ALL 18 CHAPTERS COVERED
SOLUTIONS MANUAL
, Contents
Foreword ix
1 Introduction 1
2 An Overview 3
3 Mathematical Preliminaries 20
4 Dynamic Programming under Certainty 42
5 Applications of Dynamic Programming under Certainty 53
6 Deterministic Dynamics 80
7 Measure Theory and Integration 98
8 Markov Processes 131
9 Stochastic Dynamic Programming 147
10 Applications of Stochastic Dynamic Programming 171
11 Strong Convergence of Markov Processes 190
12 Weak Convergence of Markov Processes 198
13 Applications of Convergence Results for Markov Processes 212
14 Laws of Large Numbers 234
15 Pareto Optima and Competitive Equilibria 240
,viii Contents
16 Applications of Equilibrium Theory 253
17 Fixed-Point Arguments 270
18 Equilibria in Systems with Distortions 284
,1 Introduction
In the preface to Recursive Methods in Economic Dynamics, the authors stated that their
aim was to make recursive methods accessible to the wider economics profession.
They succeeded. Since RMED appeared in 1989, the use of recursive methods
in economics has boomed. And what was once as much a research monograph
as a textbook has now been adopted in first-year graduate courses around the
world.
The best way for students to learn these techniques is to work problems.
And toward this end, RMED contains over two hundred problems, many with
multiple parts. The present book aims to assist students in this process by
providing answers and hints to a large subset of these questions.
At an early stage, we were urged to leave some of the questions in the book
unanswered, so as to be available as a “test bank” for instructors. This raises
the question of which answers to include and which to leave out. As a guiding
principle, we have tried to include answers to those questions that are the most
instructive, in the sense that the techniques involved in their solution are the
most useful later on in the book. We have also tried to answer all of the questions
whose results are integral to the presentation of the core methods of the book.
Exercises that involve particularly difficult reasoning or mathematics have also
been solved, although no doubt our specific choices in this regard are subject to
criticism.
As a result, the reader will find that we have provided an answer to almost every
question in the core “method” chapters (that is, Chapters 4, 6, 9, 15, 17, and 18),
as well as to most of the questions in the chapters on mathematical background
(Chapters 3, 7, 8, 11, 12, and 14). However, only a subset of the questions in the
“application” chapters (2, 5, 10, 13, and 16) have been answered.
It is our hope that this selection will make the assimilation of the material
easier for students. At the same time, instructors should be comforted to find
that they still have a relatively rich set of questions to assign from the application
1
,2 1 / Introduction
chapters. Instructors should also find that, because much of the material in the
method and mathematical background chapters appears repeatedly, there are
many opportunities to assign this material to their students.
Despite our best efforts, errors no doubt remain. Furthermore, it is to be
expected (and hoped) that readers will uncover more elegant, and perhaps more
instructive, approaches to answering the questions than those provided here. The
authors would appreciate being notified of any errors and, as an aid to readers,
commit to maintaining a website where readers can post corrections, comments
and alternative answers. This website is currently hosted at:
http://www.stanford.edu/~mlwright/RMEDSolutions
In the process of completing this project we have incurred various debts. A
number of people provided us with their own solutions to problems in the text,
including Xavier Gine, Ivan Werning and Rui Zhao. Others, including Vadym
Lepetyuk and Joon Hyuk Song, pointed out sins of commission and omission
in earlier drafts. Christine Groeger provided extensive comments, and lent her
LaTEX expertise to the production of the manuscript. At Harvard University
Press, Elizabeth Gilbert and Benno Weisberg made substantial improvements
to the manuscript’s style and logic. We thank all of these people, together with
Robert E. Lucas, Jr., and reserve a special thanks for Nancy Stokey, whose insight
and enthusiasm were invaluable in seeing the project through to its conclusion.
,2 An Overview
Exercise 2.1
The fact that f : R+ → R+ is continuously differentiable, strictly increasing
and strictly concave comes directly from the definition of f as
f (k) = F (k, 1) + (1 − δ)k,
with 0 < δ < 1, and F satisfying the properties mentioned above. In particular,
the sum of two strictly increasing functions is strictly increasing, and continuous
differentiability is preserved under summation. Finally, the sum of a strictly
concave and a linear function is strictly concave.
Also,
f (0) = F (0, 1) = 0,
f ′(k) = Fk (k, 1) + (1 − δ) > 0,
lim f ′(k) = lim Fk (k, 1) + lim (1 − δ) = ∞,
k→0 k→0 k→0
′
lim f (k) = lim Fk (k, 1) + lim (1 − δ) = (1 − δ).
k→∞ k→∞ k→∞
Exercise 2.2
a. With the given functional forms for the production and utility function we
can write (5) as
αβktα−1 1
= α
,
ktα − kt+1 kt−1 − kt
which can be rearranged as
αβktα−1(kt−1
α
− kt ) = (ktα − kt+1).
