# Mathematical methods for economic theory

Martin J. Osborne

## 9.1 First-order difference equations

Definition
A first-order difference equation is an equation
xt = f(txt−1),
where f is a function of two variables. A solution of the first-order difference equation xt = f(txt−1) is a function x of a single variable whose domain is the set of integers such that xt = f(txt−1) for every integer t, where xt denotes the value of x at t.
When studying differential equations, we denote the value at t of a solution x by x(t). I follow convention and use the notation xt for the value at t of a solution x of a difference equation. In both cases, x is a function of a single variable, and we could equally well use the notation x(t) rather than xt when studying difference equations.

We can find a solution of a first-order difference equation by successive calculation: given the value of x for some value of t, we can use the equation to find the value of x at the next value of t, and then use the equation again to find the value of x at the following value of t, and so forth. For example, given the value x0 of x at 0, we have

 x1 = f(1, x0) x2 = f(2, x1) = f(2, f(1, x0)) and so on.
In fact, this argument establishes that every first-order difference equation has a unique solution: given any value of x0, there exists a unique solution x, with values x1, x2, ....
Proposition
For every number x0, every first-order difference equation xt = f(txt−1) has a unique solution in which the value of x is x0 at 0.
However, calculating the solution in this way doesn't tell us much about the properties of the solution. We would like to have a formula for the solution. For some functions f, such formulas exist.

### First-order linear difference equations with constant coefficient

Definition
A linear first-order difference equation with constant coefficient is a first-order difference equation for which
f(txt−1) = axt−1 + bt
for numbers a and bt, t = 1, ....
Notice that in this definition the number a that multiplies xt−1 is independent of t, whereas bt may depend on t.

If you use the method of successive calculation, you notice a pattern; you can guess that the solution takes the form

xt = atx0 + ∑t
k=1
atkbk for all t.
To check that the function x defined in this way is indeed the unique solution, it is enough to verify that it satisfies the equation. We have
 axt−1 + bt = a(at−1x0 + ∑t−1k=1at−1−kbk) + bt = atx0 + ∑t−1k=1at−kbk + bt = atx0 + ∑tk=1at−kbk = xt,
so that the solution is correct.
Proposition
For any given value of x0, the unique solution of the linear first-order difference equation with constant coefficient
xt = axt−1 + bt
is given by
xt = atx0 + ∑t
k=1
atkbk for all t
if a ≠ 0 and xt = bt for all t if a = 0.
If bt = b for all t = 1,... we have
xt = atx0 + bt−1
j=0
aj for all t.
Now, the sum of any finite geometric series 1 + a + a2 + ... + an−1 is given by
1 + a + a2 + ... + an−1 = (1 − an)/(1 − a)
if a ≠ 1. Thus
xt = atx0 + b·(1 − at)/(1 − a)
if a ≠ 1, so that we have the following result.
Proposition
For any given value of x0, the unique solution of the linear first-order difference equation with constant coefficient
xt = axt−1 + b
is given by
xt = at(x0 − b/(1 − a)) + b/(1 − a) for all t
if a ≠ 0 and a ≠ 1, by xt = b for all t if a = 0, and by xt = x0 + tb if a = 1.

#### Equilibrium and stability

An equilibrium of a first-order difference equilibrium is defined in the same way as an equilibrium of a first-order initial value problem: a value x* of the variable such that the equation has a solution in which the variable is equal to x* for all t.
Definition
If for some value x* of x0 the first-order difference equation
xt = f(txt−1)
has a solution that is equal to x* for all t, x* is an equilibrium of the equation.
Given the formula for a solution of the equation xt = axt−1 + b, the following result is immediate.
Proposition
The linear first-order difference equation with constant coefficient
xt = axt−1 + b
with a ≠ 1 has a unique equilibrium, b/(1 − a).
Given this result, we can rewrite the solution of the equation xt = axt−1 + b for a ≠ 1 as
xt = at(x0 − x*) + x*,
where x* = b/(1 − a).

An equilibrium of a first-order difference equation, like an equilibrium of a first-order differential equation, is stable if the solution of the equation converges to the equilibrium for all initial conditions.

Definition
If, for every value of x0, the solution of the linear first-order difference equation with constant coefficient
xt = axt−1 + b
with a ≠ 1 converges to the equilibrium b/(1 − a) as t increases without bound, then the equilibrium is (globally) stable.

By a previous result, the solution of a first-order difference equation of the form xt = axt−1 + b is

xt = at(x0 − b/(1 − a)) + b/(1 − a) for all t.
If |a| < 1 the first term converges to zero, so the equilibrium b/(1 − a) is stable. If 0 < a < 1 then xt is monotone—if x0 < b/(1 − a) it increases, and if x0 > b(/1 − a) it decreases—and if −1 < a < 0 then xt oscillates (because at is negative if t is odd and positive if t is even).

If |a| > 1, the absolute value of xt increases without bound, so that the equilibrium is not stable: the system “explodes”. If a > 1 then xt is monotone—if x0 < b/(1 − a) it decreases, and if x0 > b(/1 − a) it increases—and if a < −1 then xt oscillates.

