nt+1,Kt+1)). (57c)

Conditions (57a) and (57b) imply that the marginal rate of substitution betweenconsumption and leisure, u2/u1, equals the marginal product of labor:

u2(Ct , 1− Nt)

u1(Ct , 1− Nt)= AtF1(AtNt ,Kt). (58)

Conditions (57a) and (57c) yield

u1(Ct , 1− Nt)

u1(Ct+1, 1− Nt+1)= β(1− δ + F2(At+1Nt+1,Kt+1)). (59)




Transformation to Stationary Variables

Assume η 6= 1 in Example (55) and deterministic growth of the efficiency level of labor:

At+1 = aAt

The static labor supply condition can be written as

v ′(1− Nt)

(1− η)v(1− Nt)

Ct

At= F1(Nt ,Kt/At) (60)

and the intertemporal condition:

C−ηt v(1− Nt)

C−ηt+1v(1− Nt+1)

=(aAt)

ηC−ηt v(1− Nt)

Aηt+1C−η

t+1v(1− Nt+1)

=aη(Ct/At)

−ηv(1− Nt)

(Ct+1/At+1)−ηv(1− Nt+1)

= β(1− δ + F2(Nt+1,Kt+1/At+1))

(61)




Since F is homogenous of degree one, we can transform the resource constraint to

Kt+1

At+1/a= F (Nt ,Kt/At) + (1− δ)(Kt/At)− (Ct/At). (62)

Equations (60) through (62) constitute a dynamical system in the new variables Nt ,ct := Ct/At , and kt = Kt/At . Their stationary values N, c and k are found as solution tothe system of three equations

c =(1− η)v(1− N)

v ′(1− N)F1(N, k),

1 = βa−η(1− δ + F2(N, k)),

0 = F (N, k)− (1− δ − a)k − c.




Note, that we can get the efficiency conditions (60) through (62) from solving theproblem

max{ct ,Nt}∞t=0

∞Xt=0

βtc1−ηt v(1− Nt)

s.t.

ct ≤ F (Nt , kt) + (1− δ)kt − akt+1,k0 given

with discount factor β := βa1−η in the stationary decision variables ct := Ct/At andkt+1 := Kt+1/At+1.



Real Business Cycle ModelThe decentralized economy

Firms

Assumption: n equal firms

Maximization of the firm’s present value is equivalent with maximizing one-periodprofits

Πi := ZF (ANi ,Ki)− wNi − rKi .

The first order conditions imply

w = ZAF1(ANi ,Ki) = ZAF1(A,Ki/Ni),

r = ZF2(ANi ,Ki) = ZF2(A,Ki/Ni),

due to the homogeneity of degree zero of Fi .

All firms choose the same capital-labor ratio k := Ki/Ni , output per labor is the same,yi = y = ZF (A, k):⇒ Aggregate production:

Y =X

i

Yi =X

i

Niy = Ny = NZF (A, k) = ZF (AN,K ),

Profits are zero (Euler theorem).Burkhard Heer (University of Bolzano-Bozen, Italy) Computational Economics 4. September 2006 191 / 393



Households

Mass 1

or equivalently: representative household

maximizes intertemporal utility

supplies labor Nt with efficiency At

supplies capital services Kt

saves in terms of capital Kt

Budget constraint:

Kt+1 − Kt ≤ wtNt + (r − δ)Kt − Ct . (63)




maximize its life-time utility

∞Xt=0

βtu(Ct , 1− Nt), β ∈ (0, 1) (64)

subject to (63) and the given initial stock of capital K0. From the Lagrangean of thisproblem,

L =∞Xt=0

βthu(Ct , 1− Nt) + Λt(wtNt + (1− δ + rt)Kt

− Ct − Kt+1)i

we derive the following first order conditions:

u1(Ct , 1− Nt) = Λt , (65a)

u2(Ct , 1− Nt) = Λtwt , (65b)

Λt = βΛt+1(1− δ + rt+1). (65c)




Using the factor market equilibrium conditions to substitute for wt and rt+1 and applyingthe Euler theorem to F ,

Yt = ZF (AtNt ,Kt) = ZAtF1(AtNt ,Kt)Nt + ZF2(AtNt ,Kt)Kt

equations (65) reduce to

u2(Ct , 1− Nt)

u1(Ct , 1− Nt)= ZAtF1(AtNt ,Kt), (66a)

u1(Ct , 1− Nt)

u1(Ct+1, 1− Nt+1)= β(1− δ + ZF2(At+1Nt+1,Kt+1)), (66b)

Kt+1 = ZF (AtNt ,Kt) + (1− δ)Kt − Ct . (66c)

This system is identical to the first order conditions that we derived for the Ramseymodel of the central planner.


Real Business Cycle Model The benchmark real business cycle model










Real Business Cycle ModelThe benchmark model

Our benchmark model

Example

Our benchmark real-business cycle model:

maxC0,N0

E0

"∞Xt=0

βt C1−ηt (1− Nt)

θ(1−η)

1− η

#,

β ∈ (0, 1), η ≥ 0, θ ≥ 0, η > θ/(1 + θ),

s.t.

Kt+1 + Ct ≤ Zt(AtNt)αK 1−α

t + (1− δ)Kt , α ∈ (0, 1),At+1 = aAt , a ≥ 1,Zt+1 = Z %

t eεt , % ∈ (0, 1), εt ∼ N(0, σ2),0 ≤ Ct ,0 ≤ Kt+1,

9>>>>=>>>>;∀t ,

K0, Z0 given.




From the Lagrangean

L := E0

n ∞Xt=0

βth C1−η

t (1− Nt)θ(1−η)

1− η

+ Λt

“Zt(AtNt)

αK 1−αt + (1− δ)Kt − Ct − Kt+1

” iowe derive the following first order conditions:

θCt

1− Nt= αZtAt(AtNt)

α−1K 1−αt ,

Ct = Zt(AtNt)αK 1−α

t + (1− δ)Kt − Kt+1,

1 = βEt

„Ct

Ct+1

«η „1− Nt+1

1− Nt

«θ(1−η)

×`1− δ + (1− α)Zt+1(At+1Nt+1)

αK−αt+1

´




In terms of stationary variables ct := Ct/At and kt := Kt/At this system is:

θct

1− Nt= αZtNα−1

t k1−αt , (68a)

ct = ZtNαt k1−α

t + (1− δ)kt − akt+1, (68b)

1 = βa−ηEt

„ct

ct+1

«η „1− Nt+1

1− Nt

«θ(1−η)

× (68c)`1− δ + (1− α)Zt+1Nα

t+1k−αt+1

´.


Real Business Cycle Model Calibration of the benchmark model










Real Business Cycle ModelCalibration

Basic idea of calibration: Real economic data is generated by the model.

We choose the German economy 1975.i-1989.iv

Assumption: Germany is on a balanced growth path during this period. ⇒ In order

to select values like k/y we use long-run averages

Evidence from time-series: {α, δ, ρ, σ, β}Microeconomic studies: {η}




Calibration of the parameters: svalue2.mWe choose a Cobb-Douglas production function in accordance with empiricalevidence.Labor augmenting technical progress a− 1 = 0.005: linear time trend of GDPα = 0.73, average wage share in total income (includes hypothetical income ofself-employed)Depreciation rate δ = 0.011Markov process for Zt

ln Zt = % ln Zt−1 + εt ,

where ln Zt ≈ (Zt − Z )/Z and

Zt =Yt

((1.005)tHt)0.72K 0.28t

.

We fit an AR(1) process to the percentage deviation of Zt from its mean Z ≡ 1.This delivers our estimate of % = 0.90 and of σ = 0.0072.η = 2 from micro-studiesβ = 0.994 with the help of the FOC and an average yearly return on capital equalto 6.5%.θ = 5.659: steady state labor supply N = 0.13.




Computation of the steady state values and the grid

Given a, β, α, δ and η, we can solve the steady state for y/k :

1 = βa−η(1− δ + (1− α)Nαk−α) = βa−η(1− δ + (1− α)(y/k))

In the next step, we compute c/k :

yk

=Nαk1−α

k=

ck

+ (a + δ − 1).

And, hence, y/c = y/k · k/c. With N = 0.13, we can compute θ:

N1− N

=α

θ

Nαk1−α

c=α

θ

yc

Finally, we can solve for k :

yk

= Nαk−α

⇒ we compute a grid around the steady state of k .


Real Business Cycle Model Computation of the benchmark model 1: Value function iteration










Real Business Cycle Model IComputation

Computation: Value function iteration with binary search

Code: svalue2.m

1 First, we calibrate the model and choose a grid over k .2 In a second step we approximate the continuous valued Markov process by a

Markov chain using Tauchen’s Algorithm.3 Given the realizations z = [z1, z2, ..., zm] and the associated transition matrix

P = (pij) we must iterate over

v s+1(k , zj)

= maxN,k′

(zjNαk1−α + (1− δ)k − ak ′)1−η(1− N)θ(1−η)

1− η

+ β

mXl=1

pjlv s(k ′, zl), β := βa1−η.



Real Business Cycle Model IIComputation

4 The maximization problem is split into two steps. Given a pair (z, k , k ′), we firstsolve for the optimal N using the first order condition

θziNαk1−α + (1− δ)k − ak ′

1− N= αziNα−1k1−α. (69)

We solve this non-linear equation with Broyden’s algorithm. In a second step, wesearch for the optimal k ′ using binary search.

