Chapter 2 pertubation

Perturbation Methods

Jesús Fernández-Villaverde

University of Pennsylvania

July 10, 2011

Jesús Fernández-Villaverde (PENN) Perturbation Methods July 10, 2011 1 / 91

Introduction

Introduction

Remember that we want to solve a functional equation of the form:

H (d ) = 0

for an unknown decision rule d.

Perturbation solves the problem by specifying:

n
d n (x, θ ) = ∑ θ i (x x0 )i
i =0

We use implicit-function theorems to …nd coe¢ cients θ i ’s.

Inherently local approximation. However, often good global properties.

Introduction

Motivation
Many complicated mathematical problems have:

1 either a particular case

2 or a related problem.

that is easy to solve.

Often, we can use the solution of the simpler problem as a building
block of the general solution.

Very successful in physics.

Sometimes perturbation is known as asymptotic methods.

Introduction

The World Simplest Perturbation
p
What is 26?

Without your Iphone calculator, it is a boring arithmetic calculation.

But note that:
p q p
26 = 25 (1 + 0.04) = 5 1.04 5 1.02 = 5.1

Exact solution is 5.099.
p
We have solved a much simpler problem ( 25) and added a small
coe¢ cient to it.

More in general
q p
p
y= x 2 (1 + ε ) = x 1 + ε

where x is an integer and ε the perturbation parameter.

Introduction

Applications to Economics

Judd and Guu (1993) showed how to apply it to economic problems.

Recently, perturbation methods have been gaining much popularity.

In particular, second- and third-order approximations are easy to
compute and notably improve accuracy.

A …rst-order perturbation theory and linearization deliver the same
output.

Hence, we can use much of what we already know about linearization.


Introduction

Regular versus Singular Perturbations

Regular perturbation: a small change in the problem induces a small
change in the solution.

Singular perturbation: a small change in the problem induces a large
change in the solution.

Example: excess demand function.

Most problems in economics involve regular perturbations.

Sometimes, however, we can have singularities. Example: introducing
a new asset in an incomplete markets model.


Introduction

References
General:
1 A First Look at Perturbation Theory by James G. Simmonds and
James E. Mann Jr.
2 Advanced Mathematical Methods for Scientists and Engineers:
Asymptotic Methods and Perturbation Theory by Carl M. Bender,
Steven A. Orszag.

Economics:
1 Perturbation Methods for General Dynamic Stochastic Models” by
Hehui Jin and Kenneth Judd.
2 Perturbation Methods with Nonlinear Changes of Variables” by
Kenneth Judd.
3 A gentle introduction: “Solving Dynamic General Equilibrium Models
Using a Second-Order Approximation to the Policy Function” by
Martín Uribe and Stephanie Schmitt-Grohe.

A Baby Example

A Baby Example: A Basic RBC

Model:
∞
max E0 ∑ βt log ct
t =0

s.t. ct + kt +1 = e zt ktα + (1 δ) kt , 8 t > 0
zt = ρzt 1 + σεt , εt N (0, 1)

Equilibrium conditions:
1 1
= βEt 1 + αe zt +1 ktα+1 δ
1
ct ct + 1
ct + kt +1 = e zt ktα + (1 δ) kt
zt = ρzt 1 + σεt


A Baby Example

Computing a Solution
The previous problem does not have a known “paper and pencil”
solution except when (unrealistically) δ = 1.
Then, income and substitution e¤ect from a technology shock cancel
each other (labor constant and consumption is a …xed fraction of
income).
Equilibrium conditions with δ = 1:
1
1 αe zt +1 ktα+1
= βEt
ct ct + 1
ct + kt +1 = e zt ktα
zt = ρzt 1 + σεt
By “guess and verify”:
ct = ( 1 αβ) e zt ktα
kt +1 = αβe zt ktα

A Baby Example

Another Way to Solve the Problem

Now let us suppose that you missed the lecture when “guess and
verify” was explained.

You need to compute the RBC.

What you are searching for? A decision rule for consumption:

ct = c (kt , zt )

and another one for capital:

kt +1 = k (kt , zt )

Note that our d is just the stack of c (kt , zt ) and k (kt , zt ).

A Baby Example

Equilibrium Conditions

We substitute in the equilibrium conditions the budget constraint and
the law of motion for technology.

And we write the decision rules explicitly as function of the states.

Then:

1 αe ρzt +σεt +1 k (kt , zt )α 1
= βEt
c (kt , zt ) c (k (kt , zt ) , ρzt + σεt +1 )
c (kt , zt ) + k (kt , zt ) = e zt ktα

System of functional equations.


A Baby Example

Main Idea

Transform the problem rewriting it in terms of a small perturbation
parameter.

Solve the new problem for a particular choice of the perturbation
parameter.

This step is usually ambiguous since there are di¤erent ways to do so.

Use the previous solution to approximate the solution of original the
problem.


A Baby Example

A Perturbation Approach

Hence, we want to transform the problem.

Which perturbation parameter? Standard deviation σ.

Why σ? Discrete versus continuous time.

Set σ = 0 )deterministic model, zt = 0 and e zt = 1.

We know how to solve the deterministic steady state.


A Baby Example

A Parametrized Decision Rule

We search for decision rule:

ct = c (kt , zt ; σ)

and
kt +1 = k (kt , zt ; σ)

Note new parameter σ.

We are building a local approximation around σ = 0.


A Baby Example

Taylor’ Theorem
s

Equilibrium conditions:

!
1 αe ρzt +σεt +1 k (kt , zt ; σ)α 1
Et β =0
c (kt , zt ; σ) c (k (kt , zt ; σ) , ρzt + σεt +1 ; σ)
c (kt , zt ; σ) + k (kt , zt ; σ) e zt ktα = 0

We will take derivatives with respect to kt , zt , and σ.

