Jacques I. Mathematics for Economics and Business

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If we assume that investment is the same in all time periods then

= I*

for some constant, I*. Finally, if the ﬂow of money is in balance in each time period, we also

have

= C

+ I

Substituting the expressions for C

and I

into this gives

= aY

t−1

+ b + I*

which we recognize as a difference equation of the standard form given in this section. This

equation can therefore be solved and the time path analysed.

Dynamics

560

Example

Consider a two-sector model:

= C

+ I

= 0.8Y

t−1

+ 100

= 200

Find an expression for Y

when Y

= 1700. Is this system stable or unstable?

Solution

Substituting the expressions for C

and I

into

= C

+ I

gives

= (0.8Y

t−1

+ 100) + 200

= 0.8Y

t−1

+ 300

The complementary function is given by

CF = A(0.8)

and for a particular solution we try

= D

for some constant D. Substituting this into the difference equation gives

D = 0.8D + 300

which has solution D = 1500. The general solution is therefore

= A(0.8)

+ 1500

The initial condition,

= 1700

gives

1700 = A(0.8)

+ 1500 = A + 1500

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and so A is 200. The solution is

= 200(0.8)

+ 1500

As t increases, (0.8)

converges to zero and so Y

eventually settles down at the equilibrium level of 1500. The

system is therefore stable. Note also that because 0.8 lies between 0 and 1, the time path displays uniform

convergence.

9.1 • Difference equations

561

Practice Problem

3 Consider the two-sector model:

= C

+ I

= 0.9Y

t−1

+ 250

= 350

Find an expression for Y

when Y

= 6500. Is this system stable or unstable?

In the previous example, and again in Practice Problem 3, we noted that the model is stable

and that it displays uniform convergence. If we return to the general equation

= aY

t−1

+ b + I*

it is easy to see that this is always the case for the simple two-sector model because the coefﬁci-

ent of Y

t−1

is the marginal propensity to consume, which is known to lie between 0 and 1.

9.1.2 Supply and demand analysis

In Section 1.3 we introduced a simple model of supply and demand for a single good in an iso-

lated market. If we assume that the supply and demand functions are both linear then we have

the relations

= aP − b

=−cP + d

for some positive constants, a, b, c and d. (Previously, we have written P in terms of Q and have

sketched the supply and demand curves with Q on the horizontal axis and P on the vertical axis.

It turns out that it is more convenient in the present context to work the other way round and

to write Q as a function of P.) In writing down these equations, we are implicitly assuming that

only one time period is involved, that supply and demand are dependent only on the price in

this time period, and that equilibrium values are attained instantaneously. However, for cer-

tain goods, there is a time lag between supply and price. For example, a farmer needs to decide

precisely how much of any crop to sow well in advance of the time of sale. This decision is made

on the basis of the price at the time of planting and not on the price prevailing at harvest time,

which is unknown. In other words, the supply, Q

, in period t depends on the price, P

t−1

, in the

preceding period t − 1. The corresponding time-dependent supply and demand equations are

= aP

t−1

− b

=−cP

+ d

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If we assume that, within each time period, demand and supply are equal, so that all goods are

sold, then

= Q

that is,

−cP

+ d = aP

t−1

− b

This equation can be rearranged as

−cP

= aP

t−1

− b − d (subtract d from both sides)

=− P

t−1

+ (divide both sides by −c)

which is a difference equation of the standard form. The equation can therefore be solved in

the usual way and the time path analysed. Once a formula for P

is obtained, we can use the

demand equation

=−cP

+ d

to deduce a corresponding formula for Q

by substituting the expression for P

into the right-

hand side.

b + d

Dynamics

562

Example

Consider the supply and demand equations

= 4P

t−1

− 10

=−5P

+ 35

Assuming that the market is in equilibrium, ﬁnd expressions for P

and Q

when P

= 6. Is the system stable

or unstable?

Solution

= Q

then

−5P

+ 35 = 4P

t−1

− 10

which rearranges to give

−5P

= 4P

t−1

− 45 (subtract 35 from both sides)

=−0.8P

t−1

+ 9 (divide both sides by −5)

The complementary function is given by

CF = A(−0.8)

and for a particular solution we try

= D

for some constant D. Substituting this into the difference equation gives

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D =−0.8D + 9

which has solution D = 5. The general solution is therefore

= A(−0.8)

+ 5

The initial condition, P

= 6, gives

6 = A(−0.8)

+ 5 = A + 5

and so A is 1. The solution is

= (−0.8)

+ 5

An expression for Q

can be found by substituting this into the demand equation

=−5P

+ 35

to get

=−5[(−0.8)

+ 5] + 35

=−5(−0.8)

+ 10

As t increases, (−0.8)

converges to zero and so P

and Q

eventually settle down at the equilibrium levels

of 5 and 10 respectively. The system is therefore stable. Note also that because −0.8 lies between −1 and 0,

the time paths display oscillatory convergence.

9.1 • Difference equations

563

Practice Problems

4 Consider the supply and demand equations

= P

t−1

− 8

=−2P

+ 22

Assuming equilibrium, find expressions for P

and Q

when P

= 11. Is the system stable or unstable?