5 We stop when the value function does not change by more than ε or the policyfunction is unchanged for a number n of iterations.

6 We simulate the economy with the help of the solution.

Run time on Pentium(R), 319 MHz: approximately 12 hours

Remarks:The grid can be re-specified by trial and error. In the simulation, k should not hitthe boundaries.We may need to re-specify the non-linear equation for the optimal labor supply,e.g. by multiplying through with N1−α(1− N) in order to improve convergence ofBroyden’s algorithm.



Real Business Cycle ModelComputation

1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2−520

−510

−500

−490

−480

−470

−460

−450

−440

capital stock

v




1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.20.12

0.125

0.13

0.135

0.14

0.145

0.15

capital stock

labo

r su

pply




1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.20.21

0.22

0.23

0.24

0.25

0.26

0.27

0.28

0.29

capital stock

cons

umpt

ion




0 500 1000 1500 2000 2500 30002.6

2.65

2.7

2.75

2.8

2.85

2.9

2.95

3

3.05

3.1

time

capi

tal s

tock


Real Business Cycle Model Evaluation of the Model










Real Business Cycle ModelModel Performance

Evaluation of the model with respect to business cycle features

Second moments of time series⇒ Trend has to be removed from time series data.

The HP-filterLet (yt)

Tt=1 denote the log of a time series that may be considered as realization of a

non-stationary stochastic process. The growth component (gt)Tt=1 of this series as

defined by the HP-Filter is the solution to the following minimization problem:

min(gt )

Tt=1

TXt=1

(yt − gt) + λ

T−1Xt=2

[(gt+1 − gt)− (gt − gt−1)]2. (70)

λ = 0: yt = gt .

λ→∞: constant growth rate g

For quarterly data: λ = 1600 based on the observation that with this choice thefilter removes all cycles longer than 32 quarters leaving shorter cycles unchanged.



Real Business Cycle ModelHP filter

The first order conditions of the minimization problem imply the following system oflinear equations:

Ag = y, (71)

where g = [g1, g2, . . . , gT ]′, y¯

= [y1, y2, . . . , yT ]′, and A is the tridiagonal matrix266666666664

1 + λ −2λ λ 0 0 . . . 0 0 0−2λ 1 + 5λ −4λ λ 0 . . . 0 0 0

λ −4λ 1 + 6λ −4λ λ . . . 0 0 00 λ −4λ 1 + 6λ −4λ . . . 0 0 00 0 λ −4λ 1 + 6λ . . . 0 0 0...

......

......

. . ....

......

0 0 0 0 0 . . . 1 + 6λ −4λ λ0 0 0 0 0 . . . −4λ 1 + 5λ −2λ0 0 0 0 0 . . . λ −2λ 1 + λ

377777777775




Note, that A can be factored in1

A = I + λK ′K ,

K =

26666641 −2 1 0 0 . . . 0 0 00 1 −2 0 0 . . . 0 0 00 0 1 −2 1 . . . 0 0 0...

......

......

. . ....

......

0 0 0 0 0 . . . 1 −2 1

3777775

MATLAB program: hpfilter.m, returns the growth component. Do not use time seriesthat have many more than 1000 observations with this routine (speed).

1See BRANDNER and NEUSSER (1990), p. 5.Burkhard Heer (University of Bolzano-Bozen, Italy) Computational Economics 4. September 2006 221 / 393



The cyclical component of y,

c = y− g = [I − A−1]y,

remains unchanged if a linear time trend

a :=

26664a1 + a2

a1 + 2a2...

a1 + Ta2

37775is added to the time series (yt)

Tt=1. To see this, note that

c = [I − A−1][y + a] = [I − A−1]y + [I − A−1]a| {z }=0

.




This statement can be proven by noting that

[I − A−1]a = 0 ⇔ A−1[A− I]a = 0 ⇔ [A− I]a = 0,

and considering the product on the rightmost side of this statement.

We can use the cyclical components of the log of the time series data for per capitaoutput etc and compare it to the cyclical component of the logs of the stationary valuesk = K/A from our model since

ln Kt = lnkt + ln At = ln kt + ln A0 + at .

⇒ cyclical components of ln Kt and ln kt are equal.



Real Business Cycle ModelModel evaluation

Standard deviations in %

Variable σx

Output y 1.78Investment i 7.95

Labor n 0.94Consumption c 0.72

Correlation with output

Close to one for i (0.99), n (0.99), and c (0.98)

Autocorrelation of output: 0.68




Reading Assignment

Heer, Maussner, Chapter 1.4, 9.4

Cooley, Hansen, Chapter 1

Problems1 Use the calibration of Cooley and Prescott to parameterize the real business cycle

benchmark model for the US economy: a = 1.0055, β = 0.99, α = 0.64, δ = 0.25,ρ = 0.95, σ = 0.0072, N = 1/3, η = 2.0. Compute the model with the help ofvalue function iteration for the following three utility functions:

u(ct , nt) =

8>><>>:1

1−η

hc1−η

t (1− nt)θ(1−η) − 1

iI

11−η

h`ct − θ

1+νAtn1+ν

t

´1−η − 1i

II

ln ct + θ(1−nt )1−γ

1−γIII

(72)

Calibrate θ in each case so that N = 1/3 in steady state. For the utility functions IIand III, set ν = 3.33 and γ = 7.0, respectively.

1 Show that the values of ν and γ imply a Frisch labor supply elasticity ηn,w = 0.3.



Reading Assignments and Problems II

2 Compute the models with value function iteration using binary search and 1000 gridpoints for k . Discretize the AR(1) process for technology using Tauchen’s method and 9gridpoints for the grid over technology.

3 Report your results for the standard deviations and correlations with output of thevariables output, labor, consumption, investment, and wages. Also, report first-orderautocorrelation coefficients. Compare the different specifications. For this reasonsimulate the economy 100 times for 100 periods and report averages. Also reportstandard errors.

4 Accuracy 1: Compute the maximum Euler equation residual e = (c)/c − 1 over thegrid for {k , z}, where c is consumption in period t and c is the consumption that isnecessary to equate λt = uc(c, 1− nt ) withβa−ηEtλt+1(1− δ + (1− α)Zt+1Nα

t+1k−αt+1 ). Compute the expectation Et () with the

help of the transition matrix and the policy functions. Report the average and themaximum values of the residuals for each utility function.

5 Accuracy 2: For the simulated time series compute the risk-free rate of return:

λt

βa−ηEtλt+1− 1.

Again, compute the expectation Et () with the help of the transition matrix and the policyfunctions. Compute the percentage differences between the averages of a time seriesof 10,000 points and the stationary rate aη/β − 1.


Loglinear Approximation Illustrative Example










Loglinear approximationAn Illustrative Example

maxC0

E0

∞Xt=0

βt ln(Ct), β ∈ (0, 1),

s.t. Kt+1 = ZtK αt − Ct , α ∈ (0, 1), t = 0, 1, . . . ,

K0,Z0 given.

Ct denotes consumption at time t , Kt is the stock of capital and Zt the productivityshock.

FOCs:

1Ct

= λt , (73a)

λt = βEtλt+1αZt+1K α−1t+1 , (73b)

Kt+1 = ZtK αt − Ct . (73c)




Loglinearization at the steady state

K = (αβ)1/(1−α) , (74a)

C = K α − K . (74b)

Numerical example with Matlab: α = 0.3, β = 0.99, ρ = 0.95

ks=(alpha*beta)^(1/(1-alpha))

ks =

0.1765

>> cs=ks^alpha-ks

cs =

0.4178




Total differential of (73a):

− 1C2

dCt = dλt ,

where the partial derivatives are evaluated at the stationary solution.

Note that for any variable x , the expression

x := (x − x)/x ≡ dxx' ln(x/x) (75)

is the percentage deviation of x from the point x (loglinear approximation).




⇒

− ct = λt . (76a)

Similarly (see Heer/Maussner, p. 101),

λt = βEt λt+1 − (1− α)Et Kt+1 + Et Zt+1, (77a)

Kt+1 =1β

Kt −1− αβ

αβCt +

1αβ

Zt . (77b)



Loglinear approximation IAn Illustrative Example

More convenient matrix notation

x: stacked (n × 1) vector of endogenous variables with initial conditions (statevariables). In our case: k and n = 1

z: stacked (l × 1) vector of purely exogenous shocks. In our case: z and l = 1

u: stacked (m × 1) vector of control variables. In our case: c and m = 1.

We can write the dynamics of linearized system in the following form:

Cuut = Cxλ

»xt

λt

–+ Czzt , (78a)

DxλEt

»xt+1

λt+1

–+ Fxλ

»xt

λt

–= DuEtut+1 + Fuut (78b)

+ DzEtzt+1 + Fzzt .

with



Loglinear approximation IIAn Illustrative Example

Cu=−1, Cxλ=[0 1], Cz=0,

Dxλ=

»1− α −1

1 0

–, Fxλ=

»0 1− 1

β0

–, Du=

»00

–,

Fu=

»0

− 1−αβαβ

–, Dz=

»10

–, Fz=

»01

αβ

–.




Solving the system

Zt = %Zt−1 + εt , % ∈ (0, 1), εt ∼ N(0, σ2). (79)

implies

Et Zt+1 = %Zt .

Furthermore, we replace Ct in equation (77b) by −λt , and, thirdly, we use this newequation to eliminate Kt+1 from (77a).

