CHAPTER 1
DIFFERENCE EQUATIONS
1. Time-Series Models
2. Difference Equations and Their Solutions
3. Solution by Iteration
4. An Alternative Solution Methodology
5. The Cobweb Model
6. Solving Homogeneous Difference Equations
7. Particular Solutions for Deterministic Processes
8. The Method of Undetermined Coefficients
9. Lag Operators
10. Summary
Questions and Exercises
APPENDIX 1.1 Imaginary Roots and de Moivre’s Theorem
APPENDIX 1.2 Characteristic Roots in Higher-Order Equations
Lecture Suggestions
Nearly all students will have some familiarity with the concepts developed in the chapter. A
first course in integral calculus makes reference to convergent versus divergent solutions. I draw
the analogy between the particular solution to a difference equation and indefinite integrals.
It is essential to convey the fact that difference equations are capable of capturing the types
of dynamic models used in economics and political science. Towards this end, I computer-generate
a number of simulated series and discuss how their dynamic properties depend on the parameters
of the data-generating process. Next, I show the students a number of macroeconomic variables—
such as real GDP, real exchange rates, interest rates, and rates of return on stock prices—and ask
them to think about the underlying dynamic processes that might be driving each variable. I ask
them to think about the economic theory that bears on each of the variables. For example, the figure
below shows the three real exchange rate series used in Figure 3.5. Some students see a tendency
for the series to revert to a long-run mean value. Nevertheless, the statistical evidence that real
exchange rates are actually mean reverting is debatable. Moreover, there is no compelling
theoretical reason to believe that purchasing power parity holds as a long-run phenomenon. The
classroom discussion might center on the appropriate way to model the tendency for the levels to
meander. At this stage, the precise models are not important. The objective is for students to
conceptualize economic data in terms of difference equations.
, It is also important to stress the distinction between convergent and divergent solutions. Be
sure to emphasize the relationship between characteristic roots and the convergence or divergence
of a sequence. Much of the current time-series literature focuses on the issue of unit roots. It is wise
to introduce students to the properties of difference equations with unitary characteristic roots at
this early stage in the course. Question 5 at the end of this chapter is designed to preview this
important issue. The tools to emphasize are the method of undetermined coefficients and lag
operators. Few students will have been exposed to these methods in other classes.
Figure 3.5 Indices of Real Effective Exchange Rates
150.00
140.00
130.00
120.00
Indices for year 2000 = 100
110.00
100.00
90.00
80.00
70.00
60.00
8 0 9 8 1 9 82 9 83 9 8 4 9 8 5 9 86 9 87 9 8 8 9 8 9 9 9 0 9 91 9 92 9 93 9 9 4 9 9 5 9 96 9 97 9 9 8 9 9 9 0 00 0 01 0 0 2 0 0 3 0 04 0 05 0 06 0 0 7
19 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2
Canada
Answers to Questions
1. Consider the difference equation: yt = a0+ a1yt-1 with the initial condition y0. Jill solved the
difference equation by iterating backwards:
yt = a0 + a1yt-1
= a0 + a1[a0 + a1yt-2 ]
= a0 + a0a1 + a0(a1)2 + .... + a0(a1)t-1 + (a1)ty0
Bill added the homogeneous and particular solutions to obtain: yt = a0/(1 - a1) + (a1)t[y0 - a0/(1 -
a1)].
,a. Show that the two solutions are identical for a1 < 1.
Answer: The key is to demonstrate:
a0 + a0a1 + a0(a1)2 + .... + a0(a1)t-1 + (a1)ty0 = a0/(1 - a1) + (a1)t[y0 - a0/(1 - a1)]
First, cancel (a1)ty0 from each side and then divide by a0. The two sides of the equation are
identical if:
1 + a1 + (a1)2 + .... + (a1)t-1 = 1/(1 - a1) - (a1)t/(1 - a1)
Now, multiply each side by (1 - a1) to obtain:
(1 - a1)[1 + a1 + (a1)2 + .... + (a1)t-1] = 1 - (a1)t
Multiply the two expressions on the left-hand side to obtain:
1 - (a1)t = 1 - (a1)t
The two sides of the equation are identical. Hence, Jill and Bob obtained the identical
answer.
b. Show that for a1 = 1, Jill's solution is equivalent to: yt = a0t + y0. How would you use Bill's method
to arrive at this same conclusion in the case a1 = 1.