Apply Taylor’ theorem to build solution around deterministic steady
s
state. How?


A Baby Example

Asymptotic Expansion I

ct = c (kt , zt ; σ)jk ,0,0 = c (k, 0; 0)
+ck (k, 0; 0) (kt k ) + cz (k, 0; 0) zt + cσ (k, 0; 0) σ
1 1
+ ckk (k, 0; 0) (kt k )2 + ckz (k, 0; 0) (kt k ) zt
2 2
1 1
+ ck σ (k, 0; 0) (kt k ) σ + czk (k, 0; 0) zt (kt k )
2 2
1 2 1
+ czz (k, 0; 0) zt + cz σ (k, 0; 0) zt σ
2 2
1 1
+ cσk (k, 0; 0) σ (kt k ) + cσz (k, 0; 0) σzt
2 2
1 2
+ cσ2 (k, 0; 0) σ + ...
2


A Baby Example

Asymptotic Expansion II

kt +1 = k (kt , zt ; σ)jk ,0,0 = k (k, 0; 0)
+kk (k, 0; 0) (kt k ) + kz (k, 0; 0) zt + kσ (k, 0; 0) σ
1 1
+ kkk (k, 0; 0) (kt k )2 + kkz (k, 0; 0) (kt k ) zt
2 2
1 1
+ kk σ (k, 0; 0) (kt k ) σ + kzk (k, 0; 0) zt (kt k )
2 2
1 2 1
+ kzz (k, 0; 0) zt + kz σ (k, 0; 0) zt σ
2 2
1 1
+ kσk (k, 0; 0) σ (kt k ) + kσz (k, 0; 0) σzt
2 2
1 2
+ kσ2 (k, 0; 0) σ + ...
2


A Baby Example

Comment on Notation

From now on, to save on notation, I will write
" #
1 αe ρzt +σεt +1 k (k ,zt ;σ)α 1
β c (k (kt ,zt ;σ),ρztt+σεt +1 ;σ) 0
F (kt , zt ; σ) = Et c (kt ,zt ;σ ) =
c (kt , zt ; σ) + k (kt , zt ; σ) e zt ktα 0

Note that:

F (kt , zt ; σ) = H (ct , ct +1 , kt , kt +1 , zt ; σ)
= H (c (kt , zt ; σ) , c (k (kt , zt ; σ) , zt +1 ; σ) , kt , k (kt , zt ; σ) , zt ; σ)

I will use Hi to represent the partial derivative of H with respect to
the i component and drop the evaluation at the steady state of the
functions when we do not need it.


A Baby Example

Zeroth-Order Approximation

First, we evaluate σ = 0:

F (kt , 0; 0) = 0

Steady state:
1 αk α 1
=β
c c
or
1
1 = αβk α
Then: α 1
c = c (k, 0; 0) = (αβ) 1 α (αβ) 1 α

1
k = k (k, 0; 0) = (αβ) 1 α


A Baby Example

First-Order Approximation

We take derivatives of F (kt , zt ; σ) around k, 0, and 0.

With respect to kt :
Fk (k, 0; 0) = 0

With respect to zt :
Fz (k, 0; 0) = 0

With respect to σ:
Fσ (k, 0; 0) = 0


A Baby Example

Solving the System I

Remember that:

F (kt , zt ; σ)
= H (c (kt , zt ; σ) , c (k (kt , zt ; σ) , zt +1 ; σ) , kt , k (kt , zt ; σ) , zt ; σ) = 0

Because F (kt , zt ; σ) must be equal to zero for any possible values of
kt , zt , and σ, the derivatives of any order of F must also be zero.

Then:

Fk (k, 0; 0) = H1 ck + H2 ck kk + H3 + H4 kk = 0
Fz (k, 0; 0) = H1 cz + H2 (ck kz + ck ρ) + H4 kz + H5 = 0
Fσ (k, 0; 0) = H1 cσ + H2 (ck kσ + cσ ) + H4 kσ + H6 = 0


A Baby Example

Solving the System II
A quadratic system:
Fk (k, 0; 0) = H1 ck + H2 ck kk + H3 + H4 kk = 0
Fz (k, 0; 0) = H1 cz + H2 (ck kz + ck ρ) + H4 kz + H5 = 0
of 4 equations on 4 unknowns: ck , cz , kk , and kz .
Procedures to solve quadratic systems:
1 Blanchard and Kahn (1980).
2 Uhlig (1999).
3 Sims (2000).
4 Klein (2000).

All of them equivalent.
Why quadratic? Stable and unstable manifold.

A Baby Example

Solving the System III
Also, note that:
Fσ (k, 0; 0) = H1 cσ + H2 (ck kσ + cσ ) + H4 kσ + H6 = 0
is a linear, and homogeneous system in cσ and kσ .
Hence:
cσ = kσ = 0

This means the system is certainty equivalent.
Interpretation)no precautionary behavior.
Di¤erence between risk-aversion and precautionary behavior. Leland
(1968), Kimball (1990).
Risk-aversion depends on the second derivative (concave utility).
Precautionary behavior depends on the third derivative (convex
marginal utility).

A Baby Example

Comparison with LQ and Linearization
After Kydland and Prescott (1982) a popular method to solve
economic models has been to …nd a LQ approximation of the
objective function of the agents.

Close relative: linearization of equilibrium conditions.

When properly implemented linearization, LQ, and …rst-order
perturbation are equivalent.

Advantages of perturbation:

1 Theorems.

2 Higher-order terms.

A Baby Example

Some Further Comments

Note how we have used a version of the implicit-function theorem.

Important tool in economics.

Also, we are using the Taylor theorem to approximate the policy
function.