Thus we have the following result.

Proposition
The solution of the linear first-order difference equation with constant coefficient
xt = axt−1 + b
with a ≠ 1 behaves as follows.
|a| < 1
Solution converges to the equilibrium b/(1 − a) (equilibrium is stable)
0 < a < 1
Solution is monotonic
−1 < a < 0
Solution oscillates, with decreasing amplitude
|a| > 1
Solution diverges
a > 1
Divergence is monotonic
a < −1
Solution oscillates, with increasing amplitude.
Example
Suppose that demand depends upon the current price
Dt = max{γ − δpt, 0},
where γ > 0 and δ > 0, whereas supply depends on the previous price
St = max{(pt−1 − α)/β, 0},
where α > 0 and β > 0. (Maybe the lag exists because production takes time—think of agricultural production.) Demand and supply are each specified not simply as a linear function, but the maximum of the value of a linear function and zero, because neither of them can be negative.

The following argument assumes that pt starts in the range for which the both demand and supply are positive, and remains in this range, so that only the linear parts of the specifications are relevant.

For equilibrium in period t we need Dt = St, so that

pt = −(1/(βδ))pt−1 + (α + βγ)/(βδ),
a first-order difference equation with constant coefficient. The equilibrium price is
p* = (α + βγ)/(1 + βδ),
so from a previous result, we can write the solution as
pt = p* + (−1/(βδ))t(p0 − p*).
From another result, the equilibrium is stable if
βδ > 1.
A solution path for such parameters is shown in the following figure.

If the system is unstable, then eventually a boundary is reached, and the linear parts of the specifications of Dt and St are no longer relevant.

In the previous example bt is independent of t. In the next example it depends on t.
Example
You have assets of \$z0. You can earn a constant rate of return r on these assets. That is, after t years you will have (1 + r)tz0 if you do not consume any assets. The rate of inflation is i, where i < r. In each period t ≥ 1 you withdraw an amount of money equivalent in purchasing power to y in period 1. (That is, you withdraw y(1 + i)t−1 in each period t.)
• How long will your assets last if r = 0.08, i = 0.04, y = \$50,000, and z0 = 1,000,000?
• How large do you assets need to be to last for 30 years of retirement if you want to withdraw \$80,000 per year during retirement (with r = 0.08 and i = 0.04)?
We have
zt = (1 + r)zt−1 − (1 + i)t−1y,
a first-order difference equation with constant coefficient. So by a previous result we have
 zt = (1 + r)tz0 − ∑tk=1(1 + r)t−k(1 + i)k−1y = (1 + r)tz0 − y(1 + r)t−1[1 − ((1 + i)/(1 + r))t]/[1 − (1 + i)/(1 + r)],
 zt = (1 + r)tz0 − ∑tk=1(1 + r)t−k(1 + i)k−1y = (1 + r)tz0 − y(1 + r)t−1 × [1 − ((1 + i)/(1 + r))t]/[1 − (1 + i)/(1 + r)],
using the formula for the sum of a finite geometric series, so
 zt = (1 + r)tz0 − y(1 + r)t[1 − ((1 + i)/(1 + r))t]/(r − i) = (1 + r)t[z0 − y(1 − ((1 + i)/(1 + r))t)/(r − i)].
Thus your assets last until the period t for which
z0 = (y/(ri))(1 − ((1 + i)/(1 + r))t).
Hence
t =
 log(1−(r−i)z0/y) log((1 + i)/(1 + r))
.
Thus in the example we have t = 42.6 years. The amount of money you need to last 30 years at \$80,000 per year is z0 = \$1,355,340. (Your total assets first increase, then gradually decrease.)

### First-order linear difference equations with variable coefficient

The solution of the equation
xt = atxt−1 + bt
where the coefficient of xt−1 depends on t, can be found, as in the case in which the coefficient is independent of t, by successive calculation. In the following result, Πt
s=r
as denotes the product arar+1···at if r ≤ t and is taken to be 1 if r > t. The result can be proved by checking that the solution satisfies the equation.
Proposition
For any given value x0, the unique solution of the first-order difference equation
xt = atxt−1 + bt
is given by
xt = (Πt
s=1
as)x0 + ∑t
k=1
t
s=k+1
as)bk for all t
if at ≠ 0 for all t.
Example
Modify the previous example by allowing the rate of return and rate of inflation to depend on t. Denoting the rate of return and rate of inflation in period t by rt and it respectively, the relation between assets in periods t and t − 1 is
zt = (1 + rt)zt−1 − Πt−1
s=1
(1 + is)y.
(The second term on the right-hand side is the amount of money in period t that has the same purchasing power as y in period 1.)

By the previous result, the solution of this difference equation is given by

 zt = (Πts=1(1 + rs))z0 − ∑tk=1(Πts=k+1(1 + rs))(Πk−1s=1(1 + is))y for all t.