Mathematical methods for economic theory: 9.1 First-order difference equations

9.1 First-order difference equations

Definition: A first-order difference equation is an equation
x_t = f(t, x_t−1),
where f is a function of two variables. A solution of the first-order difference equation x_t = f(t, x_t−1) is a function x of a single variable whose domain is the set of integers such that x_t = f(t, x_t−1) for every integer t, where x_t 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 x_t 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 x_t 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 9.1.1: For every number x₀, every first-order difference equation x_t = f(t, x_t−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(t, x_t−1) = ax_t−1 + b_t
for numbers a and b_t, t = 1, ....

Notice that in this definition the number a that multiplies x_t−1 is independent of t, whereas b_t 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

x_t = a^tx₀ + ∑t
k=1a^t−kb_k 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

ax_t−1 + b_t	=	a(a^t−1x₀ + ∑t−1 k=1a^t−1−kb_k) + b_t
	=	a^tx₀ + ∑t−1 k=1a^t−kb_k + b_t
	=	a^tx₀ + ∑t k=1a^t−kb_k
	=	x_t,

so that the solution is correct.

Proposition 9.1.2: For any given value of x₀, the unique solution of the linear first-order difference equation with constant coefficient
x_t = ax_t−1 + b_t
is given by
x_t = a^tx₀ + ∑t
k=1a^t−kb_k for all t
if a ≠ 0 and x_t = b_t for all t if a = 0.

If b_t = b for all t = 1,... we have

x_t = a^tx₀ + b∑t−1
j=0a^j 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

x_t = a^tx₀ + b·(1 − a^t)/(1 − a)

if a ≠ 1, so that we have the following result.

Proposition 9.1.3: For any given value of x₀, the unique solution of the linear first-order difference equation with constant coefficient
x_t = ax_t−1 + b
is given by
x_t = a^t(x₀ − b/(1 − a)) + b/(1 − a) for all t
if a ≠ 0 and a ≠ 1, by x_t = b for all t if a = 0, and by x_t = 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
x_t = f(t, x_t−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 x_t = ax_t−1 + b, the following result is immediate.

Proposition 9.1.4: The linear first-order difference equation with constant coefficient
x_t = ax_t−1 + b
with a ≠ 1 has a unique equilibrium, b/(1 − a).

Given this result, we can rewrite the solution of the equation x_t = ax_t−1 + b for a ≠ 1 as

x_t = a^t(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
x_t = ax_t−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 x_t = ax_t−1 + b is

x_t = a^t(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 x_t is monotone—if x₀ < b/(1 − a) it increases, and if x₀ > b(/1 − a) it decreases—and if −1 < a < 0 then x_t oscillates (because a^t is negative if t is odd and positive if t is even).

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

Thus we have the following result.

Proposition 9.1.5

The solution of the linear first-order difference equation with constant coefficient

x_t = ax_t−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 9.1.1

Suppose that demand depends upon the current price

D_t = max{γ − δp_t, 0},

where γ > 0 and δ > 0, whereas supply depends on the previous price

S_t = max{(p_t−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 p_t 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 D_t = S_t, so that

p_t = −(1/(βδ))p_t−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

p_t = 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 D_t and S_t are no longer relevant.

In the previous example b_t is independent of t. In the next example it depends on t.

Example 9.1.2

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

z_t = (1 + r)z_t−1 − (1 + i)^t−1y,

a first-order difference equation with constant coefficient. So by a previous result we have

z_t	=	(1 + r)^tz₀ − ∑t k=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)],

z_t	=	(1 + r)^tz₀ − ∑t k=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

z_t	=	(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/(r−i))(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

x_t = a_tx_t−1 + b_t

where the coefficient of x_t−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=ra_s denotes the product a_ra_r+1···a_t 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 9.1.6: For any given value x₀, the unique solution of the first-order difference equation
x_t = a_tx_t−1 + b_t
is given by
x_t = (Πt
s=1a_s)x₀ + ∑t
k=1(Πt
s=k+1a_s)b_k for all t
if a_t ≠ 0 for all t.

Example 9.1.3

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 r_t and i_t respectively, the relation between assets in periods t and t − 1 is

z_t = (1 + r_t)z_t−1 − Πt−1
s=1(1 + i_s)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

z_t

(Πt
s=1(1 + r_s))z₀ − ∑t
k=1(Πt
s=k+1(1 + r_s))(Πk−1
s=1(1 + i_s))y for all t.

Your first name*
Your last name*
Your email address*
Comment*
Enter the first six letters of the alphabet*	(to help establish that you are human)