5 Consider the supply and demand equations

= 3P

t−1

− 20

=−2P

+ 80

Assuming equilibrium, find expressions for P

and Q

when P

= 8. Is the system stable or unstable?

Two features emerge from the previous example and Practice Problems 4 and 5. Firstly,

the time paths are always oscillatory. Secondly, the system is not necessarily stable and so equi-

librium might not be attained. These properties can be explained if we return to the general

equation

=− P

t−1

The coefﬁcient of P

t−1

is −a/c. Given that a and c are both positive, it follows that −a/c is

negative and so oscillations will always be present. Moreover,

 if a > c then −a/c <−1 and P

diverges

 if a < c then −1 <−a/c < 0 and P

converges.

b + d

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We conclude that stability depends on the relative sizes of a and c, which govern the slopes of

the supply and demand curves. Bearing in mind that we have chosen to consider supply and

demand equations in which Q is expressed in terms of P, namely

= aP − b

=−cP + d

we deduce that the system is stable whenever the supply curve is ﬂatter than the demand curve

when P is plotted on the horizontal axis.

Throughout this section we have concentrated on linear models. An obvious question to ask

is whether we can extend these to cover the case of non-linear relationships. Unfortunately, the

associated mathematics gets hard very quickly, even for mildly non-linear problems. It is usu-

ally impossible to ﬁnd an explicit formula for the solution of such difference equations. Under

these circumstances, we fall back on the tried and trusted approach of actually calculating the

ﬁrst few values until we can identify its behaviour. A spreadsheet provides an ideal way of doing

this, since the parameters in the model can be easily changed.

Dynamics

564

Example

Consider the supply and demand equations

= P

0.8

t−1

= 12 − P

(a) Assuming that the market is in equilibrium, write down a difference equation for price.

(b) Given that P

= 1, ﬁnd the values of the price, P

for t = 1, 2,..., 10 and plot a graph of P

against t.

Describe the qualitative behaviour of the time path.

Solution

(a) If

= Q

then

12 − P

= P

0.8

t−1

which rearranges to give

= 12 − P

0.8

t−1

Notice that this difference equation is not of the form considered in this section, so we cannot obtain

an explicit formula for P

in terms of t.

(b) We are given that P

= 1, so we can compute the values of P

, P

,...in turn. Setting t = 1 in the differ-

ence equation gives

= 12 − P

0.8

= 12 − 1

0.8

= 11

This number can now be substituted into the difference equation, with t = 2, to get

= 12 − P

0.8

= 12 − 11

0.8

= 5.190 52

and so on.

EXCEL

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Excel provides an easy way to perform the calculations. We type in the values, 0, 1, 2, ..., 10 for each

time period down the ﬁrst column, type in the value of the initial price, P

, in the second column, and then

copy the relevant formula down the second column to generate successive values of P

. Once this has been

done, Chart Wizard can be used to draw the time path.

The most appropriate way of representing the results graphically is to use a bar chart. We would like the

numbers in the ﬁrst column of the spreadsheet to act as labels for the bars on the horizontal axis.

Unfortunately, unless we tell Excel that we want to do this, it will actually produce two sets of bars on the

same diagram, using the numbers in column A as heights for the ﬁrst set of bars, and the numbers in col-

umn B as heights for the second set. This can be avoided by entering the values down the ﬁrst column as

text. This is done by ﬁrst highlighting column A and then selecting Format: Cells from the menu bar. We

choose the Number tab, and click on Text and OK. We can now ﬁnally enter the values of 0, 1, 2, ..., 10 in

the ﬁrst column, together with the headings and numerical value of P

in column 2 as shown in Figure 9.3.

The remaining entries in column B are worked out using the difference equation

= 12 − P

0.8

t−1

For example, the entry in cell B6 is P

which is calculated from

= 12 − P

0.8

The value of P

is located in cell B5, so we need to type the formula

=12−B5^0.8

into B6. Subsequent values are worked out in the same way, so all we need do is drag the formula down to

B15 to complete the table.

Chart Wizard can now be used to plot this time path. We ﬁrst highlight both columns and then click on

Chart Wizard. The bar chart that we want is the one that is automatically displayed, so we just press the

Finish button. To close the gaps between the bars, click on any one bar. A square dot will appear on each

bar to indicate that all bars have been selected. We then choose Format: Selected data series for the menu

bar. Finally, click on the Options tab, reduce the Gap Width to zero and click OK. The ﬁnal spreadsheet is

shown in Figure 9.4. It illustrates the oscillatory convergence and shows that the price eventually settles

down to a value just greater than 7.

9.1 • Difference equations

565



Figure 9.3

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Dynamics

566

Complementary function of a difference equation The solution of the difference equa-

tion, Y

= bY

t−1

+ c when the constant c is replaced by zero.

Difference equation An equation that relates consecutive terms of a sequence of numbers.

Dynamics Analysis of how equilibrium values vary over time.

Equilibrium value of a difference equation A solution of a difference equation that does

not vary over time; it is the limiting value of Y

as n tends to inﬁnity.