3
,4 2 / An Overview
α
Dividing both sides by ktα and using the change of variable zt = kt /kt−1 we
obtain
1
αβ( − 1) = 1 − zt+1,
zt
or
αβ
zt+1 = 1 + αβ − ,
zt
which is the equation represented in Figure 2.1.
As can be seen in the figure, the first-order difference equation has two steady
states (that is, z’s such that zt+1 = zt = z), which are the two solutions to the
characteristic equation
z2 − (1 + αβ)z + αβ = 0.
These are given by z = 1 and αβ.
z
z
Figure 2.1
, 2 / An Overview 5
b. Using the boundary condition zT +1 = 0 we can solve for zT as
αβ
zT = .
1 + αβ
Substituting recursively into the law of motion for zt derived above we can solve
for zT −1 as
αβ
zT −1 =
1 + αβ − zT
αβ
= αβ
1 + αβ − 1+αβ
αβ(1 + αβ)
= ,
1 + αβ + (αβ)2
and in general,
αβ[1 + αβ + . . . + (αβ)j ]
zT −j = .
1 + αβ + . . . + (αβ)j +1
Hence for t = T − j ,
αβ[1 + αβ + . . . + (αβ)T −t ]
zt =
1 + αβ + . . . + (αβ)T −t+1
αβsT −t
=
sT −t+1
where si = 1 + αβ + . . . + (αβ)i . In order to solve for the series, take for instance
the one in the numerator,
sT −t = 1 + αβ + . . . + (αβ)T −t ,
multiply both sides by αβ to get
αβsT −t = αβ + . . . + (αβ)T −t+1,
and substract this new expression from the previous one to obtain
(1 − αβ)sT −t = 1 − (αβ)T −t+1.
,6 2 / An Overview
Hence
1 − (αβ)T −t+1
sT −t = ,
1 − αβ
1 − (αβ)T −t+2
sT −t+1 = ,
1 − αβ
and therefore
1 − (αβ)T −t+1
zt = αβ ,
1 − (αβ)T −t+2
for t = 1, 2, . . . , T + 1, as in the text. Notice also that
1 − (αβ)T −(T +1)+1
zT +1 = αβ
1 − (αβ)T −(T +1)+2
= 0.
c. Plugging (7) into the right-hand side of (5) we get
−1
1 − (αβ)T −t+1 1 − (αβ)T −t+2
α
kt−1 − αβ kα = .
T −t+2 t−1 α
kt−1 (1 − αβ)
1 − (αβ)
Similarly, by plugging (7) into the left-hand side of (5) we obtain
α−1
1−(αβ)T −t+1
α
αβ αβ kt−1
1−(αβ)T −t+2
α
1−(αβ)T −t+1 [ 1−(αβ)T −t ]
αβ kα 1 − αβ
1−(αβ)T −t+2 t−1 1−(αβ)T −t+1
−1
T −t+1 T −t
1 1 − (αβ) − αβ 1 − (αβ)
= α
kt−1 1 − (αβ) T −t+2
1 − (αβ)T −t+2
= α
.
kt−1 (1 − αβ)
Hence, the law of motion for capital given by (7) satisfies (5).
, 2 / An Overview 7
Evaluating (7) for t = T yields
1 − (αβ)T −T α
kT +1 = αβ kT
1 − (αβ)T −T +1
= 0,
so (7) satisfies (6) too.
Exercise 2.3
a. We can write the value function using the optimal path for capital given by
(8) as
∞
υ(k0) = β t log(ktα − αβktα )
t=0
∞
log(1 − αβ)
= +α β t log(kt ).
(1 − β) t=0
The optimal policy function, written (by recursive substitution) as a function of
the initial capital stock is (in logs)
t−1
log kt = α i log(αβ) + α t log k0.