Answer: When a1 = 1, Jill's solution can be written as:
yt = a0(10 + 11 + 12 + ... + 1t-1) + y0
= a0t + y0
To use Bill's method, find the homogeneous solution from the equation yt = yt-1. Clearly, the
homogeneous solution is any arbitrary constant A. The key in finding the particular solution
is to realize that the characteristic root is unity. In this instance, the particular solution has
the form a0t. Adding the homogeneous and particular solutions, the general solution is:
yt = a0t + A
To eliminate the arbitrary constant, impose the initial condition. The general solution must
hold for all t including t = 0. Hence, at t = 0, y0 = a0t + A so that A = y0. Hence, Bill's method
yields:
yt = a0t + y0
, 2. The Cobweb model in section 5 assumed static price expectations. Consider an alternative
¿
p
formulation called adaptive expectations. Let the expected price in t (denoted by t ) be a weighted
average of the price in t-1 and the price expectation of the previous period. Formally:
¿ ¿
pt p
= pt-1 + (1 - ) t−1 0 < 1.
Clearly, when = 1, the static and adaptive expectations schemes are equivalent. An interesting
feature of this model is that it can be viewed as a difference equation expressing the expected price
as a function of its own lagged value and the forcing variable pt-1.
¿
a.
p
Find the homogeneous solution for t
¿ ¿
p p
Answer: Form the homogeneous equation t - (1 - ) t−1 = 0.
The homogeneous solution is:
¿
pt = A(1-)t
where A is an arbitrary constant and (1-) is the characteristic root.
b. Use lag operators to find the particular solution. Check your answer by substituting your answer
into the original difference equation.
Answer: The particular solution can be written as:
¿
[ 1 - (1-)L ]
pt = p
t-1
or
¿
pt = p /[ 1 - (1-)L ] so that:
t-1
¿
pt = [pt-1 + (1-)pt-2 + (1-)2pt-3 + ... ]
To check the answer, substitute the particular solution into the original difference equation
[pt-1 + (1-)pt-2 + (1-)2pt-3 + ... ] = pt-1 + (1-)[pt-2 + (1-)pt-3 + (1-)2pt-4 + ... ]
It should be clear that the equation holds as an identity.
DIFFERENCE EQUATIONS
1. Time-Series Models
2. Difference Equations and Their Solutions
3. Solution by Iteration
4. An Alternative Solution Methodology
5. The Cobweb Model
6. Solving Homogeneous Difference Equations
7. Particular Solutions for Deterministic Processes
8. The Method of Undetermined Coefficients
9. Lag Operators
10. Summary
Questions and Exercises
APPENDIX 1.1 Imaginary Roots and de Moivre’s Theorem
APPENDIX 1.2 Characteristic Roots in Higher-Order Equations
Lecture Suggestions
Nearly all students will have some familiarity with the concepts developed in the chapter. A
first course in integral calculus makes reference to convergent versus divergent solutions. I draw
the analogy between the particular solution to a difference equation and indefinite integrals.
It is essential to convey the fact that difference equations are capable of capturing the types
of dynamic models used in economics and political science. Towards this end, I computer-generate
a number of simulated series and discuss how their dynamic properties depend on the parameters
of the data-generating process. Next, I show the students a number of macroeconomic variables—
such as real GDP, real exchange rates, interest rates, and rates of return on stock prices—and ask
them to think about the underlying dynamic processes that might be driving each variable. I ask
them to think about the economic theory that bears on each of the variables. For example, the figure
below shows the three real exchange rate series used in Figure 3.5. Some students see a tendency
for the series to revert to a long-run mean value. Nevertheless, the statistical evidence that real
exchange rates are actually mean reverting is debatable. Moreover, there is no compelling
theoretical reason to believe that purchasing power parity holds as a long-run phenomenon. The
classroom discussion might center on the appropriate way to model the tendency for the levels to
meander. At this stage, the precise models are not important. The objective is for students to
conceptualize economic data in terms of difference equations.
, It is also important to stress the distinction between convergent and divergent solutions. Be
sure to emphasize the relationship between characteristic roots and the convergence or divergence
of a sequence. Much of the current time-series literature focuses on the issue of unit roots. It is wise
to introduce students to the properties of difference equations with unitary characteristic roots at
this early stage in the course. Question 5 at the end of this chapter is designed to preview this
important issue. The tools to emphasize are the method of undetermined coefficients and lag
operators. Few students will have been exposed to these methods in other classes.