Alternatives?


A Baby Example

Second-Order Approximation
We take second-order derivatives of F (kt , zt ; σ) around k, 0, and 0:

Fkk (k, 0; 0) = 0
Fkz (k, 0; 0) = 0
Fk σ (k, 0; 0) = 0
Fzz (k, 0; 0) = 0
Fz σ (k, 0; 0) = 0
Fσσ (k, 0; 0) = 0
Remember Young’ theorem!
s
We substitute the coe¢ cients that we already know.
A linear system of 12 equations on 12 unknowns. Why linear?
Cross-terms kσ and zσ are zero.
Conjecture on all the terms with odd powers of σ.

A Baby Example

Correction for Risk

We have a term in σ2 .

Captures precautionary behavior.

We do not have certainty equivalence any more!

Important advantage of second-order approximation.

Changes ergodic distribution of states.


A Baby Example

Higher-Order Terms

We can continue the iteration for as long as we want.

Great advantage of procedure: it is recursive!

Often, a few iterations will be enough.

The level of accuracy depends on the goal of the exercise:

1 Welfare analysis: Kim and Kim (2001).

2 Empirical strategies: Fernández-Villaverde, Rubio-Ramírez, and Santos
(2006).


A Numerical Example

A Numerical Example

Parameter β α ρ σ
Value 0.99 0.33 0.95 0.01

Steady State: c = 0.388069 k = 0.1883
First-order terms:
ck (k, 0; 0) = 0.680101 kk (k, 0; 0) = 0.33
cz (k, 0; 0) = 0.388069 kz (k, 0; 0) = 0.1883
Second-order terms:
ckk (k, 0; 0) = 2.41990 kkk (k, 0; 0) = 1.1742
ckz (k, 0; 0) = 0.680099 kkz (k, 0; 0) = 0.33
czz (k, 0; 0) = 0.388064 kzz (k, 0; 0) = 0.1883
cσ2 (k, 0; 0) ' 0 kσ2 (k, 0; 0) ' 0
cσ (k, 0; 0) = kσ (k, 0; 0) = ck σ (k, 0; 0) = kk σ (k, 0; 0) =
cz σ (k, 0; 0) = kz σ (k, 0; 0) = 0.

A Numerical Example

Comparison

ct = 0.6733e zt kt0.33
ct ' 0.388069 + 0.680101 (kt k ) + 0.388069zt
2.41990 0.388064 2
(kt k )2 + 0.680099 (kt k ) zt + zt
2 2
and:

kt +1 = 0.3267e zt kt0.33
kt +1 ' 0.1883 + 0.33 (kt k ) + 0.1883zt
1.1742 0.1883 2
(kt k )2 + 0.33 (kt k ) zt + zt
2 2


A Numerical Example


A Numerical Example

A Computer

In practice you do all this approximations with a computer:

1 First-, second-, and third-order: Matlab and Dynare.

2 Higher-order: Mathematica, Dynare++, Fortran code by Jinn and
Judd.

Burden: analytical derivatives.

Why are numerical derivatives a bad idea?

Alternatives: automatic di¤erentiation?


A Numerical Example

Local Properties of the Solution
Perturbation is a local method.
It approximates the solution around the deterministic steady state of
the problem.
It is valid within a radius of convergence.
What is the radius of convergence of a power series around x? An
r 2 R∞ such that 8x 0 , jx 0 z j < r , the power series of x 0 will
+
converge.

A Remarkable Result from Complex Analysis
The radius of convergence is always equal to the distance from the center
to the nearest point where the policy function has a (non-removable)
singularity. If no such point exists then the radius of convergence is in…nite.

Singularity here refers to poles, fractional powers, and other branch
powers or discontinuities of the functional or its derivatives.

A Numerical Example

Remarks

Intuition of the theorem: holomorphic functions are analytic.

Distance is in the complex plane.

Often, we can check numerically that perturbations have good non
local behavior.

However: problem with boundaries.


A Numerical Example

Non Local Accuracy Test
Proposed by Judd (1992) and Judd and Guu (1997).

Given the Euler equation:
1 αe zt +1 k i (kt , zt )α 1
i (k , z )
= Et
c t t c i (k i (kt , zt ), zt +1 )
we can de…ne:
αe zt +1 k i (kt , zt )α 1
EE i (kt , zt ) 1 c i (kt , zt ) Et
c i (k i (kt , zt ), zt +1 )

Units of reporting.

Interpretation.

A Numerical Example


The General Case

The General Case

Most of previous argument can be easily generalized.

The set of equilibrium conditions of many DSGE models can be
written as (note recursive notation)

Et H(y , y 0 , x, x 0 ) = 0,

where yt is a ny 1 vector of controls and xt is a nx 1 vector of
states.

De…ne n = nx + ny .

Then H maps R ny R ny R nx R nx into R n .

The General Case

Partitioning the State Vector

The state vector xt can be partitioned as x = [x1 ; x2 ]t .

x1 is a (nx n ) 1 vector of endogenous state variables.

x2 is a n 1 vector of exogenous state variables.

Why do we want to partition the state vector?


The General Case

Exogenous Stochastic Process

0
x2 = Λx2 + ση 0

Process with 3 parts:
1 The deterministic component Λx2 :
1 Λ is a n n matrix, with all eigenvalues with modulus less than one.
2 More general: x2 = Γ(x2 ) + ση 0 , where Γ is a non-linear function
0
satisfying that all eigenvalues of its …rst derivative evaluated at the
non-stochastic steady state lie within the unit circle.
2 The scaled innovation η 0 where:
1 η is a known n n matrix.
2 is a n 1 i.i.d innovation with bounded support, zero mean, and
variance/covariance matrix I .
3 The perturbation parameter σ.
We can accommodate very general structures of x2 through changes
in the de…nition of the state space: i.e. stochastic volatility.
Note we do not impose Gaussianity.