i=0
Using the optimal policy function we can break up the last summation to get
∞ ∞
t−1
t log(k0)
t
β log(kt ) = + log(αβ) β αi
t=0
(1 − αβ) t=1 i=0
log(k0) log(αβ)
= +β ,
(1 − αβ) [(1 − β)(1 − αβ)]
where we have used the fact that the solution to a series of the form st = ti=0λi
is 1 − λt+1 / (1 − λ) , as shown in Exercise 2.2b. Hence, we obtain a log linear
expression for the value function
υ(k0) = A + B log(k0),
where
αβ log(αβ)
A = log(1 − αβ) + (1 − β)−1,
(1 − αβ)
SOLUTIONS MANUAL
, Contents
Foreword ix
1 Introduction 1
2 An Overview 3
3 Mathematical Preliminaries 20
4 Dynamic Programming under Certainty 42
5 Applications of Dynamic Programming under Certainty 53
6 Deterministic Dynamics 80
7 Measure Theory and Integration 98
8 Markov Processes 131
9 Stochastic Dynamic Programming 147
10 Applications of Stochastic Dynamic Programming 171
11 Strong Convergence of Markov Processes 190
12 Weak Convergence of Markov Processes 198
13 Applications of Convergence Results for Markov Processes 212
14 Laws of Large Numbers 234
15 Pareto Optima and Competitive Equilibria 240
,viii Contents
16 Applications of Equilibrium Theory 253
17 Fixed-Point Arguments 270
18 Equilibria in Systems with Distortions 284
,1 Introduction
In the preface to Recursive Methods in Economic Dynamics, the authors stated that their
aim was to make recursive methods accessible to the wider economics profession.
They succeeded. Since RMED appeared in 1989, the use of recursive methods
in economics has boomed. And what was once as much a research monograph
as a textbook has now been adopted in first-year graduate courses around the
world.
The best way for students to learn these techniques is to work problems.
And toward this end, RMED contains over two hundred problems, many with
multiple parts. The present book aims to assist students in this process by
providing answers and hints to a large subset of these questions.
At an early stage, we were urged to leave some of the questions in the book
unanswered, so as to be available as a “test bank” for instructors. This raises
the question of which answers to include and which to leave out. As a guiding
principle, we have tried to include answers to those questions that are the most
instructive, in the sense that the techniques involved in their solution are the
most useful later on in the book. We have also tried to answer all of the questions
whose results are integral to the presentation of the core methods of the book.
Exercises that involve particularly difficult reasoning or mathematics have also
been solved, although no doubt our specific choices in this regard are subject to
criticism.
As a result, the reader will find that we have provided an answer to almost every
question in the core “method” chapters (that is, Chapters 4, 6, 9, 15, 17, and 18),
as well as to most of the questions in the chapters on mathematical background
(Chapters 3, 7, 8, 11, 12, and 14). However, only a subset of the questions in the
“application” chapters (2, 5, 10, 13, and 16) have been answered.
It is our hope that this selection will make the assimilation of the material
easier for students. At the same time, instructors should be comforted to find
that they still have a relatively rich set of questions to assign from the application
1
,2 1 / Introduction
chapters. Instructors should also find that, because much of the material in the
method and mathematical background chapters appears repeatedly, there are
many opportunities to assign this material to their students.
Despite our best efforts, errors no doubt remain. Furthermore, it is to be
expected (and hoped) that readers will uncover more elegant, and perhaps more
instructive, approaches to answering the questions than those provided here. The
authors would appreciate being notified of any errors and, as an aid to readers,
commit to maintaining a website where readers can post corrections, comments
and alternative answers. This website is currently hosted at:
http://www.stanford.edu/~mlwright/RMEDSolutions
In the process of completing this project we have incurred various debts. A
number of people provided us with their own solutions to problems in the text,
including Xavier Gine, Ivan Werning and Rui Zhao. Others, including Vadym
Lepetyuk and Joon Hyuk Song, pointed out sins of commission and omission
in earlier drafts. Christine Groeger provided extensive comments, and lent her
LaTEX expertise to the production of the manuscript. At Harvard University
Press, Elizabeth Gilbert and Benno Weisberg made substantial improvements
to the manuscript’s style and logic. We thank all of these people, together with
Robert E. Lucas, Jr., and reserve a special thanks for Nancy Stokey, whose insight
and enthusiasm were invaluable in seeing the project through to its conclusion.
,2 An Overview
Exercise 2.1
The fact that f : R+ → R+ is continuously differentiable, strictly increasing
and strictly concave comes directly from the definition of f as
f (k) = F (k, 1) + (1 − δ)k,
with 0 < δ < 1, and F satisfying the properties mentioned above. In particular,
the sum of two strictly increasing functions is strictly increasing, and continuous
differentiability is preserved under summation. Finally, the sum of a strictly
concave and a linear function is strictly concave.
Also,
f (0) = F (0, 1) = 0,
f ′(k) = Fk (k, 1) + (1 − δ) > 0,
lim f ′(k) = lim Fk (k, 1) + lim (1 − δ) = ∞,
k→0 k→0 k→0
′
lim f (k) = lim Fk (k, 1) + lim (1 − δ) = (1 − δ).
k→∞ k→∞ k→∞
Exercise 2.2
a. With the given functional forms for the production and utility function we
can write (5) as
αβktα−1 1
= α
,
ktα − kt+1 kt−1 − kt
which can be rearranged as
αβktα−1(kt−1
α
− kt ) = (ktα − kt+1).