Et

»Kt+1

λt+1

–=

"1β

1−αβαβ

1−αβ

1−α(1−αβ)αβ

# »Kt

λt

–+

"1

αβ1−ααβ

− %

#Zt ,

⇔ Et

»Kt+1

λt+1

–= W

»Kt

λt

–+ RZt .

(80)



LoglinearapproximationAn Illustrative Example

W =

1.0101 2.36700.7071 2.6569

>> R=[1/(alpha*beta); (1-alpha)/(alpha*beta)-rho]

R =

3.36701.4069




Method of King-Watson to solve (80)

Schur factorization of W :

W = TST−1,

where S is a (possibly complex) upper triangular matrix, whose diagonal elements arethe eigenvalues of W .

W from equation (80) has two real eigenvalues so that S ∈ R2×2 is real, as is thetransformation matrix T satisfying T ′ = T−1.




W has two real roots 0 < µ1 < 1 and µ2 > 1:the determinant of W equals the product of the roots, i.e.,

µ1µ2 = |W | = 1− α(1− αβ)

αβ2 − (1− α)(1− αβ)

αβ2 = (1/β).

the trace of W equals the sum of the roots, i.e.,

µ1 + µ2 = (1/β) +1− α(1− αβ)

αβ=

1 + α2β

αβ.

Therefore, the roots of W solve

φ(µ) := µ+1/βµ

=1 + α2β

αβ.

⇒ the first root µ1 is located to the left of 1 and the second root must lie to the right of 1.

⇒

T−1WT = S =

»µ1 s12

0 µ2

–(81)




>> [T,S]=schur(W)

T =

-0.9578 -0.28730.2873 -0.9578

S =

0.3000 1.65990 3.3670

>> inv(T)*W*T

ans =

0.3000 1.65990.0000 3.3670




We compute W , T , T−1, µ1, µ2 numerically.

Transformation of variables:»Kt

λt

–:= T−1

»Kt

λt

–, ⇔ T

»Kt

λt

–:=

»Kt

λt

–(82)

⇒

Et

»Kt+1

λt+1

–=

»µ1 s12

0 µ2

–| {z }

S

»Kt

λt

–+

»q1

q2

–| {z }

:=Q=T−1R

Zt . (83)

→ we solve the second equation forward, the first backward.




>> Q=inv(T)*R

Q =

-2.8207-2.3151

>>>> [eigv,mu]=eig(W)

eigv =

-0.9578 -0.70860.2873 -0.7056

mu =

0.3000 00 3.3670




Computation of λt and Ct

Et λt+1 = µ2λt + q2Zt . (84)

We can solve this equation for λt via repeated substitution:

λt =1µ2

Et λt+1 −q2

µ2Zt . (85)

Shifting the time index one period into the future and taking expectations conditional oninformation as of period t yields:

Et λt+1 =1µ2

Et λt+2 −q2

µ 2Et Zt+1 =

1µ2

Et λt+2 −q2

µ 2%Zt , (86)

due to (79). Substitution into (85) gives:

λt =1µ2

2Et λt+2 −

»q2

µ2+

q2

µ2

%

µ2

–Zt .



Loglinear approximation IIAn Illustrative Example

We can use (86) to get an expression for λt+3 and so on up to period t + τ :

λt =

»1µ2

–τ

Et λt+τ −q2

µ2

τ−1Xi=0

»%

µ2

–i

Zt . (87)

Suppose that the sequence1µτ

2Et λt+τ

ff∞

τ=0

converges towards zero for τ →∞.

In this case we can compute the limit of (87) for τ →∞:

λt = − q2/µ2

1− (%/µ2)Zt . (88)

We substitute this solution into the second equation of (82),2

λt = t21Kt + t22λt ,



Loglinear approximation IIIAn Illustrative Example

to get the solution for λt in terms of Kt and Zt :

λt = − t21

t22|{z}=:Lux

Kt −q2/µ2

t22(1− (%/µ2))| {z }=:Luz

Zt . (89)

Thus, the policy function that gives Ct in terms of the (deviations) of the state variablesis

Ct = Lux Kt + Luz Zt .

2We denote the elements of T−1 by (t ij ).Burkhard Heer (University of Bolzano-Bozen, Italy) Computational Economics 4. September 2006 245 / 393



Computation of kt+1

From the first equation of (80),

Kt+1 =1β

Kt +1− αβ

αβλt +

1αβ

Zt ,

we can derive the solution for Kt :

Kt+1 =

„1β− 1− αβ

αβ

t21

t22

«| {z }

=:Lxx

Kt

+

„1αβ

− 1− αβ

αβ

q2/µ2

t22(1− (%/µ2))

«| {z }

=:Lxz

Zt .

Thus, given a sequence of shocks {Zt}Tt=0 and an initial K0 we may compute the entire

time path of consumption and the stock of capital by iteration over

Ct = Lux Kt + Luz Zt ,

Kt+1 = Lxx Kt + Lxz Zt .




>> Lux=-T(2,1)/T(2,2)Lux =

0.3000

>> Llx=-Q(2)/mu(2,2)*1/(T(2,2)*(1-rho/mu(2,2)))Llx =

-1

>> Lxx=1/beta-(1-alpha*beta)/(alpha*beta)*T(2,1)/T(2,2)Lxx =

0.3000

>> Lxz=1/(alpha*beta)-(1-alpha*beta)/(alpha*beta)*Q(2)/mu(2,2)*1/(T(2,2)*(1-rho/mu(2,2)))

Lxz =1




Comparison of loglinear and exact solution

>> Lxx=0.30;>> ks=(alpha*beta)^(1/(1-alpha))

ks =

0.1765

>> kgrid=0.8*ks:0.01:1.2*ks;>> kp=alpha*beta*kgrid.^alpha;>> kplog=ks+Lxx*(kgrid-ks);>> plot(kgrid,kp, kgrid, kplog)



Loglinear approximationComparison: exact and log-linear solution

0.15 0.16 0.17 0.18 0.19 0.2 0.21 0.22 0.230.165

0.17

0.175

0.18

0.185

0.19


Loglinear Approximation The General Method










Loglinear approximation

The General Method

Cuut = Cxλ

»x¯ tλt

–+ Czzt , (90a)

DxλEt

»xt+1

λt+1

–+ Fxλ

»xt

λt

–= DuEtut+1 + Fuut (90b)

+ DzEtzt+1 + Fzzt .

We assume that zt is governed by a stable vector autoregressive process of first orderwith normally distributed innovations εt :

zt = Πzt−1 + εt , εt ∼ N(0,Σ). (91)

Stability requires that the eigenvalues of the matrix Π lie within the unit circle.



Loglinear approximation I

System Reduction:

We assume that the first equation can be solved for the vector ut :

ut = C−1u Cxλ

»xt

λt

–+ C−1

u Czzt . (92)

Shifting the time index one period into the future and taking expectations conditional oninformation as of period t yields:

Etut+1 = C−1u CxλEt

»xt+1

λt+1

–+ C−1

u CzEtzt+1. (93)

The solutions (92) and (93) allow us to eliminate ut and Etut+1 from (90b):“Dxλ − DuC−1

u Cxλ

”Et

»xt+1

λt+1

–= −

“Fxλ − FuC−1

u Cxλ

” »xt

λt

–+

“Dz + DuC−1

u Cz

”Etzt+1

+“

Fz + FuC−1u Cz

”zt



Loglinear approximation II

Assume that this system can be solved for Et(xt+1, λt+1)′. In other words, the matrix

Dxλ − DuC−1u Cxλ must be invertible. Using Etzt+1 = Πzt , which is implied by (91), we

get the following reduced dynamic system:

Et

»xt+1

λt+1

–= W

»xt

λt

–+ Rzt ,

W =“

Dxλ − DuC−1u Cxλ

”−1 “Fxλ − FuC−1

u Cxλ

”,

R =“

Dxλ − DuC−1u Cxλ

”−1

×h“

Dz + DuC−1u Cz

”Π +

“Fz − FuC−1

u Cz

”i.

(94)



Loglinear approximation I

Change of Variables:

Multiple possibilities to factorize the matrix W :

1 Schur decomposition: S = T−1WT , S upper triangular, T ′ = T−1

Algorithm of King/Watson, more robust, a little bit more complicated → Heer,Maussner provide programs in Gauss/Fortran

2 Diagonalization: W = PΛP−1, where Λ is a diagonal matrix with the eigenvalues,P matrix with eigenvectors

Algorithm of King, Plosser, Rebelo, adapted from Burnside (1999), MATLABroutine: SolveLA.m

Derivation of the solution: exactly the same as in the simple example.

See also: Heer, Maussner, Chapter 2.3.2 for Schur decomposition, Burnside, Chapter3, for KPR algorithm.


Loglinear Approximation Application 1: The real business cycle model










Loglinear approximationThe real business cycle model

Loglinear solution of the real business cycle model

Our starting point are the first order conditions and the associated stationary solution.At the point of the stationary solution the loglinearized equations are:

λt = −ηct − θ(1− η)N

1− NNt ,

λt + (1− α)kt + Zt = (1− η)ct + ξ1Nt ,

αξ2Et kt+1 − Et λt+1 + λt = αξ2Et Nt+1 + ξ2Et Zt+1,

aEt kt+1 − (aη/β)kt = −(c/k)ct + α(y/k)Nt + (y/k)Zt ,

ξ1 :=`θ(1− η)− 1

´ N1− N

+ 1− α,

ξ2 := 1− βa−η(1− δ).