The General Case

The Perturbation Parameter

The scalar σ 0 is the perturbation parameter.

If we set σ = 0 we have a deterministic model.

Important: there is only ONE perturbation parameter. The matrix η
takes account of relative sizes of di¤erent shocks.

Why bounded support? Samuelson (1970) and Jin and Judd (2002).


The General Case

Solution of the Model

The solution to the model is of the form:

y = g (x; σ)
0 0
x = h(x; σ) + ση

where g maps R nx R + into R ny and h maps R nx R + into R nx .

The matrix η is of order nx n and is given by:

∅
η=
η


The General Case

Perturbation

We wish to …nd a perturbation approximation of the functions g and
h around the non-stochastic steady state, xt = x and σ = 0.
¯

We de…ne the non-stochastic steady state as vectors (x, y ) such that:
¯ ¯

H(y , y , x, x ) = 0.
¯ ¯ ¯ ¯

Note that y = g (x; 0) and x = h(x; 0). This is because, if σ = 0,
¯ ¯ ¯ ¯
then Et H = H.


The General Case

Plugging-in the Proposed Solution
Substituting the proposed solution, we de…ne:

F (x; σ) Et H(g (x; σ), g (h(x; σ) + ησ 0 , σ), x, h(x; σ) + ησ 0 ) = 0

Since F (x; σ) = 0 for any values of x and σ, the derivatives of any
order of F must also be equal to zero.

Formally:

Fx k σj (x; σ) = 0 8x, σ, j, k,

where Fx k σj (x, σ) denotes the derivative of F with respect to x taken
k times and with respect to σ taken j times.

The General Case

First-Order Approximation
We look for approximations to g and h around (x, σ) = (x, 0):
¯
g (x; σ) = g (x; 0) + gx (x; 0)(x
¯ ¯ x ) + gσ (x; 0)σ
¯ ¯
h(x; σ) = h(x; 0) + hx (x; 0)(x
¯ ¯ x ) + hσ (x; 0)σ
¯ ¯
As explained earlier,
g (x; 0) = y
¯ ¯
and
h(x; 0) = x.
¯ ¯
The four unknown coe¢ cients of the …rst-order approximation to g
and h are found by using:
Fx (x; 0) = 0
¯
and
Fσ (x; 0) = 0
¯
Before doing so, I need to introduce the tensor notation.

The General Case

Tensors
General trick from physics.
An nth -rank tensor in a m-dimensional space is an operator that has n
indices and mn components and obeys certain transformation rules.
[Hy ]iα is the (i, α) element of the derivative of H with respect to y :
1 The derivative of H with respect to y is an n ny matrix.
2 Thus, [Hy ]iα is the element of this matrix located at the intersection of
the i-th row and α-th column.
n ∂ H i ∂g α ∂h β
Thus, [Hy ]iα [gx ]α [hx ]j = ∑α=1 ∑nx 1
β y
∂y α ∂x β ∂x j .
3
β β=

[Hy 0 y 0 ]iαγ :
1 Hy 0 y 0 is a three dimensional array with n rows, ny columns, and ny
pages.
2 Then [Hy 0 y 0 ]iαγ denotes the element of Hy 0 y 0 located at the
intersection of row i, column α and page γ.

The General Case


gx and hx can be found as the solution to the system:

[Fx (x; 0)]ij = [Hy 0 ]iα [gx ]α [hx ]j + [Hy ]iα [gx ]jα + [Hx 0 ]iβ [hx ]j + [Hx ]ij =
β β
¯ β
i = 1, . . . , n; j, β = 1, . . . , nx ; α = 1, . . . , ny

Note that the derivatives of H evaluated at (y , y 0 , x, x 0 ) = (y , y , x, x )
¯ ¯ ¯ ¯
are known.

Then, we have a system of n nx quadratic equations in the n nx
unknowns given by the elements of gx and hx .

We can solve with a standard quadratic matrix equation solver.


The General Case

gσ and hσ are identi…ed as the solution to the following n equations:
[Fσ (x; 0)]i =
¯
Et f[Hy 0 ]iα [gx ]α [hσ ] β + [Hy 0 ]iα [gx ]α [η ]φ [ 0 ]φ + [Hy 0 ]iα [gσ ]α
β
β β

+[Hy ]iα [gσ ]α + [Hx 0 ]iβ [hσ ] β + [Hx 0 ]iβ [η ]φ [ 0 ]φ g
β

i = 1, . . . , n; α = 1, . . . , ny ; β = 1, . . . , nx ; φ = 1, . . . , n .
Then:
[Fσ (x; 0)]i = [Hy 0 ]iα [gx ]α [hσ ] β + [Hy 0 ]iα [gσ ]α + [Hy ]iα [gσ ]α + [fx 0 ]iβ [hσ ] β = 0;
¯ β
i = 1, . . . , n; α = 1, . . . , ny ; β = 1, . . . , nx ; φ = 1, . . . , n .
Certainty equivalence: this equation is linear and homogeneous in gσ
and hσ . Thus, if a unique solution exists, it must satisfy:
hσ 6 = 0
gσ = 0

The General Case

Second-Order Approximation I

The second-order approximations to g around (x; σ) = (x; 0) is
¯

[g (x; σ)]i = [g (x; 0)]i + [gx (x; 0)]ia [(x x )]a + [gσ (x; 0)]i [σ]
¯ ¯ ¯ ¯
1
+ [gxx (x; 0)]iab [(x x )]a [(x x )]b
¯ ¯ ¯
2
1
+ [gx σ (x; 0)]ia [(x x )]a [σ]
¯ ¯
2
1
+ [gσx (x; 0)]ia [(x x )]a [σ]
¯ ¯
2
1
+ [gσσ (x; 0)]i [σ][σ]
¯
2
where i = 1, . . . , ny , a, b = 1, . . . , nx , and j = 1, . . . , nx .