3
,4 2 / An Overview
α
Dividing both sides by ktα and using the change of variable zt = kt /kt−1 we
obtain
1
αβ( − 1) = 1 − zt+1,
zt
or
αβ
zt+1 = 1 + αβ − ,
zt
which is the equation represented in Figure 2.1.
As can be seen in the figure, the first-order difference equation has two steady
states (that is, z’s such that zt+1 = zt = z), which are the two solutions to the
characteristic equation
z2 − (1 + αβ)z + αβ = 0.
These are given by z = 1 and αβ.
z
z
Figure 2.1
, 2 / An Overview 5
b. Using the boundary condition zT +1 = 0 we can solve for zT as
αβ
zT = .
1 + αβ
Substituting recursively into the law of motion for zt derived above we can solve
for zT −1 as
αβ
zT −1 =
1 + αβ − zT
αβ
= αβ
1 + αβ − 1+αβ
αβ(1 + αβ)
= ,
1 + αβ + (αβ)2
and in general,
αβ[1 + αβ + . . . + (αβ)j ]
zT −j = .
1 + αβ + . . . + (αβ)j +1
Hence for t = T − j ,
αβ[1 + αβ + . . . + (αβ)T −t ]
zt =
1 + αβ + . . . + (αβ)T −t+1
αβsT −t
=
sT −t+1
where si = 1 + αβ + . . . + (αβ)i . In order to solve for the series, take for instance
the one in the numerator,
sT −t = 1 + αβ + . . . + (αβ)T −t ,
multiply both sides by αβ to get
αβsT −t = αβ + . . . + (αβ)T −t+1,
and substract this new expression from the previous one to obtain
(1 − αβ)sT −t = 1 − (αβ)T −t+1.
,6 2 / An Overview
Hence
1 − (αβ)T −t+1
sT −t = ,
1 − αβ
1 − (αβ)T −t+2
sT −t+1 = ,
1 − αβ
and therefore
1 − (αβ)T −t+1
zt = αβ ,
1 − (αβ)T −t+2
for t = 1, 2, . . . , T + 1, as in the text. Notice also that
1 − (αβ)T −(T +1)+1
zT +1 = αβ
1 − (αβ)T −(T +1)+2
= 0.
c. Plugging (7) into the right-hand side of (5) we get
−1
1 − (αβ)T −t+1 1 − (αβ)T −t+2
α
kt−1 − αβ kα = .
T −t+2 t−1 α
kt−1 (1 − αβ)
1 − (αβ)
Similarly, by plugging (7) into the left-hand side of (5) we obtain
α−1
1−(αβ)T −t+1
α
αβ αβ kt−1
1−(αβ)T −t+2
α
1−(αβ)T −t+1 [ 1−(αβ)T −t ]
αβ kα 1 − αβ
1−(αβ)T −t+2 t−1 1−(αβ)T −t+1
−1
T −t+1 T −t
1 1 − (αβ) − αβ 1 − (αβ)
= α
kt−1 1 − (αβ) T −t+2
1 − (αβ)T −t+2
= α
.
kt−1 (1 − αβ)
Hence, the law of motion for capital given by (7) satisfies (5).
, 2 / An Overview 7
Evaluating (7) for t = T yields
1 − (αβ)T −T α
kT +1 = αβ kT
1 − (αβ)T −T +1
= 0,
so (7) satisfies (6) too.
Exercise 2.3
a. We can write the value function using the optimal path for capital given by
(8) as
∞
υ(k0) = β t log(ktα − αβktα )
t=0
∞
log(1 − αβ)
= +α β t log(kt ).
(1 − β) t=0
The optimal policy function, written (by recursive substitution) as a function of
the initial capital stock is (in logs)
t−1
log kt = α i log(αβ) + α t log k0.
i=0
Using the optimal policy function we can break up the last summation to get
∞ ∞
t−1
t log(k0)
t
β log(kt ) = + log(αβ) β αi
t=0
(1 − αβ) t=1 i=0
log(k0) log(αβ)
= +β ,
(1 − αβ) [(1 − β)(1 − αβ)]
where we have used the fact that the solution to a series of the form st = ti=0λi
is 1 − λt+1 / (1 − λ) , as shown in Exercise 2.2b. Hence, we obtain a log linear
expression for the value function
υ(k0) = A + B log(k0),
where
αβ log(αβ)
A = log(1 − αβ) + (1 − β)−1,
(1 − αβ)