The matrices Cu , Cxλ, and Cz from the static equations (78a) are:

Cu:=

»−η −θ((1− η) N

1−N1− η ξ1

–, Cxλ:=

»0 1

1− α 1

–, Cz :=

»01

–.

The matrices of the dynamic part of the system are:

Dxλ:=

»αξ2 −1a 0

–, Fxλ:=

»0 1

−βa−η 0

–,

Du:=

»0 αξ2

0 0

–, Fu:=

»0 0

−(c/k) α(y/k)

–,

Dz :=

»ξ2

0

–, Fz :=

»0

y/k

–.




The matlab program ramsey4d.m computes the policy functions Lxx , Lxz , Lux , and Luz :

xt+1 = Lxx xt + Lxz zt ,

zt+1 = %zt + εt+1,»ct

Nt

–= Lux xt + Luz zt

Output from ramsey4d.g: Lxx = 0.9614, Lxz = 0.1299, Llx = −0.4635,Llz = −0.2014, Lux = (0.2512, 0.0448)′, Luz = (0.5581, 1.0535)′




Time series simulation: 500 simulations of 50 periods

Iteration over the dynamics to get time series for kt , zt , ct , nt

For production, investment, and wages, the time series can be computed with the helpof:

yt = αNt + (1− α)kt + Zt ,

it = (y/i)yt − [(y/i)− 1]ct , y/i = (y/k)/(a + δ − 1)

wt = yt − Nt .



Loglinear approximationPercentage deviation in simulation: y, c, i, n, w

0 10 20 30 40 50 60−0.25

−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

0.2



Real Business Cycle ModelLoglinear Approximation: ramsey4d.m

Standard deviations in %

Variable σx

Output y 1.53Investment i 7.07

Labor n 0.91Consumption c 0.49

Wage w 0.62

Correlation with output

Close to one for i (0.99), n (1.0), c (1.00), and w (1.0)

Autocorrelation of output: 0.63



Loglinear approximationImpulse response functions

Impulse response functions

percentage deviations of model variables following a technology shock Z0 = 1 inperiod 0 and zero thereafter

illustrate the dynamics of the model in response to a shock

comparable to the empirical observations: impulse responses from VAR



Loglinear approximationImpulse response functions

0 10 20 30 400

0.2

0.4

0.6

0.8

1z

0 10 20 30 400

0.5

1

1.5

2k and y

0 10 20 30 400

2

4

6

8

10c and i

0 10 20 30 400

0.5

1

1.5n and w


Loglinear Approximation Application 2: A monetary business cycle model










Loglinear approximationThe New Keynesian Phillips Curve

A monetary business cycle model

Money demand motive:1 Money-in-the-utility: Sidrauski2 Cash-in-advance: consumption, investment3 transaction costs4 store of value

Role of frictions: sticky prices/wages necessary for significant real effects: Calvo pricestaggering

→ microfoundation for Phillips curve: New Keynesian Phillips curve (forward-looking)




1. Households

The household maximizes the expected discounted stream of future utility

E0

∞Xt=0

βtu(Ct , 1− Nt)

with

u(Ct , 1− Nt) :=C1−η

t (1− Nt)θ(1−η)

1− η. (96)

subject to the budget constraint

Wt

PtNt + rtKt + Dt + Tt − TCt − Ct ≥

Mt+1 −Mt

Pt+ Kt+1 − (1− δ)Kt . (97)

Dt — dividendsMt — beginning of period t money stockTt — government transfers




Transaction costs:

TCt = γ

„Ct

Mt+1/Pt

«κ

Ct , γ, κ > 0. (98)




2. Monetary Authority

exogenous money growth µt := Mt+1/Mt :

µt+1 = ρµµt + εµt , µt := lnµt − lnµ, εµt ∼ N(0, σµ). (99)

In the stationary equilibrium money grows at the rate µ− 1.

The government transfers seignorage to the households:

Tt =Mt+1 −Mt

Pt. (100)

With this specification, money is neutral, but not super-neutral.




3. Final goods producers

assemble the output of a large number Jt of intermediary producers to the single goodYt according to

Yt =

24J−1/εt

JtXj=1

Y (ε−1)/εjt

35ε/(ε−1)

, ε > 1. (101)

Pjt – price of the intermediary good jPt – price of the final good

Maximizing its profits PtYt −PJt

j=1 PjtYj subject to (101) produces the following demandfor good j :

Yjt =

„Pjt

Pt

«−ε Yt

Jt. (102)

ε is the price elasticity of demand for good j .




No profit condition (perfect competition and free entry):

Pt =

24 1Jt

JtXj=1

P1−εjt

351/(1−ε)

. (103)




4. Intermediary goods’ producers

J firmsmonopolistic competitionCalvo price setting: in each period, a fraction (1− ϕ) can adjust their pricesoptimally (type A), the remaining ϕ producers increase their prices by the averageinflation π (type N).

Production of good j :

Yjt = Zt(AtNjt)1−αK α

jt − F , α ∈ (0, 1),F > 0 (104)

with a deterministic process for labor augmenting technical progress,

At+1 = aAt , a ≥ 1,

and Zt is a stationary, stochastic process for total factor productivity that follows

Zt+1 = ρZ Zt + εZt , Zt = ln Zt , ε

Zt ∼ N(0, σZ ).




Price setting of type N firms:

PNt = πPNt−1. (105)

Type A are free to adjust their price to optimize discounted real profits until the firmreset its price optimally again:

maxPAt

Et

∞Xτ=t

ϕτ−t%τ

»„πτ−tPAt

Pτ

«YAτ − gτ (YAτ + F )

–

s.t. YAτ =

„πτ−tPAt

Pτ

«−ε Yτ

Jτ.

(106)

with the discount factor depending on marginal utility of wealth:

%τ = βτ−t Λτ

Λt. (107)

gτ – the firm’s variable unit costs




Intermediary producers distribute their profits to the household sector. Thus,

Dt :=

JtXj=1

Pjt

PtYjt −

Wt

PtNjt − rtKjt . (108)

The Calvo price setting implies the following dynamics of the optimal price:

(1− ϕ)(1− ϕβa−η)

ϕgt = −βa−ηEt πt+1 + πt . (109)

→ NEW KEYNESIAN PHILLIPS CURVE

gt is a measure of market tension (deviation of the marginal costs from its steady statelevel)

derivation: see Heer/Maussner, Appendix 4




Stationary equilibrium

Stationary variables: ct := Ct/At , yt := Yt/At , kt := Kt/At , λt := ΛtAηt .

In addition, we define the inflation factor πt := Pt/Pt−1 and real end-of-period moneybalances mt := Mt+1/(AtPt).

Properties:

(1) The productivity shock and the money supply shock equal their respective meansZt = Z ≡ 1 and µt = µ for all t .

(2) Inflation is constant: π = PtPt−1

for all t .

(3) All (scaled) variables are constant.

(4) All firms in the intermediary sector earn zero profits.



Loglinear approximation IThe New Keynesian Phillips Curve

Implications:

inflation is directly proportional to the growth rate of money supply µ− 1:

π =µ

a.

The optimal relative price of type A firms is a markup on the firm’s marginal cost g:

PA

P=

ε

ε− 1g,

Therefore, PA = P = PA, and all firms are equal Nj = N/J, and Kj = K/J,

Profits amount to

Dj =YJ− g

„YJ

+ F«.

Imposing Dj = 0 for all j and using Y/J = (AN/J)1−α(K/J)α − (F/J) yields

j :=Jt

At=

N1−αkα

εF.



Loglinear approximation IIThe New Keynesian Phillips Curve

Production is given by

y =ε− 1ε

N1−αkα.

Cost minimization implies

r = α(y/k).

implying

yk

=aη − β(1− δ)

αβ. (110a)

The former result allows to solve for the consumption-output ratio via theeconomy’s resource constraint (see (??)):

cy

=

„1 +

1− a− δ

y/k

« »1 + γ

„C

µ(M/P)

«κ–−1

.



Loglinear approximation IIIThe New Keynesian Phillips Curve

The stationary version of the Euler condition for money balances delivers:

βa1−η

µ= 1− κγ

„C

µ(M/P)

«1+κ

. (110b)

The labor supply is given by the foc of the household:

N1− N

=1− α

θ

„1 +

1− a− δ

y/k

«−1

h(c/x), (110c)

h(c/x) :=1 + γ(c/x)κ

1 + γ(1 + κ)(c/x)κ,

cx

:=PCµM

.

⇒ k/y , and therefore k/N and y/N are independent of the money growth rate µ!!⇒ velocity of money c/x increases with money growth⇒ steady-state working hours are a decreasing function of the money growth rate




Loglinearization: see Appendix 4.

− (η + ξ1)ct − ξ2Nt = λt − ξ1xt , (111)

(1− η)ct − ξ3Nt − wt = λt , (112)

πt = mt − xt + µt , (113)

ξ1 :=κγ(1 + κ)(c/x)κ

1 + γ(1 + κ)(c/x)κ,

cx

=C

µ(M/P),

ξ2 := θ(1− η)N

1− N,

ξ3 := [θ(1− η)− 1]N

1− N.