Figure 3.5 Indices of Real Effective Exchange Rates
150.00
140.00
130.00
120.00
Indices for year 2000 = 100
110.00
100.00
90.00
80.00
70.00
60.00
8 0 9 8 1 9 82 9 83 9 8 4 9 8 5 9 86 9 87 9 8 8 9 8 9 9 9 0 9 91 9 92 9 93 9 9 4 9 9 5 9 96 9 97 9 9 8 9 9 9 0 00 0 01 0 0 2 0 0 3 0 04 0 05 0 06 0 0 7
19 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2
Canada
Answers to Questions
1. Consider the difference equation: yt = a0+ a1yt-1 with the initial condition y0. Jill solved the
difference equation by iterating backwards:
yt = a0 + a1yt-1
= a0 + a1[a0 + a1yt-2 ]
= a0 + a0a1 + a0(a1)2 + .... + a0(a1)t-1 + (a1)ty0
Bill added the homogeneous and particular solutions to obtain: yt = a0/(1 - a1) + (a1)t[y0 - a0/(1 -
a1)].
,a. Show that the two solutions are identical for a1 < 1.
Answer: The key is to demonstrate:
a0 + a0a1 + a0(a1)2 + .... + a0(a1)t-1 + (a1)ty0 = a0/(1 - a1) + (a1)t[y0 - a0/(1 - a1)]
First, cancel (a1)ty0 from each side and then divide by a0. The two sides of the equation are
identical if:
1 + a1 + (a1)2 + .... + (a1)t-1 = 1/(1 - a1) - (a1)t/(1 - a1)
Now, multiply each side by (1 - a1) to obtain:
(1 - a1)[1 + a1 + (a1)2 + .... + (a1)t-1] = 1 - (a1)t
Multiply the two expressions on the left-hand side to obtain:
1 - (a1)t = 1 - (a1)t
The two sides of the equation are identical. Hence, Jill and Bob obtained the identical
answer.
b. Show that for a1 = 1, Jill's solution is equivalent to: yt = a0t + y0. How would you use Bill's method
to arrive at this same conclusion in the case a1 = 1.
Answer: When a1 = 1, Jill's solution can be written as:
yt = a0(10 + 11 + 12 + ... + 1t-1) + y0
= a0t + y0
To use Bill's method, find the homogeneous solution from the equation yt = yt-1. Clearly, the
homogeneous solution is any arbitrary constant A. The key in finding the particular solution
is to realize that the characteristic root is unity. In this instance, the particular solution has
the form a0t. Adding the homogeneous and particular solutions, the general solution is:
yt = a0t + A
To eliminate the arbitrary constant, impose the initial condition. The general solution must
hold for all t including t = 0. Hence, at t = 0, y0 = a0t + A so that A = y0. Hence, Bill's method
yields:
yt = a0t + y0
, 2. The Cobweb model in section 5 assumed static price expectations. Consider an alternative
¿
p
formulation called adaptive expectations. Let the expected price in t (denoted by t ) be a weighted
average of the price in t-1 and the price expectation of the previous period. Formally:
¿ ¿
pt p
= pt-1 + (1 - ) t−1 0 < 1.
Clearly, when = 1, the static and adaptive expectations schemes are equivalent. An interesting
feature of this model is that it can be viewed as a difference equation expressing the expected price
as a function of its own lagged value and the forcing variable pt-1.
¿
a.
p
Find the homogeneous solution for t
¿ ¿
p p
Answer: Form the homogeneous equation t - (1 - ) t−1 = 0.
The homogeneous solution is:
¿
pt = A(1-)t
where A is an arbitrary constant and (1-) is the characteristic root.
b. Use lag operators to find the particular solution. Check your answer by substituting your answer
into the original difference equation.
Answer: The particular solution can be written as:
¿
[ 1 - (1-)L ]
pt = p
t-1
or
¿
pt = p /[ 1 - (1-)L ] so that:
t-1
¿
pt = [pt-1 + (1-)pt-2 + (1-)2pt-3 + ... ]
To check the answer, substitute the particular solution into the original difference equation
[pt-1 + (1-)pt-2 + (1-)2pt-3 + ... ] = pt-1 + (1-)[pt-2 + (1-)pt-3 + (1-)2pt-4 + ... ]
It should be clear that the equation holds as an identity.