General solution of a difference equation The solution of a difference equation that

contains an arbitary constant. It is the sum of the complementary function and particular

solution.

Initial condition The value of Y

that needs to be speciﬁed to obtain a unique solution of

a difference equation.

Particular solution of a difference equation Any one solution of a difference equation

such as Y

= bY

t−1

+ c.

Recurrence relation An alternative phrase for a difference equation. It is an expression for

in terms of Y

n−1

(and possibly Y

n−2

, Y

n−3

etc).

Stable (unstable) equilibrium An economic model in which the solution of the associated

difference equation converges (diverges).

Uniformly convergent sequence A sequence of numbers which progressively increases (or

decreases) to a ﬁnite limit.

Uniformly divergent sequence A sequence of numbers which progressively increases (or

decreases) without a ﬁnite limit.

Key Terms

Figure 9.4

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9.1 • Difference equations

567

Practice Problems

6 Calculate the first four terms of the sequences defined by the following difference equations. Hence

write down a formula for Y

in terms of t. Comment on the qualitative behaviour of the solution in

each case.

(a)

= Y

t−1

+ 2; Y

= 0 (b) Y

=−Y

t−1

+ 6; Y

= 4 (c) Y

= 0Y

t−1

+ 3; Y

= 3

7 Solve the following difference equations with the specified initial conditions:

(a)

= Y

t−1

+ 6; Y

= 1 (b) Y

=−4Y

t−1

+ 5; Y

= 2

Comment on the qualitative behaviour of the solution as t increases.

8 Show, by substituting into the difference equation, that

= A(b

) + D where D =

is a solution of

= bY

t−1

+ c (b ≠ 1)

9 Consider the two-sector model:

= C

+ I

= 0.7Y

t−1

+ 400

= 0.1Y

t−1

+ 100

Given that Y

= 3000, find an expression for Y

. Is this system stable or unstable?

10 Consider the supply and demand equations

= 0.4P

t−1

− 12

=−0.8P

+ 60

Assuming that equilibrium conditions prevail, find an expression for P

when P

= 70. Is the system

stable or unstable?

11 The Harrod–Domar model of the growth of an economy is based on three assumptions.

(1) Savings, S

, in any time period are proportional to income, Y

, in that period, so that

=αY

(α>0)

(2) Investment, I

, in any time period is proportional to the change in income from the previous

period to the current period so that

=β(Y

− Y

t−1

)(β>0)

(3) Investment and savings are equal in any period so that

= S

Use these assumptions to show that

= Y

t−1

and hence write down a formula for Y

in terms of Y

. Comment on the stability of the system in the

case when α=0.1 and β=1.4, and write down expressions for S

and I

in terms of Y

β−α

1 − b



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12 Consider the difference equation

= 0.1Y

t−1

+ 5(0.6)

(a) Write down the complementary function.

(b) By substituting Y

= D(0.6)

into this equation, find a particular solution.

cific solution that satisfies the initial condition, Y

= 9.

(d) Is the solution in part (c) stable or unstable?

13 Consider the difference equation

= 0.2Y

t−1

+ 0.8t + 5

(a) Write down the complementary function.

(b) By substituting Y

= Dt + E into this equation, find a particular solution.

cific solution that satisfies the initial condition, Y

= 10.

(d) Is the solution in part (c) stable or unstable?

14 (Excel) Consider the two-sector model

= 100 + 0.6Y

0.8

t−1

= C

+ 60

(a) Write down a difference equation for Y

(b) Given that Y

= 10, calculate the values of Y

for t = 1, 2,...8 and plot these values on a dia-

gram. Is this system stable or unstable?

15 (Excel) An economic growth model is based on three assumptions:

(1) Aggregate output, y

, in time period t depends on capital stock, k

, according to

= k

0.6

(2) Capital stock in time period t + 1 is given by

t+1

= 0.99k

+ s

where the first term reflects the fact that capital stock has depreciated by 1%, and the second

term denotes the output that is saved during period t.

(3) Savings during period t are one-fifth of income so that

= 0.2y

(a) Use these assumptions to write down a difference equation for k

(b) Given that k

= 7000, find the equilibrium level of capital stock and state whether k

displays

uniform or oscillatory convergence. Do you get the same behaviour for other initial values of

capital stock?

Dynamics

568

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section 9.2

Differential equations

A differential equation is an equation that involves the derivative of an unknown function.

Several examples have already been considered in Chapter 6. For instance, in Section 6.2 we

noted the relationship

= I

where K and I denote capital stock and net investment, respectively. Given any expression for

I(t), this represents a differential equation for the unknown function, K(t). In such a simple

case as this, we can solve the differential equation by integrating both sides with respect to t.

For example, if I(t) = t the equation becomes

= t

and so

K(t) =



t dt =+c

where c is a constant of integration. The function K(t) is said to be the general solution of the

differential equation and c is referred to as an arbitrary constant. Some additional information

Objectives

At the end of this section you should be able to:

 Find the complementary function of a differential equation.

 Find the particular solution of a differential equation.

 Analyse the stability of economic systems.

 Solve continuous time national income determination models.

 Solve continuous time supply and demand models.

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