The loglinear cost-minimizing conditions deliver two further equations:

αNt + wt = αkt + gt + Zt , (114)

(α− 1)Nt + rt = (α− 1)kt + gt + Zt . (115)

and

yt − ϑ(1− α)Nt = ϑαkt + ϑZt . (116)




aEt kt+1 − (1− δ)kt − ξ4xt =yk

yt − ξ5ct , (117)

−Et λt+1 + λt = ξ6Et rt+1, (118)

Et λt+1 − λt − ξ7xt = −ξ7ct + Et πt+1, (119)

Etmt+1 − xt = 0, (120)

ξ4 := κγ(c/x)κ(c/k),

ξ5 := (1 + γ(1 + κ)(c/x)κ)(c/k),

ξ6 := 1− βa−η(1− δ),

ξ7 :=κγ(1 + κ)(c/x)1+κ

1− γκ(c/x)1+κ.

and the New Keynesian Phillips curve (109).




Control variables:

ut := [ct , Nt , yt , πt , wt , rt ]′.

State variables:

xt = [kt , mt ]′.

Exogenous variables:

zt = [Zt , µt ]′.

Costate:

λt = [λt , gt , xt ]′.

Why is m′ = mt+1 a costate and not a state?




Calibration

β, α, a, δ and θ as usual from the steady state conditions

Problem: we cannot identify Zt easily:

yt = ϑ(1− α)Nt + ϑαkt + ϑZt(1− ϑ)jt , ϑ =ε

ε− 1. (121)

In our benchmark real business cycle model:

yt = (1− α)Nt + αkt + Zt .

Since ϑ > 1 we overstate the size of Zt .

Money supply in Germany: we fit an AR(1) process for µt

We use the average velocity of money 0.84 to fix γ

We fix κ with the help of the interest elasticity of money demand -0.2:

d(M/P)/(M/P)

dq/q=

−1(1 + κ)π(1− δ + r)

.




Preference parameters:

β = 0.994, η = 2.0, N = 0.13

Production:

a = 1.005, δ = 0.011, σZ = 0.0072 ρZ = 0.90 , α = 0.27

Money supply:

µ = 1.0167, ρµ = 0.0, σµ = 0.0173

Transaction costs:

C/(M/P) = 0.84, κ = 4.0

Market structure:

ϕ = 0.25, ε = 6.0




0 5 10 15 200

0.5

1technology shock z

0 5 10 15 200

0.5

1k and m

0 5 10 15 20−0.5

0

0.5

1c and N

0 5 10 15 200

0.5

1

1.5y and w

0 5 10 15 20−0.5

0

0.5Pi

0 5 10 15 20−0.5

0

0.5

1r




0 5 10 15 200

0.5

1money shock mu

0 5 10 15 20−1

−0.5

0

0.5k and m

0 5 10 15 20−0.5

0

0.5

1c and N

0 5 10 15 200

0.2

0.4y and w

0 5 10 15 200

1

2Pi

0 5 10 15 20−0.5

0

0.5

1r




Observations and problems:1 expansionary effect of higher money growth2 spike-like behavior: no persistence of inflation!3 no liquidity effect4 profits decline following a monetary expansion




Reading Assignment

Heer, Maussner, Chapter 2.3, 2.4.1., 2.4.3., Appendix A.4.

Problems1 Simulate time series and compute second moments for the New Keynesian model.

Compute the impulse responses for the mark-up ratio.2 Solve problems 2.5 and 2.6 in Heer/Maussner, pp 150-153.


More Numerical Methods Function Approximation










Function approximation


local approximation, e.g. log-linear approximation

approximation over an interval [a, b]

Taylor’s theorem

TheoremLet f : [a, b] → R be a n + 1 times continuously differentiable function on (a, b), let x bea point in (a, b). Then

f (x + h) = f (x) + f (1)(x)h + f (2)(x)h2

2+ · · ·+ f (n)(x)

hn

n!

+ f (n+1)(ξ)hn+1

(n + 1)!, ξ ∈ (x , x + h).




Local approximation in Rn

e.g. quadratic approximation:

f (x + h) ≈ f (x) + [∇f (x)]′h +12

h′H(x)h, (122)

where the approximation error φ(h), with φ(0¯) = 0, has the property

limh→0

¯h 6=0

φ(h)

‖h‖2 = 0,

∇f (x) := [D1f (x),D2f (x), . . . ,Dnf (x)]′

and the Hesse matrix

H(x) :=

26664D1D1f (x) D1D2f (x) . . . D1Dnf (x)D2D1f (x) D2D2f (x) . . . D2Dnf (x)

......

. . ....

DnD1f (x) DnD2f (x) . . . DnDnf (x)

37775Burkhard Heer (University of Bolzano-Bozen, Italy) Computational Economics 4. September 2006 298 / 393



Linear interpolation

Suppose we want to approximate f (x) at a given point x with the property x1 < x < x2.Linear interpolation uses the point

f (x) := f (x1) +f (x2)− f (x1)

x2 − x1(x − x1). (123)

Thus, f is approximated by the line through (x1, f (x1)) and (x2, f (x2)).

MATLAB command for one-dimensional interpolation:y1=interp1(x,y,xi,’linear’)x – vector of nodesy – vector of function valuesxi – vector containing the points at which to interpolate



Function approximationLinear interpolation



Function approximationcubic spline

Splines: smooth function that is piecewise polynomial

Cubic spline

Suppose the data set {(xi , yi)|i = 0, . . . , n}.xi are called the nodes of the spline.On each subinterval [xi−1, xi ], the spline s(x) is cubic:

s(x) = ai + bix + cix2 + dix3

How do we determine the 4n parameters ai , bi , ci and ci?




1 2n interpolating conditions

yi = ai + bixi + cix2i + dix3

i

yi = ai+1 + bi+1xi + ci+1x2i + di+1x3

i

2 The first and second derivatives must agree at the interior nodes i = 1, . . . , n − 1providing 2n − 2 conditions:

bi + 2cixi + 3dix2i = bi+1 + 2ci+1xi + 3di+1x2

i

2ci + 6dixi = 2ci+1 + 6di+1xi

→ linear equations in the unknown parameters, 2 conditions are missing




1 Natural spline: imposes

s′(x0) = 0 = s′(xn)

2 Hermite spline: we know the true slopes

s′(x0) = y ′0, s′(xn) = y ′n

3 secant Hermit spline: we choose s(x) to make

s′(x0) =s(x1)− s(x0)

x1 − x0, s′(xn) =

s(xn)− s(xn−1)

xn − xn−1



Function approximationLinear versus cubic spline interpolation

An example

>> x=0.1:0.1:1;>> y=log(x);>> xi=0.1:0.01:1;>> yi=log(xi);>> yi1=interp1(x,y,xi,’linear’);>> yi2=interp1(x,y,xi,’cubic’);>> plot(xi,yi,’k-’, xi,yi1, ’k:’, xi,yi2, ’k’);>> error1=yi-yi1;>> error2=yi-yi2;



Function approximationOrthogonal polynomials

Orthogonal polynomials

Weierstrass Theorem: Any continuous real-valued function f defined on a boundedinterval [a, b] of the real line can be approximated to any degree of accuracy using apolynomial

Approximation of a function on f (x) on [a, b] by a family of polynomial:

f (x) =nX

i=0

αiϕi(x),

The monomials x i , i = 0, 1, . . . build a base B, for this space, i.e., every member ofC[a, b] can be represented by

∞Xi=0

αix i .

For this reason it is common to use a linear combination of the first p members of thisbase to approximate a continuous function f (x) ∈ C[a, b]:

f (x) ' α0 + α1x + α2x2 + · · ·+ αpxp.



Function approximationLinear versus cubic spline interpolation

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−2.5

−2

−1.5

−1

−0.5

0

0.5



Function approximationLinear versus cubic spline interpolation: error

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.01

0

0.01

0.02

0.03

0.04

0.05

0.06




Orthogonal polynomials

Weierstrass Theorem: Any continuous real-valued function f defined on a boundedinterval [a, b] of the real line can be approximated to any degree of accuracy using apolynomial

Approximation of a function on f (x) on [a, b] by a family of polynomial:

f (x) =nX

i=0

αiϕi(x),

The monomials x i , i = 0, 1, . . . build a base B, for this space, i.e., every member ofC[a, b] can be represented by

∞Xi=0

αix i .




Approximation by a linear combination of the first p members of this base:

f (x) ' α0 + α1x + α2x2 + · · ·+ αpxp.

Problem: multicollinearity of xp−1, xp

→ orthogonal polynomials




Assume we want to approximate f (x) ∈ C[a, b] by

f (x) :=nX

i=1

αiϕi(x),

where P := (ϕ0(x), ϕ1(x), . . . ) is some family of polynomials.

Let w(x) be a weight function on [a, b], i.e. a function that has a finite integralZ b

aw(x)dx <∞

and that is positive almost everywhere on [a, b]




Problem: Continuous least-squares approximation of f (x):

Choose the parameters αi , i = 0, 1, . . . n such that the weighted sum of squared errorsR(α, x) := f (x)− f (α, x) over all x ∈ [a, b] attains a minimum:

minα

Z b

aw(x)

"f (x)−

nXi=0

αiϕi(x)

#2

dx . (124)




The first order conditions for this problem are

0 = 2Z b

aw(x)

24f (x)−nX

j=0

αjϕj(x)

35ϕi(x)dx , i = 0, 1, . . . , n,

which may be rewritten asZ b

aw(x)f (x)ϕi(x)dx =

nXj=0

αj

Z b

aw(x)ϕj(x)ϕi(x)dx ,

i = 0, 1, . . . , n.

