Discrete time: First order difference equations
Introduction
When dealing with discrete time, the value of variable y will change only when the
variable changes from one integer to the next.
The pattern of change is now represented by difference quotient
counterpart of
𝑑𝑦
𝑑𝑡
Δ𝑦
Δ𝑡
which is the discrete
. t can only now take on integer values.
Therefore, comparing the value of y in two consecutive periods, we must have ∆t = 1.
This implies that the difference quotient
Δ𝑦
Δ𝑡
can be simplified to ∆y. This is called the
first difference of y.
∆ means difference
In our case, we define the first difference as follows.
∆yt = yt+ 1 - yt
yt – means value of y in tth period
yt +1 – means value y in t+1 period.
With this then we can describe the pattern of change of y by an equation such as,
∆yt = 2 …………………. (i)
OR
∆yt = -0.1yt ……………... (ii)
Such equations are known as difference equations. We can rewrite equation (i) and (ii)
as follows
yt +1 - yt = 2 ………………(iii)
OR
yt+1 = yt + 2
yt+1 – yt = -0.1yt ---------------------(iv)
OR
yt+1 – 0.9yt = 0
1
yt+1 = 0.9yt
Difference equations can be either linear or nonlinear, homogenous or nonhomogeneous
and of the first or second (or higher) orders.
Equations (i) for instance can be classified as linear for there is no y term raised to the
second or higher power. Nonhomogeneous because the right-hand side is non zero. The
first order because there exists only a first difference ∆yt.
Equations (ii) can be characterized as having constant coefficients and constant term.
We only consider difference equations with constant terms.
Solving a first order difference equation.
To solve a first order difference equation involves finding a time path yt. We can solve
the difference equation using iterative method. This method can only be understood by
working out examples.
Example 1
Find the solution to difference equation
yt+1 – yt = 2
(assuming yo initial value) = 15
It is more convenient to use the alternative
𝑦𝑡+1 = 𝑦𝑡 + 2
𝑦1 = 𝑦0 + 2
𝑦2 = 𝑦1 + 2 = (𝑦0 + 2) + 2 = 𝑦0 + 2(2)
𝑦3 = 𝑦2 + 2 = (𝑦0 + 2(2)) + 2 = 𝑦0 + 3(2)
In general
𝑦𝑡 = 𝑦0 + 𝑡(2) = 15 + 2𝑡
This is the solution as it can be used to find the value of yt for any time period.
Solve the difference equation
∆𝑦𝑡 = −0.1𝑦𝑡 , with the initial value of y being yo
2
𝑦𝑡+1 − 𝑦𝑡 = −0.1𝑦𝑡
𝑦𝑡+1 − 0.9𝑦𝑡 = 0
𝑦𝑡+1 = 0.9𝑦𝑡
𝑦1 = 0.9𝑦0
𝑦2 = 0.9𝑦1 = 0.9(0.9)𝑦0 = (0.9)2 𝑦0
𝑦3 = 0.9𝑦2 = 0.9(0.9)2 𝑦0 = (0.9)3 𝑦0
In general the solution can be written as
𝑦𝑡 = (0.9)𝑡 𝑦0
Solve the homogeneous difference equation
𝑀𝑦𝑡+1 − 𝑛𝑦𝑡 = 0
𝑀𝑦𝑡+1 = 𝑛𝑦𝑡
𝑛
𝑦𝑡+1 = 𝑚 𝑦𝑡
The solution is
𝑛
𝑦𝑡 = (𝑚)𝑡 𝑦0
𝑛
It is through ( )𝑡 that various values of t will lead to their corresponding values of y. It
𝑚
therefore corresponds to 𝑒 −𝑎𝑡 in the solution of the differential equations. If we write it
more generally as 𝑏𝑡 (b for base) and attach the more general multiplicative constant A
(instead of yo) it gives the general solution for the homogenous difference equation.