The General Case

Second-Order Approximation II

The second-order approximations to h around (x; σ) = (x; 0) is
¯

[h(x; σ)]j = [h(x; 0)]j + [hx (x; 0)]ja [(x x )]a + [hσ (x; 0)]j [σ]
¯ ¯ ¯ ¯
1
+ [hxx (x; 0)]jab [(x x )]a [(x x )]b
¯ ¯ ¯
2
1
+ [hx σ (x; 0)]ja [(x x )]a [σ]
¯ ¯
2
1
+ [hσx (x; 0)]ja [(x x )]a [σ]
¯ ¯
2
1
+ [hσσ (x; 0)]j [σ][σ],
¯
2
where i = 1, . . . , ny , a, b = 1, . . . , nx , and j = 1, . . . , nx .


The General Case

Second-Order Approximation III

The unknowns of these expansions are [gxx ]iab , [gx σ ]ia , [gσx ]ia , [gσσ ]i ,
[hxx ]jab , [hx σ ]ja , [hσx ]ja , [hσσ ]j .

These coe¢ cients can be identi…ed by taking the derivative of F (x; σ)
with respect to x and σ twice and evaluating them at (x; σ) = (x; 0).
¯

By the arguments provided earlier, these derivatives must be zero.


The General Case

We use Fxx (x; 0) to identify gxx (x; 0) and hxx (x; 0):
¯ ¯ ¯

[Fxx (x; 0)]ijk =
¯
[Hy 0 y 0 ]iαγ [gx ]γ [hx ]k + [Hy 0 y ]iαγ [gx ]k + [Hy 0 x 0 ]iαδ [hx ]k + [Hy 0 x ]iαk [gx ]α [hx ]j
δ γ δ β
δ β

+[Hy 0 ]iα [gxx ]α [hx ]k [hx ]j + [Hy 0 ]iα [gx ]α [hxx ]jk
δ β β
βδ β

+ [Hyy 0 ]iαγ [gx ]γ [hx ]k + [Hyy ]iαγ [gx ]k + [Hyx 0 ]iαδ [hx ]k + [Hyx ]iαk [gx ]jα
δ
δ γ δ

+[Hy ]iα [gxx ]jk
α

+ [Hx 0 y 0 ]iβγ [gx ]γ [hx ]k + [Hx 0 y ]iβγ [gx ]k + [Hx 0 x 0 ]iβδ [hx ]k + [Hx 0 x ]iβk [hx ]j
δ γ δ β
δ

+[Hx 0 ]iβ [hxx ]jk
β

+[Hxy 0 ]ij γ [gx ]γ [hx ]k + [Hxy ]ij γ [gx ]k + [Hxx 0 ]ij δ [hx ]k + [Hxx ]ijk = 0;
δ
δ γ δ

i = 1, . . . n, j, k, β, δ = 1, . . . nx ; α, γ = 1, . . . ny .


The General Case


We know the derivatives of H.

We also know the …rst derivatives of g and h evaluated at
(y , y 0 , x, x 0 ) = (y , y , x, x ).
¯ ¯ ¯ ¯

Hence, the above expression represents a system of n nx nx linear
equations in then n nx nx unknowns elements of gxx and hxx .


The General Case

Solving the System III
Similarly, gσσ and hσσ can be obtained by solving:

[Fσσ (x; 0)]i
¯ = [Hy 0 ]iα [gx ]α [hσσ ] β
β

+[Hy 0 y 0 ]iαγ [gx ]γ [η ]δ [gx ]α [η ]φ [I ]ξ
β φ
δ ξ β

+[Hy 0 x 0 ]iαδ [η ]δ [gx ]α [η ]φ [I ]ξ
β φ
ξ β

+[Hy 0 ]iα [gxx ]α [η ]δ [η ]φ [I ]ξ + [Hy 0 ]iα [gσσ ]α
β φ
βδ ξ

+[Hy ]iα [gσσ ]α + [Hx 0 ]iβ [hσσ ] β
+[Hx 0 y 0 ]iβγ [gx ]γ [η ]δ [η ]φ [I ]ξ
β φ
δ ξ

+[Hx 0 x 0 ]iβδ [η ]δ [η ]φ [I ]ξ = 0;
β φ
ξ
i = 1, . . . , n; α, γ = 1, . . . , ny ; β, δ = 1, . . . , nx ; φ, ξ = 1, . . . , n

a system of n linear equations in the n unknowns given by the elements of
gσσ and hσσ .

The General Case

Cross Derivatives
The cross derivatives gx σ and hx σ are zero when evaluated at (x, 0).
¯
Why? Write the system Fσx (x; 0) = 0 taking into account that all
¯
terms containing either gσ or hσ are zero at (x, 0).
¯
Then:
[Fσx (x; 0)]ij = [Hy 0 ]iα [gx ]α [hσx ]j + [Hy 0 ]iα [gσx ]α [hx ]jγ +
β
¯ β γ

[Hy ]iα [gσx ]jα + [Hx 0 ]iβ [hσx ]j = 0;
β

i = 1, . . . n; α = 1, . . . , ny ; β, γ, j = 1, . . . , nx .
a system of n nx equations in the n nx unknowns given by the
elements of gσx and hσx .
The system is homogeneous in the unknowns.
Thus, if a unique solution exists, it is given by:
gσx = 0
hσx = 0

The General Case

Structure of the Solution

The perturbation solution of the model satis…es:

gσ (x; 0) = 0
¯
hσ (x; 0) = 0
¯
gx σ (x; 0) = 0
¯
hx σ (x; 0) = 0
¯

Standard deviation only appears in:
1 A constant term given by 1 gσσ σ2 for the control vector yt .
2
2 The …rst nx 1
n elements of 2 hσσ σ2 .