(125)

If the integral on the rhs of (125) vanishes for i 6= j and equals a constant ζj for i = j , itwill be easy to compute αi from

αi =1ζi

Z b

aw(x)f (x)ϕi(x)dx .

This motivates the following definition of orthogonal polynomials: A set of functions P iscalled orthogonal with respect to the weight function w(x) if and only if:Z b

aw(x)ϕi(x)ϕj(x)dx =

0 if i 6= j,ζi if i = j. (126)

If in addition ζi = 1 ∀i , the set of functions is said to be orthonormal.Burkhard Heer (University of Bolzano-Bozen, Italy) Computational Economics 4. September 2006 312 / 393



Chebyshev polynomials

The domain of Chebyshev polynomials is the interval [−1, 1], and the i-th member ofthis family is defined by

Ti(x) = cos(i arccos x). (127)

The weight function for whichR 1−1 w(x)Ti(x)Tj(x)dx = 0 for i 6= j is given by

w(x) :=1√

1− x2. (128)

In particular, the following holds:

Z 1

−1

Ti(x)Tj(x)√1− x2

dx =

8<:0 if i 6= j,π2 if i = j ≥ 1,π if i = j = 0.

(129)




Approximation of g(x) ∈ C[−1, 1]

g(α, x) :=12α0 +

nXi=1

αiTi(x) (130)

The coefficients of the continuous least squares approximation are given by

αi =2π

Z 1

−1

g(x)Ti(x)√1− x2

dx . (131)

→ α0 is multiplied by the factor 1/2 so that (131) also holds for i = 0 where the integral(129) for i = j = 0 is equal to ζ0 = π.




Approximation of f (z), f : [a, b] → R

Use the following transformation of z:

X (z) =2z

b − a− a + b

b − a, z ∈ [a, b] (132)

and the reverse transformation

Z (x) =(x + 1)(b − a)

2+ a, x ∈ [−1, 1]. (133)

With these transformations, we can define the function g(x) = f (Z (x)) on the interval[−1, 1] with approximation

f (z;α) =nX

i=0

αiTi (X (z)) . (134)

The coefficients of the continuous least squares approximation are then given by

αj =2π

Z b

a

f (z)Tj(X (z))p1− (X (z))2

dz. (135)




Properties of Chebyshev polynomials

We can compute the Chebyshev polynomials recursively:

Ti+1(x) = 2xTi(x)− Ti−1(x). (136)

The first four Chebyshev polynomials are

T0(x) = cos(0 · arccos x) = 1,

T1(x) = cos(1 · arccos x) = x ,

T2(x) = 2xT1(x)− T0(x) = 2x2 − 1,

T3(x) = 2xT2(x)− T1(x) = 4x3 − 3x .



Function approximationChebyshev polynomials




Evaluation of a n-th degree Chebyshev polynomial at x

Algorithm: Chebyshev Evaluation

Purpose: Evaluate a n-th degree Chebyshev polynomial at x

Step 1: Initialize: use the n + 1-vector y to store the values of Ti(x) for i = 0, 1, . . . , n.Put y [1] = 1 and y [2] = x .

Step 2: For i = 2, 3, . . . , n − 1 compute:

yi+1 = 2xyi − yi−1.

Step 3: ReturnPn

i=0 aiyi .

Matlab program: chebeval.m




Zeros of the Chebyshev polynomials

xk := cos„

2k − 12n

π

«, k = 1, 2, . . . , n. (137)

Useful property: Chebychev node polynomial interpolants are very nearly optimalpolynomial approximations

mXk=1

Ti(xk )Tj(xk ) =

8<:0 i 6= j,m/2 i = j 6= 0,m i = j = 0.

(138)




Computation of the Chebyshev coefficients a ∈ Rn

1 Computation of the integral (135)2 Interpolation at the n nodes: f and the approximation coincide at n nodes3 Regression on m > n nodes




Assume we want to approximate f (x) ∈ C[a, b] by a n-th degree Chebyshevpolynomial.

Let x = [x1, x2, . . . , xm] denote the m zeros of Tm(x).

The corresponding points in the interval [a, b] are zk := Z (xk ) with yk = f (zk ).

We choose α = [α0, α1, . . . , αn−1] to solve

minα

mXk=1

24yk −n−1Xj=0

αjTj(X (zk ))

352

.




The respective first order conditions are:

mXk=1

T0yk =n−1Xj=0

αj

mXk=1

T0Tj(xk ) = mα0,

mXk=1

yk T1(xk ) =n−1Xj=0

αj

mXk=1

T1(xk )Tj(xk ) = (m/2)α1,

... =...,

mXk=1

yk Tn−1(xk ) =n−1Xj=0

αj

mXk=1

Tn−1(xk )Tj(xk ) = (m/2)αn−1,

(139)




Algorithm: Chebyshev Regression

Purpose: Approximate f (z) ∈ C[a, b] withPn−1

j=0 αjTj(x).

Step 1: Choose the degree n − 1 of the approximating Chebyshev polynomial.Compute m ≥ n Chebyshev interpolation nodes xk from (137) and adjust thenodes to the interval [a, b] using (133).

Step 2: For k = 1, 2, . . . ,m, compute yk = f (zk ).

Step 3: Compute the Chebyshev coefficients: α0 = (1/m)Pm

k=1 yk . Fori = 1, 2, . . . , n − 1 the coefficients are given by:

αi =2m

mXk=1

yk Ti(xk ).

Matlab program: Chebcoef.m



Function approximationChebyshev approximation of y = ex , n = 2, m = 10

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 11

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

x

y

approximationexact solution




Chebyshev coefficients for approximation of f (x) = ex on [0, 1]

Coefficient n = 2, m = 10 n = 5, m = 10α0 1.753 1.753α1 0.850 0.850α2 0.105 0.105α3 0.0087α4 0.0005α5 0.000027

Matlab program: fig83.m




Theorem

If f ∈ Ck [−1, 1] has a Chebyshev representation f (x) =P∞

i=1 αiTi(x), then there is aconstant c such that

|αi | ≤cik , i ≥ 1. (140)



Function approximation IMultidimensional Approximation

Multidimensional Approximation

Choice of Bases

1 tensor product; e.g.ϕk (xi) ≡ xki . The set

T :=

(nY

i=1

ϕki (xi)|ki = 0, 1, . . . p

)

In the stochastic growth model with x = (K ,Z ), p = 2, T is given by

{K 0Z 0,K 0Z 1,K 0Z 2,K 1Z 0,K 1Z 1,K 1Z 2,K 2Z 0,K 2Z 1,K 2Z 2}

≡ {1,Z ,Z 2,K ,KZ ,KZ 2,K 2,K 2Z ,K 2Z 2}.



Function approximation IIMultidimensional Approximation

2 Complete set of polynomials:

Pnp =

((xk1

1 xk22 · · · xkn

n )

˛ nXi=1

ki = j, ki ≥ 0, j = 0, 1, 2, . . . , p

)

Thus, for x = (K ,Z ) and p = 2, the set is

P22 ={K 0Z 0| {z }

j=0

,K 1Z 0,K 0Z 1| {z }j=1

,K 2Z 0,K 1Z 1,K 0Z 2| {z }j=2

},

= {1,K ,Z ,K 2,KZ ,Z 2}.



Function approximationMultidimensional Approximation

Chebyshev approximation in two dimensions

Let f (z1, z2) be a function on [a, b]× [c, d ]:

n1Xj1=0

n2Xj2=0

αj1 j2 Tj1(X (z1))Tj2(X (z2)). (141)

Choose m1 ≥ n1 + 1 and m2 ≥ n2 + 1 and compute the zeros of the m1-dimensionaland m2-dimensional Chebyshev polynomial, adjusted to the interval [a, b] and [c, d ],respectively:

yk1k2 := f (Z (x1k1),Z (x2k2)),




The least squares criterion

minm1X

k1=1

m2Xk2=1

24yk1k2 −n1X

j1=0

n2Xj2=0

αj1 j2 Tj1(x1k1)Tj2(x2k2)

352

.

yields the following first-order condition with respect to αj1 j2 :

m1Xk1=1

m2Xk2=1

yk1k2 Tj1(x1k1)Tj2(x2k2)

=

m1Xk1=1

m2Xk2=1

n1Xi1=0

n2Xi2=0

αi1 i2 Ti1(x1k1)Ti2(x2k2)Tj1(x1k1)Tj2(x2k2)

=

n1Xi1=0

n2Xi2=0

αi1 i2

m1Xk1=1

Ti1(x1k1)Tj1(x1k1)| {z }= 0 if j1 6= i1= m1/2 if j1 = i1 6= 0= m1 if j1 = i1 = 0

m2Xk2=1

Ti2(x2k2)Tj2(x2k2)| {z }= 0 if j2 6= i2= m2/2 if j2 = i2 6= 0= m2 if j2 = i2 = 0

.




Therefore, we get the following estimator:

α00 =1

m1m2

m1Xk1=1

m2Xk2=1

yk1k2 ,

α0j2 =1

m1

2m2

m1Xk1=1

m2Xk2=1

yk1k2 Tj2(x2k2),

αj10 =2

m1

1m2

m1Xk1=1

m2Xk2=1

yk1k2 Tj1(x1k1),

αj1 j2 =2

m1

2m2

m1Xk1=1

m2Xk2=1

yk1k2 Tj1(x1k1)Tj2(x2k2).