𝑦𝑡 = 𝐴𝑏𝑡
This expression 𝐴𝑏𝑡 plays the same important role in difference equations as the
expression 𝐴𝑒 −𝑎𝑡 in the differential equation.
The general method
Suppose that we are seeking the solution to the first difference equation
𝑦𝑡+1 + 𝑎𝑦𝑡 = 𝑐
3
Where a and c are constants. The general solution will consist of the sum of two
components.
1. A particular integral yp – which is any solution of the complete non homogeneous
equation.
2. A complementary function yc – which is the general solution of the reduced
equation
𝑦𝑡+1 + 𝑎𝑦𝑡 = 0
yp represents intertemporal equilibrium level of y.
yc represents deviations for the time path from that equilibrium.
𝑦𝑐 + 𝑦𝑝 constitute the general solution due to the presence of the arbitrary constants.
First, find complementary function
𝑦𝑡 = 𝐴𝑏𝑡
𝑦𝑡+1 = 𝐴𝑏𝑡+1
So that
𝑦𝑡+1 + 𝑎𝑦𝑡 = 0
𝐴𝑏 𝑡+1 + 𝑎𝐴𝑏𝑡 = 0
So that
𝐴𝑏𝑡 𝑏1 + 𝑎𝐴𝑏 𝑡 = 0
(𝐴𝑏1 + 𝐴𝑎)𝑏𝑡 = 0
Divide throughout by 𝑏𝑡
𝐴𝑏 + 𝐴𝑎 = 0
𝑏+𝑎 =0
hence
𝑏 = −𝑎
Therefore set 𝑏 = −𝑎
𝑦𝑐 = 𝐴𝑏𝑡 = 𝐴(−𝑎)𝑡
The particular integral
Assume that 𝑦𝑡 = 𝑘 (constant)
𝑦𝑡+1 = 𝑘
4
𝑘 + 𝑎𝑘 = 𝑐
𝑐
→
𝑘 = 1+𝑎
Therefore, the particular integral can be written as
𝑐
𝑦𝑝 = 𝑘 = 1+𝑎
(𝑎 ≠ −1)
If it happens that 𝑎 = −1, then 𝑦𝑝 is not defined and therefore another solution must be
sought. In this case let,
𝑦𝑡 = 𝑘𝑡
𝑦𝑡+1 = 𝑘(𝑡 + 1)
𝑘 (𝑡 + 1) + 𝑎𝑘𝑡 = 𝑐
𝑘=
𝑐
because a = -1
𝑡+1+𝑎𝑡
𝑦𝑝 = 𝑘𝑡 = 𝑐𝑡 is a non-constant function of t. it represents the moving equilibrium.
Adding 𝑦𝑐 𝑎𝑛𝑑 𝑦𝑝 gives the general solution
𝑐
𝑦𝑡 = 𝐴(−𝑎)𝑡 + 1+𝑎
(general solution a ≠ -1)
𝑦𝑡 = 𝐴(−𝑎)𝑡 + 𝑐𝑡 = 𝐴 + 𝑐𝑡 (general solution a = -1)
Corresponding definite solutions
Let 𝑡 = 0
𝑐
𝑦0 = 𝐴 + 1+𝑎
𝑦𝑡 = [𝑦0 −
𝑐
1+𝑎
→
] (−𝑎)𝑡 +
𝑦𝑡 = 𝐴 + 𝑐𝑡
𝑦𝑡 = 𝑦0 + 𝑐𝑡
→
𝑐
𝐴 = 𝑦0 − 1+𝑎
𝑐
1+𝑎
(𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑎 ≠ −1)
𝑦0 = 𝐴
(𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑎 = −1)
Example
Solve 𝑦𝑡+1 − 5𝑦𝑡 = 1
7
𝑦0 = 4
5
𝑦𝑐 = 𝐴(−𝑎)𝑡 = 𝐴(− − 5)𝑡 = 𝐴(5)𝑡
𝑐
1
1
𝑦𝑝 = 1+𝑎 = 1−5 = − 4
1
𝑦𝑡 = 𝑦𝑐 + 𝑦𝑝 = 𝐴(5)𝑡 − 4
(𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛)
When t = 0
1
𝑦0 = 𝐴 − 4
1
→
𝐴 = 𝑦0 + 4
1
1
4
4
𝑦𝑡 = (𝑦0 + ) (5)𝑡 −
7
1
1
= (4 + 4) (5)𝑡 − 4
1
= 2(5)𝑡 − 4
(𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛)
The dynamic stability of equilibrium
In the continuous time case, the dynamic stability of equilibrium depends on the 𝐴𝑒 −𝑎𝑡
term in the complementary function. In period analysis, it depends on 𝐴𝑏𝑡 the
complementary function. However, its interpretation is more complicated than in the
case of 𝐴𝑒 −𝑎𝑡 .