Correction for risk.
Quadratic terms in endogenous state vector x1 .
Those terms capture non-linear behavior.

The General Case

Higher-Order Approximations

We can iterate this procedure as many times as we want.

We can obtain n-th order approximations.

Problems:

1 Existence of higher order derivatives (Santos, 1992).

2 Numerical instabilities.

3 Computational costs.


Change of Variables

Erik Eady
It is not the process of linearization that limits insight.
It is the nature of the state that we choose to linearize about.


Change of Variables

Change of Variables

We approximated our solution in levels.

We could have done it in logs.

Why stop there? Why not in powers of the state variables?

Judd (2002) has provided methods for changes of variables.

We apply and extend ideas to the stochastic neoclassical growth
model.


Change of Variables

A General Transformation

We look at solutions of the form:

µ ζ
cµ c0 = a kζ k0 + bz

k 0γ
γ ζ
k0 = c kζ k0 + dz

Note that:

1 If γ, ζ, and µ are 1, we get the linear representation.

2 As γ, ζ and µ tend to zero, we get the loglinear approximation.


Change of Variables

Theory
The …rst-order solution can be written as
f (x ) ' f (a ) + (x a ) f 0 (a )

Expand g (y ) = h (f (X (y ))) around b = Y (a), where X (y ) is the
inverse of Y (x ).

Then:
g (y ) = h (f (X (y ))) = g (b ) + gα (b ) (Y α (x ) bα )
where gα = hA fi A Xα comes from the application of the chain rule.
i

From this expression it is easy to see that if we have computed the
values of fi A , then it is straightforward to …nd gα .

Change of Variables

Coe¢ cients Relation

Remember that the linear solution is:

k0 k0 = a1 ( k k0 ) + b1 z
(l l0 ) = c1 (k k0 ) + d1 z

Then we show that:

γ ζ γ 1
a3 = γ k0
ζ a1 b3 = γk0 b1
µ µ 1 1 ζ µ 1
c3 = ζ l0 k0 c1 d3 = µl0 d1


Change of Variables

Finding the Parameters

Minimize over a grid the Euler Error.

Some optimal results

Euler Equation Errors
γ ζ µ SEE
1 1 1 0.0856279
0.986534 0.991673 2.47856 0.0279944


Change of Variables

Sensitivity Analysis

Di¤erent parameter values.

Most interesting …nding is when we change σ:

Optimal Parameters for di¤erent σ’s
σ γ ζ µ
0.014 0.98140 0.98766 2.47753
0.028 1.04804 1.05265 1.73209
0.056 1.23753 1.22394 0.77869

A …rst-order approximation corrects for changes in variance!


Change of Variables


Change of Variables

A Quasi-Optimal Approximation
Sensitivity analysis reveals that for di¤erent parametrizations

γ'ζ

This suggests the quasi-optimal approximation:

k 0γ
γ γ
k0 = a3 k γ k0 + b3 z
µ γ
lµ l0 = c3 k γ k0 + d3 z
b
If we de…ne k = k γ k0 and b = l µ
γ
l
µ
l0 we get:
b b
k 0 = a3 k + b3 z
b = c3 k + d3 z
l b

Linear system:
1 Use for analytical study.
2 Use for estimation with a Kalman Filter.

Perturbing the Value Function

We worked with the equilibrium conditions of the model.

Sometimes we may want to perform a perturbation on the value
function formulation of the problem.

Possible reasons:

1 Gain insight.

2 Di¢ culty in using equilibrium conditions.

3 Evaluate welfare.

4 Initial guess for VFI.


Basic Problem

Imagine that we have:
" #
1 γ
c
V (kt , zt ) = max (1 β) t + βEt V (kt +1 , zt +1 )
ct 1 γ
s.t. ct + kt +1 = e zt ktθ + (1 δ) kt
zt = λzt 1 + σεt , εt N (0, 1)

Write it as:
" #
1 γ
c
V (kt , zt ; χ) = max (1 β) t + βEt V (kt +1 , zt +1 ; χ)
ct 1 γ
s.t. ct + kt +1 = e zt ktθ + (1 δ) kt
zt = λzt 1 + χσεt , εt N (0, 1)



Alternative

Another way to write the value function is:

V (kt , zt ; χ) =
" 1 γ
#
max
(1 β ) c t γ +
1
ct βEt V e zt ktθ + (1 δ) kt ct , λzt + χσεt +1 ; χ

This form makes the dependences in the next period states explicit.

The solution of this problem is value function V (kt , zt ; χ) and a
policy function for consumption c (kt , zt ; χ).



Expanding the Value Function
The second-order Taylor approximation of the value function around the
deterministic steady state (kss , 0; 0) is:
V (kt , zt ; χ) '
Vss + V1,ss (kt kss ) + V2,ss zt + V3,ss χ
1 1 1
+ V11,ss (kt kss )2 + V12,ss (kt kss ) zt + V13,ss (kt kss ) χ
2 2 2
1 1 1
+ V21,ss zt (kt kss ) + V22,ss zt2 + V23,ss zt χ
2 2 2
1 1 1
+ V31,ss χ (kt kss ) + V32,ss χzt + V33,ss χ2
2 2 2
where
Vss = V (kss , 0; 0)
Vi ,ss = Vi (kss , 0; 0) for i = f1, 2, 3g
Vij ,ss = Vij (kss , 0; 0) for i, j = f1, 2, 3g


Expanding the Value Function

By certainty equivalence, we will show below that:

V3,ss = V13,ss = V23,ss = 0

Taking advantage of the equality of cross-derivatives, and setting
χ = 1, which is just a normalization:

V (kt , zt ; 1) ' Vss + V1,ss (kt kss ) + V2,ss zt
1 1
+ V11,ss (kt kss )2 + V22,ss ztt 2
2 2
1
+V12,ss (kt kss ) z + V33,ss
2
Note that V33,ss 6= 0, a di¤erence from the standard linear-quadratic
approximation to the utility functions.