(142)


More Numerical Methods Numerical Integration










Numerical integration

1. Newton-Cotes Formulas

Z b

af (x)dx ≈

nXi=1

ai f (xi). (143)

fix nodes xi , usually equispaced

approximate f (x) by piecewise polynomials: choice of ai

e.g. linear Lagrange polynomial:

P1(x) =x − ba− b

f (a) +x − ab − a

f (b) (144)

results in:Z b

af (x)dx ≈ b − a

2[f (a) + f (b)] . (145)




2. Gaussian formulas

choose both xi and ai optimally

Importance of orthogonal families of polynomials for numerical integration:

TheoremSuppose that {ϕi(x)}∞i=0 is an orthonormal family of polynomials with respect to theweight function w(x) on [a, b] with ϕi(x) = qix i + qi−1x i−1 + . . .+ q0. Let xi ,i = 1, . . . , n, be the n zeros of ϕn. Then a < x1 < · · · < xn < b, and if f ∈ C(2n)[a, b],then Z b

aw(x)f (x)dx =

nXi=1

ωi f (x i) +f 2n(ζ)

q2n(2n)!

, (146)

for some ζ ∈ [a, b], where

ωi = − qn+1/qn

ϕ′n(xi)ϕn+1(xi)> 0.



Numerical integration I

Accordingly, we can evaluate the integral of a polynomial of degree 2n − 1 exactly byapplying formula (146).

Gauss-Chebyshev quadrature

constant weights ωi

The Gauss-Chebyshev quadrature formula for a function f (x) on the interval[−1, 1] is defined by:Z 1

−1

f (x)√1− x2

dx =π

n

nXi=1

f (x i) +π

22n−1

f (2n)(ζ)

(2n)!

for some ζ ∈ [−1, 1], where the quadrature nodes x i are the zeros of theChebyshev polynomial Tn(x)



Numerical integration II

For integrals of the formR b

a f (z)dz, we use (133) to adjust [−1, 1] to [a, b]. Sincethe linear transformation (133) implies

dz =b − a

2dx ,

we can derive the following approximation:Z b

af (z)dz =

Z 1

−1f (Z (x))

b − a2

√1− x2

√1− x2

dx

≈ π(b − a)

2n

nXi=1

f„

(x i + 1)(b − a)

2+ a

« q1− x2

i ,

(147)

where the x i , again, are the Chebyshev zeros from (137).



Gauss Hermite quadrature

E(f (z)) ' π−1/2nX

i=1

ωi f“√

2σx + µ”. (148)

For n = 2, . . . , 4 the integration nodes and weights are provided in table 8.2,Heer/Maussner.

usually applied to evaluate conditional expectations where the stochastic variableis distributed normally

weight function is ω(x) = e−x2



Assume that z ∼ N(µ, σ). Then

E(f (z)) := (2πσ2)−1/2Z ∞

−∞f (z)e[(z−µ)2/2σ2]dz.

Since

x =z − µ√

2σ

has a standard normal distribution (i.e., E(x) = 0 and var(x) = 1), we get

E(f (z)) = π−1/2Z ∞

−∞f

“√2σx + µ

”e−x2

dx ,

where we used

dz =√

2σdx .




Multidimensional integration

use product rules: curse of dimensionality

monomial formulas

Computation of the integral of f (z1, . . . , zn) over the n-dimensional rectangle[a1, b1]× [a2, b2]× . . . [an, bn] the Gauss-Chebyshev quadrature (147) impliesZ b1

a1

. . .

Z bn

an

f (z1, . . . , zn)dz1 . . . dzn

'πn(b1 − a1) . . . (bn − an)

(2m)n

mXi1=1

· · ·mX

in=1

f (Z (x1), . . . ,Z (xn))

q1− x2

i1. . .

q1− x2

in .

(149)

In this formula Z (xi) denotes the linear transformation given in equation (133), and xi isthe i-th the zero of the Chebyshev polynomial of degree m.




⇒ Matlab provides only quadrature for Newton-Codes formula: Simpson rule(approximation with piecewise second-order polynomials)

quad(’f1’,0,1)

evaluate the function f1 over the interval [0, 1].

Matlab also provides routine for 2- and 3-dimensional Simpson rule integration.

See our Matlab code gc_int1.m and gh_int1.m for one-dimensional integration withGauss Chebyshev and Gauss Hermite, respectively.



Matlab Program

function v=gc_int1(f,d,n)temp=1:1:n;temp=temp’;x=cos(((2*temp-1)/(2*n))*pi);z=(x+1).*(d(2)-d(1))*0.5 + d(1);sum=0;



Matlab Program

for i=1:ny=feval(f,z(i));sum=sum+y*sqrt(1-x(i)^2);

endsum=pi*(d(2)-d(1))*sum;v=sum/(2*n);

→ evaluate f (x) = ex with Simpson rule and Gauss Chebyshev quadrature




Reading Assignment

Heer, Maussner, Chapter 8.2 and 8.3

Judd, Chapter 6.9, optional chapter 6

Miranda, Fackler, Chapter 5, optional: Chapter

Problems1 Write program codes chebeval2.m and chebcoef2.m that evaluate a complete

Chebyshev polynomial of order n1 × n2 and compute the coefficients in order toapproximate a two-dimensional function f (x) : [a, b]× [c, d ] → R. Test your codewith the help of the function f (x , y) = x ln(y) on the interval [0, 1]× [1, 5].

2 Write program codes for two-dimensional integration with the help ofGauss-Chebyshev and Gauss-Hermite quadrature.


Projection Methods A simple example










Projection methods

A simple example

Consider the ordinary differential equation

x(t) + x(t) = 0, x(0) = 1, (150)

with solution

x(t) = e−t .

Suppose we use the monomials (1, t , t2) to approximate the solution in the interval[0, 2]:

x(t) = 1 + γ1t + γ2t2. (151)

We have set γ0 = 1 to satisfy the boundary condition x(0) = 1.



Projection methods

Let us define the residual function:

R(γ, t) := γ1 + 2γ2t| {z }dx/dt

+ 1 + γ1t + γ2t2| {z }x(t)

(152)

Our aim: Choose (γ1, γ2) to make the residual small.

Least squares:

minγ1,γ2

Z 2

0R(γ, t)2dt . (153)

The first-order conditions for this problem are given by the following two equations:

0 =

Z 2

0R(γ, t)

∂R(γ, t)∂γ1

dt ,

0 =

Z 2

0R(γ, t)

∂R(γ, t)∂γ2

dt .



Projection methods

In our example, explicit analytical solution:

−4 = 823γ1 + 16γ2,

−623

= 16γ1 + 331

15γ2.



Projection methods



Projection methods

Galerkin method

multiplying the residual by the basis functions

0 =

Z 2

0R(γ, t)tdt ,

0 =

Z 2

0R(γ, t)t2dt .

(154)

Computing the integrals on the rhs of (154) gives a second set of linear equations inthe unknown parameters γ1 and γ2:

−2 = 423γ1 + 9

13γ2,

−223

= 623γ1 + 14

25γ2.



Projection methods

Collocation

Residual function is equal to zero at a given set of points.

Suppose we choose t1 = 1 and t2 = 2. This gives the linear system

−1 = 2γ1 + 3γ2,

−1 = 3γ1 + 8γ2.


Projection Methods The general framework










Projection methods I

The general framework

We want to approximate an unknown function f : X → Y , where X and Y aresubsets of Rn and Rm, respectively.

This function is implicitly defined by the functional equation F (f ) = 0, whereF : C1 → C2. C1 and C2 are given spaces of functions, e.g., the set of allcontinuously differentiable functions on [a, b].

Given a family of polynomials P := {ϕi}∞i=0, we approximate f by a finite linearcombination of the first p + 1 members of this family:

f (x) =

pXi=0

γiϕi(x), x ∈ X ⊂ Rn. (155)

The residual function is obtained by substituting f into the functional equation:

R(γ, x) := F (f (γ, x)), γ := (γ0, . . . , γp).



Projection methods II

Suppose there is a set of test functions {gi(x)}pi=0 and a weight function w(x).

Together with R they define an inner product given byZX

w(x)R(γ, x)gi(x)dx.

On a function space, this inner product induces a norm (i.e., a measure ofdistance) on this space and we choose the vector of parameters γ so thatZ

Xw(x)R(γ, x)gi(x)dx = 0, ∀i = 0, 1, . . . , n. (156)

Three possible choices of gi :1 The least squares solution puts gi ≡ ∂R/∂γi and w ≡ 1.2 The Galerkin solution chooses gi ≡ ϕi and w ≡ 1.3 The collocation method uses the Dirac delta function weight function,

w(x) =

0 if x 6= xi ,1 if x = xi ,

and puts gi ≡ 1.



Projection methods I

Algorithm: Projection Method

Prupose: Approximate the solution f : X → Y of a functional equation F (f ) = 0.

Step 1: Choose a bounded state space X ⊂ Rn and a family of functionsϕi(x) : X → Y , i = 0, 1, . . . .

Step 2: Choose a degree of approximation p and let

f (γ, x) =

pXi=0

γiϕi(x).

Step 3: Define the residual function:

R(γ, x) := F (f (γ, x)).



Projection methods II

Step 4: Choose a projection function gi , a weight function w and compute the innerproduct:

Gi :=

ZX

w(x)R(f¯γ, x)gi(x)dx, i = 0, . . . , n.

Find the value of γ that solves Gi = 0, or, in the case of least squaresprojection (gi = ∂R/∂γi and w ≡ 1), minimizeZ

XR(γ, x)2dx

with respect to γ.