Significance of b
Whether the equilibrium is dynamically stable or not is a question of whether the
complementary function will tend to zero as t → ∞. So we must analyze the path of 𝐴𝑏𝑡
as t increased indefinitely. The value of b is important in this regard. We first consider
its significance alone disregarding the coefficient A (by assuming its equal to 1). We
can divide the range of possible values of b (-∞, +∞) into seven distinct regions
1
2
3
4
5
6
7
b>1
b=1
0 < b< 1
b=0
-1< b< 0
b = -1
b < -1
|b|>1
|b|=1
|b|<1
|b|=0
|b|<1
|b| = 1
|b| > 1
𝑡
(2)
(1)𝑡
(1/2)𝑡
(−1/2)𝑡
(−1)𝑡
(−2)𝑡
t=0
1
1
1
0
1
1
1
t=1
2
1
1/2
0
- 1/2
-1
-2
6
t=2
4
1
1/4
0
1/4
1
4
t=3
8
1
1/8
0
- 1/8
-1
-8
In each region, bt generates a different type of time path.
i.
ii.
iii.
iv.
v.
vi.
vii.
When b > 1, bt must increase, t increases to ∞
When b = 1, bt will remain at unity for all values of t.
(0 < b< 1), bt represent a positive fraction raised to integer power. b t decreases
as t increases but remains positive.
When b = 0, bt remains at zero for all values of t (not of interest since b≠0)
When (-1 < b< 0), bt alternates between positive and negative values from period
to period. The alternating path tend closer and closer to the horizontal axis.
When (b = -1), bt alternate between positive and negative one.
When (b < -1), bt alternate between positive and negative values and deviates
further and further from the horizontal axis
In summary, the time path of bt (b ≠ 0) will be
Non oscillatory
if
b>0
Oscillatory
if
b<0
Divergent
if
|b| > 1
Convergent
if
|b| < 1
The convergence of the expression bt depends on the absolute value b while that of the
continuous case depends on the sign of a in 𝑒 −𝑎𝑡
The role of A
First the magnitude of A can ‘blow up’ or ‘pare down’ the value of bt. It has a scale effect
without changing the basic configuration of the time path. The sign of A does materially
affect the change of the path because if multiplied by bt it will lead to change of bt if A
is negative. Thus, a negative A produces a mirror effect as well as a scale effect.
Convergence of equilibrium
Based on the above discussion, we can interpret the Abt as the complementary function.
Abt represents the deviations from some intertemporal equilibrium level.
In summary,
𝑦𝑡 = 𝐴𝑏𝑡 + 𝑦𝑝 is convergent iff
|b| < 1
7
Economic applications
Cobweb model
Illustrates first order difference equation application in economic analysis.
In this model, Qs is a function of price of the preceding time period and not current price.
The model considers a situation in which a producer’s output decision is made one period
in advance of the actual sale – such as in agriculture production – where planting must
precede by appropriate length of time to the harvesting and sale of the output. Assume
the output decision in period t is based on the prevailing price Pt. Since the output is not
available for sale until period (t+1). Pt determines not Qst but Qst+1. So, we have a lagged
supply function.