Expanding the Consumption Function

The policy function for consumption can be expanded as:

ct = c (kt , zt ; χ) ' css + c1,ss (kt kss ) + c2,ss zt + c3,ss χ

where:

c1,ss = c1 (kss , 0; 0)
c2,ss = c2 (kss , 0; 0)
c3,ss = c3 (kss , 0; 0)

Since the …rst derivatives of the consumption function only depend on
the …rst and second derivatives of the value function, we must have
c3,ss = 0 (precautionary consumption depends on the third derivative
of the value function, Kimball, 1990).



Linear Components of the Value Function

To …nd the linear approximation to the value function, we take
derivatives of the value function with respect to controls (ct ), states
(kt , zt ), and the perturbation parameter χ.

Notation:

1 Vi ,t : derivative of the value function with respect to its i-th argument,
evaluated in (kt , zt ; χ) .

2 Vi ,ss : derivative evaluated in the steady state, (kss , 0; 0).

3 We follow the same notation for higher-order (cross-) derivatives.



Derivatives
Derivative with respect to ct :
γ
(1 β ) ct βEt V1,t +1 = 0
Derivative with respect to kt :

V1,t = βEt V1,t +1 θe zt ktθ 1
+1 δ

Derivative with respect to zt :
h i
V2,t = βEt V1,t +1 e zt ktθ + V2,t +1 λ

Derivative with respect to χ:
V3,t = βEt [V2,t +1 σεt +1 + V3,t +1 ]
In the last three derivatives, we apply the envelope theorem to
eliminate the derivatives of consumption with respect to kt , zt , and χ.


System of Equations I

Now, we have the system:

ct + kt +1 = e zt ktθ + (1 δ) kt
1 γ
ct
V (kt , zt ; χ) = (1 β) + βEt V (kt +1 , zt +1 ; χ)
1 γ
γ
(1 β ) ct βEt V1,t +1 = 0
V1,t = βEt V1,t +1 θe zt ktθ 1 + 1 δ
h i
V3,t = βEt [V2,t +1 σεt +1 + V3,t +1 ]
zt = λzt 1 + χσεt



System of Equations II
If we set χ = 0 and compute the steady state, we get a system of six
equations on six unknowns, css , kss , Vss , V1,ss , V2,ss , and V3,ss :
θ
css + δkss = kss
1 γ
css
Vss = (1 β) + βVss
1 γ
(1 β) css γ βV1,ss = 0
V1,ss = βV1,ss θkss 1 + 1 δ
θ

h i
θ
V2,ss = β V1,ss kss + V2,ss λ
V3,ss = βV3,ss
From the last equation: V3,ss = 0.
1 γ
From the second equation: Vss = css γ .
1
1 β γ
From the third equation: V1,ss = β css .


System of Equations III
After cancelling redundant terms:
θ
css + δkss = kss
1 = β θkss 1 + 1 δ
θ

h i
θ
V2,ss = β V1,ss kss + V2,ss λ

Then:
1
1 1 θ 1
kss = 1+δ
θ β
θ
css = kss δkss
1 β θ γ
V2,ss = k c
1 βλ ss ss
V1,ss > 0 and V2,ss > 0, as predicted by theory.


Quadratic Components of the Value Function
From the previous derivations, we have:
γ
(1 β) c (kt , zt ; χ) βEt V1,t +1 = 0
V1,t = βEt V1,t +1 θe zt ktθ 1 + 1 δ
h i
V3,t = βEt [V2,t +1 σεt +1 + V3,t +1 ]
where:
kt +1 = e zt ktθ + (1 δ) kt c (kt , zt ; χ)
zt = λzt 1 + χσεt , εt N (0, 1)
We take derivatives of each of the four equations w.t.r. kt , zt , and χ.
We take advantage of the equality of cross derivatives.
The envelope theorem does not hold anymore (we are taking
derivatives of the derivatives of the value function).


First Equation I
We have:
γ
(1 β) c (kt , zt ; χ) βEt V1,t +1 = 0

γ 1
(1 β) γc (kt , zt ; χ) c1,t
h i
βEt V11,t +1 e zt θktθ 1
+1 δ c1,t =0

In steady state:
h i
1 1
βV11,ss (1 β) γcss γ θ
c1,ss = β V11,ss θkss +1 δ

or
V11,ss
c1,ss = γ 1
βV11,ss (1 β) γcss
where we have used that 1 = β θkss
θ 1 +1 δ .