Step 5: Verify the quality of the candidate solution γ. If necessary, return to step 2 andincrease the degree of approximation p or even return to step 1 and choose adifferent family of basis functions.



Projection methods IBuilding blocks

1 Approximating Function

prime candidate are orthogonal polynomials: Chebyshev functionsProblem of monomials: collinearity

2 Residual Function:

Consider the Euler equation of the deterministic growth model:

0 =u′(Ct)

u′(Ct+1)− βf ′(Kt+1),

Kt+1 = f (Kt)− Ct ,

Possible residual functions:

R1(γ,K ) :=u′[C(γ,K )]

u′[C(γ, f (K )− C(γ,K ))]

− βf ′[f (K )− C(γ,K )].

(157)

R2(γ,K ) := u′(C(γ,K ))

− βu′[C(γ, f (K )− C(γ,K ))]f ′[f (K )− C(γ,K )]



Projection methods IIBuilding blocks

3 Projection:I Least squares: often fails in multi-dimensional state spaceI Galerkin: time-consuming in multi-dimensional state space due to the integration stepI Collocation: often very fast, problem to pick the nodes if polynomial is not the tensor

product, but only a complete polynomial

Problem: finding an initial value for γ.

Possible solution: coefficients from log-linear solution4 Accuracy of solution

I study the behavior of the residualI study the behavior of the γi : do they drop off nicely?I den Haan-Marcet statisticI compare a simple case to solution with another method, e.g. from value function

iteration or log-linear


Projection Methods Application: Deterministic Ramsey model










Projection methodsDeterministic Ramsey model

Application: deterministic Ramsey model

maxC0,C1,...

U0 =∞Xt=0

βt C1−ηt − 11− η

, β ∈ (0, 1) , η > 0,

s.t. (158)

Kt+1 + Ct ≤ K αt + (1− δ)Kt , α ∈ (0, 1),

0 ≤ Ct ,0 ≤ Kt+1,

9=; t = 0, 1, . . . ,

K0 given,

The Euler equation of this problem is given by:»Ct+1

Ct

–−η

β“

1− δ + αK α−1t+1

”− 1 = 0. (159)

From this equation we derive the steady state value of the capital stock:

K ∗ =

»αβ

1− β(1− δ)

–1/(1−α)

.



Projection methods IDeterministic Ramsey model

Implementation: Ramseyproj.m

Choice of the state space [K ,K ] that embeds the steady state value K ∗:

→ should contain the ergodic set→ we choose X := [0.5K ∗, 1.5K ∗]→ try others

→ sometimes it may be helpful to choose two sets X ⊂ X ′. You approximate thepolicy function over the set X ′ and compute the residual over X .

We approximate the consumption function C(K ) with the help of a Chebyshevpolynomial of order p,

C(γ,K ) :=

pXj=0

γjTj(K (K )),

where K (K ) maps the capital stock K into the interval [−1, 1]

Initial guess of γ:1 Solution from Log-Linearization2 Solution from Value Function Iteration and approximation with Chebyshev regression →

our implementation



Projection methods IIDeterministic Ramsey model

3 Genetic search: see Heer/Maussner, ramsey2b.g4 Homotopy: Start from known initial solution, e.g. with δ = 0 and increase δ step by

step and use the new solution for γ in the next iteration

Computation of the residual function:

R2(γ,K ) := u′(C(γ,K ))

− βu′[C(γ, f (K )− C(γ,K ))]f ′[f (K )− C(γ,K )]

1 Given γ and K0 ∈ [K , K ] we compute C0 := C(γ, K0).

→ C0 > 0?2 From C0 we compute the future stock of capital K1 from the resource constraint:

K1 = K α + (1− δ)K − C0.

→ K1 ∈ X?3 Given K1 we get C1 := C(γ, K1).→ C1 > 0?



Projection methods IIIDeterministic Ramsey model

4 In this final step we compute the residual from

R(γ, K0) :=

"C1

C0

#−η “1− δ + αK α−1

1

”− 1.

Galerkin projection: We use Gauss Chebyshev quadrature with 100 nodes andsolve the system of p + 1 non-linear equations:

π(K − K )

2L

LXl=1

R(γ,K (Kl))Ti(Kl)

q1− K 2

l ,

i = 0, 1, . . . , p.

(160)

Chebyshev collocation: we determine the coefficients γ0, . . . , γp from theconditions:

R(γ,K (Ki)) = 0, ∀i = 0, 2, . . . , p, (161)

where, again, Ki is the i-th zero of the Chebyshev polynomial of order p + 1.

We solve the non-linear equation problem with Broyden’s algorithm.



Projection methods

Solution from ramseyproj.m

Chebyshev coefficient for Collocation and Galerkin:

aCol =

[email protected]

−0.03470.0052

−0.0008

1CCCCA aGal =

[email protected]

−0.03460.0051

−0.0009

1CCCCA

Running time: 0.064 and 0.471 seconds



Projection methodsCollocation solution of Ramsey model

0 10 20 30 40 50 60 70 80 90

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

3

3.2

capital stock

c

value functioncollocation



Projection methodsCollocation solution of Ramsey model

0 10 20 30 40 50 60 70 80 90

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

3

3.2

capital stock

c

collocationGalerkin



Projection methodsResidual: Collocation solution

20 30 40 50 60 70 80−3

−2

−1

0

1

2

3

4

5x 10−5

capital stock

resi

dual

collocationGalerkin




Reading Assignment

Heer, Maussner, Chapter 4.1.-4.3.2.

Judd, Chapter 11, optional

Problems1 Compute the stochastic growth model with non-negative investment as in section

4.3.2 in Heer/Maussner. Write a Matlab program. Compute the second momentsof the model variables (after HP filtering) for σ ∈ {0.0072, 0.05}. Simulate themodel and compute the average error of the Euler equation. In order to computethe expectation for the residual use Gauss Hermite quadrature with 10 points.


Outlook Perturbation methods










Perturbation methods


local approximation of f (x)

higher-order polynomial

higher accuracy than loglinear or linear approximationproblems:

1 computation of the higher-order derivatives: numerical derivatives are often not exactenough

→ problem in Gauss or Fortran, Matlab is able to compute analytical derivatives→ you may want to download the code by Schmitt-Grohe and Uribe

2 solution boils down to solving the log-linearization and additional non-linear equations:easy computational problem


Outlook Parameterized expectations











Parameterized expectations: An example

Kt+1 = ZtK αt + (1− δ)Kt − Ct , (162a)

C−ηt = βEt

ˆC−η

t+1

“1− δ + αZt+1K α−1

t+1

” ˜. (162b)

Remember, Ct denotes consumption, Kt the stock of capital, and Zt is the size of theshock to total factor productivity that evolves over time according to

Zt = Z %t−1eεt , εt ∼ N(0, σ2). (162c)

→ solution is time-invariant function Kt+1 = g(Kt ,Zt).




Solution for consumption:

C(Kt ,Zt) := ZtK αt + (1− δ)Kt − g(Kt ,Zt)

and Ct+1 = C(g(Kt ,Zt),Zt+1).

Let φ(Kt ,Zt , εt+1) be the expression inside the expectations operator Et in a function:

φ(Kt ,Zt , εt+1) :=C(g(Kt ,Zt),Z %t eεt+1)−η

×“

1− δ + αZ %t eεt+1 g(Kt ,Zt)

α−1”.

Since the innovations to are normal variables with density π(ε), we get E via integrationof φ(·)π(ε):

E(Kt ,Zt) :=

Z ∞

−∞φ(Kt ,Zt , ε)π(ε)dε.




Approximation of E

Approximate E by a simple parameterized function, e.g.

ψ(γ1, γ2, γ3,Kt ,Zt) = γ1K γ2t Z γ3

t

Make an initial guess of the parameter γ.

Given K0 and Z0, we can compute the rhs of equation (162b) and solve the systemof two equations (162) for K1 and C0.

With K1 at hand, Z1 derived from Z0, and a draw from the N(0, σ2)-distribution, wecan use E(K1,Z1) and (162) again to solve for (K2,C1).

Simulate a time series: Repeating these steps over and over we can trace out anentire time path for the variables of our model economy.

Minimize the mean squared error in forecasting E with respect to the parameter γMain difference to projection methods:

1 Simulates a time series with the model2 Uses a non-linear regression step to update γ3 The residual is weighted by its probability


Outlook The curse of dimensionality










Curse of dimensionality

Curse of dimensionality

If the dimension ns of the state space is getting large, it is difficult to find a solution

Value function iteration: difficult for ns > 1

Log-linearization: no problem, but may be inaccurateProjection methods:

I ns � 15 is troublesomeI Galerkin: integration step is too time-consuming even with complete polynomialsI Chebyshev collocation: problem to fix collocation nodes for complete polynomials

(Smolyak’s Algorithm)

Projection methods: preliminary research results indicate that they may perform ina rather satisfactory manner for high dimensions


Outlook Heterogeneous agents










Heterogeneous agents I

Heterogeneous agents

Exogenous source of heterogeneity:1 innate abilities2 education3 individual productivities4 wage5 age

Endogenous distribution1 wealth2 income

Question: Does the distribution of income and wealth effect the endogenousaggregate variables like savings, total income and employment?

computation of the endogenous distribution difficult and time-consuming

computation of the dynamics of the distribution: very difficult:→ see Chapters 5-7 in Heer/Maussner


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