First Equation II

γ 1
(1 β) γc (kt , zt ; χ) c2,t
βEt V11,t +1 e zt ktθ c2,t + V12,t +1 λ = 0

In steady state:

1
βV11,ss (1 β) γcss γ c2,ss = β V11,ss ktθ + V12,ss λ

or
β θ
c2,ss = γ 1
V11,ss kss + V12,ss λ



First Equation III

γ 1
(1 β) γc (kt , zt ; χ) c3,t
βEt ( V11,t +1 c3,t + V12,t +1 σεt +1 + V13,t +1 ) = 0

In steady state:
1
βV11,ss (1 β) γcss γ c3,ss = βV13,ss

or
β
c3,ss = V13,ss
γ 1



Second Equation I
We have:
V1,t = βEt V1,t +1 θe zt ktθ 1
+1 δ

" #
V11,t +1 θe zt ktθ 1 + 1 δ c1,t θe zt ktθ 1
+1 δ
V11,t = βEt
2
+V1,t +1 θ (θ 1) e zt ktθ

In steady state:

1 θ 2
V11,ss = V11,ss c1,ss + βV1,ss θ (θ 1) kss
β
or
β θ 2
V11,ss = 1
V1,ss θ (θ 1) kss
1 β + c1,ss



Second Equation II

2 3
V11,t +1 e zt ktθ c2,t θe zt ktθ 1 + 1 δ
V12,t = βEt 4 5
+V12,t +1 λ θe zt ktθ 1 + 1 δ + V1,t +1 θe zt ktθ 1

In steady state:

θ 1
V12,ss = V11,ss kss c2,ss + V12,ss λ + βV1,ss θktθ

or h i
1 θ θ 1
V12,ss = V11,ss kss c2,ss + βV1,ss θkss
1 λ



Second Equation III


V13,t = βEt [ V11,t +1 c3,t + V12,t +1 σεt +1 + V13,t +1 ]

In steady state,

V13,ss = β [ V11,ss c3,ss + V13,ss ] )
β
V13,ss = V11,ss c3,ss
β 1

but since we know that:
β
c3,ss = V13,ss
γ 1

the two equations can only hold simultaneously if V13,ss = c3,ss = 0.


Third Equation I
We have h i


V11,t +1 e zt ktθ c2,t e zt ktθ + V12,t +1 λe zt ktθ
V22,t = βEt
+V1,t +1 e zt ktθ + V21,t +1 λ e zt ktθ c2,t + V22,t +1 λ2

In steady state:

V11,ss ktθ c2,ss kss + V12,ss λkss + V1,ss kss
θ θ θ
V22,t = β )
+V21,ss λ kss c2,ss + V22,ss λ2
θ

β V11,ss ktθ c2,ss kss + 2V12,ss λkss
θ θ
V22,ss =
1 βλ2 +V1,ss kss V12,ss λc2,ss
θ

where we have used V12,ss = V21,ss .


Third Equation II


V11,t +1 e zt ktθ c3,t + V12,t +1 e zt ktθ σεt +1 + V13,t +1 e zt ktθ
V23,t = βEt
V21,t +1 λc3,t + V22,t +1 λσεt +1 + V23,t +1 λ

In steady state:
V23,ss = 0



Fourth Equation

We have
V3,t = βEt [V2,t +1 σεt +1 + V3,t +1 ] .


V21,t +1 c3,t σεt +1 + V22,t +1 σ2 ε2+1 + V23,t +1 σεt +1
t
V33,t = βEt
V31,t +1 c3,t + V32,t +1 σεt +1 + V33,t +1

In steady state:
β
V33,ss = V22,ss
1 β



System I

V11,ss
c1,ss = γ 1
β θ
c2,ss = γ 1
V11,ss kss + V12,ss λ
β
V11,ss = 1
1) kss 2
θ
V1,ss θ (θ
1 β + c1,ss
1 h i
V12,ss = V11,ss kss c2,ss + βV1,ss θkss 1
θ θ
1 λ
β V11,ss ktθ c2,ss kss + 2V12,ss λkss
θ θ
V22,ss =
1 βλ2 +V1,ss kss V12,ss λc2,ss
θ

β
V33,ss = σ2 V22,ss
1 β
plus c3,ss = V13,ss = V23,ss = 0.


System II

This is a system of nonlinear equations.

However, it has a recursive structure.

By substituting variables that we already know, we can …nd V11,ss .

Then, using this results and by plugging c2,ss , we have a system of
two equations, on two unknowns, V12,ss and V22,ss .

Once the system is solved, we can …nd c1,ss , c2,ss , and V33,ss directly.



The Welfare Cost of the Business Cycle

An advantage of performing the perturbation on the value function is
that we have evaluation of welfare readily available.

Note that at the deterministic steady state, we have:

1
V (kss , 0; χ) ' Vss + V33,ss
2
Hence 1 V33,ss is a measure of the welfare cost of the business cycle.
2

This quantity is not necessarily negative: it may be positive. For
example, in an RBC with leisure choice (Cho and Cooley, 2000).



Our Example

1 γ
css
We know that Vss = 1 γ.
We can compute the decrease in consumption τ that will make the
household indi¤erent between consuming (1 τ ) css units per period
with certainty or ct units with uncertainty.
Thus:
1 γ
css 1 (css (1 τ ))1 γ
+ V33,ss = )
1 γ 2 1 γ
1
(1 τ )1 γ 1
1 css γ
= (1 γ) V33,ss
2
or 1
1 γ1 1 γ
τ=1 1+ 1
V
γ 2 33,ss
css



A Numerical Example
We pick standard parameter values by setting
β = 0.99, γ = 2, δ = 0.0294, θ = 0.3, and λ = 0.95.
We get:
V (kt , zt ; 1) ' 0.54000 + 0.00295 (kt kss ) + 0.11684zt
2
0.00007 (kt kss ) 0.00985zt2
0.97508σ2 0.00225 (kt kss ) zt
c (kt , zt ; χ) ' 1.85193 + 0.04220 (kt kss ) + 0.74318zt
DYNARE produces the same policy function by linearizing the
equilibrium conditions of the problem.
The welfare cost of the business cycle (in consumption terms) is
8.8475e-005, lower than in Lucas (1987) because of the smoothing
possibilities allowed by capital.
Use as an initial guess for VFI.

Chapter 2 pertubation

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Chapter 2 pertubation