𝑄𝑠,𝑡+1 = 𝑆(𝑝𝑡 )
𝑄𝑠𝑡 = 𝑆(𝑃𝑡−1 )
or
When this supply function interacts with a demand function of the form
𝑄𝑑𝑡 = 𝐷(𝑃𝑡 ),
an interesting dynamic price pattern will result. Let us assume a linear version of the (lagged)
supply and (unlagged) demand functions and assume that in each time period, the market price
is always set at a level which clears the market. The market model is given as
𝑄𝑑𝑡 = 𝑄𝑠𝑡
𝑄𝑑𝑡 = 𝛼 − 𝛽𝑃𝑡
( 𝛼, 𝛽 > 0)
𝑄𝑠𝑡 = −𝛾 + 𝛿𝑃𝑡−1
(𝛾, 𝛿 > 0)
𝛼 − 𝛽𝑃𝑡 = −𝛾 + 𝛿𝑃𝑡−1
𝛽𝑃𝑡 + 𝛿𝑃𝑡−1 = 𝛼 + 𝛾
To solve the difference equation, normalize first and shift the time subscript ahead by one
period.
𝛿
𝑃𝑡 + 𝛽 𝑃𝑡−1 =
𝛼+𝛾
𝛽
𝛿
𝛼+𝛾
𝛽
𝛽
𝑃𝑡+1 + 𝑃𝑡 =
𝑦𝑡+1 + 𝑎𝑦𝑡 = 𝑐
which is similar to the equation
𝛿
𝑦 = 𝑝, 𝑎 = 𝛽 , 𝑐 =
8
𝛼+𝛾
𝛽
Since 𝛿 𝑎𝑛𝑑 𝛽 are both positive → 𝑎 ≠ −1, therefore we apply the following solution
𝑐
𝑦𝑡 = 𝐴(−𝑎)𝑡 + 1+𝑎
𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝛼+𝛾
𝛿 𝑡
𝛽
𝑦𝑡 = 𝐴(− ) +
𝛽
1+𝑎
𝛼+𝛾
𝛿 0
𝛽
𝑦0 = 𝐴(− ) +
𝛽
1+𝑎
𝑦0 = 𝐴 +
𝛼+𝛾
𝛽+𝛿
𝛿
𝛼+𝛾
𝑦𝑡 = 𝐴(− 𝛽)𝑡 + 𝛽+𝛿
𝑐
the general solution
𝑐
𝑦𝑡 = (𝑦0 − 1+𝑎) (−𝑎)𝑡 + 1+𝑎
𝑐
1+𝑎
=
𝛼+𝛾
𝛽
𝛿
1+
𝛽
=
𝛼+𝛾
𝛽
×
𝛽
𝛽+𝛿
𝛿 𝑡
𝛼+𝛾
𝛼+𝛾
= 𝛽+𝛿
𝛼+𝛾
𝑃𝑡 = (𝑃0 − 𝛽+𝛿 ) (− 𝛽) + 𝛽+𝛿
𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
The cobwebs
Three points regarding the time path
𝛼+𝛾
1. The expression 𝛽+𝛿 which constitutes the particular integral of the difference equation can
be taken as the intertemporal equilibrium price of the model.
𝛼+𝛾
𝑃̅ = 𝛽+𝛿 it is a stationary equilibrium.
If we substitute
𝛼+𝛾
𝛽+𝛿
with 𝑃̅ in the equation, we have,
𝑡
𝛿
𝑃𝑡 = (𝑃𝑜 − 𝑃̅) (− 𝛽) + 𝑃̅
9
2. Significance of the expression (𝑃0 − 𝑃̅) – corresponds to A in term 𝐴𝑏𝑡 . It’s negative sign
will have a mirror effect and its magnitude will have scale effects.
3. The expression
−𝛿
𝛽
which corresponds to b-component of 𝐴𝑏𝑡 . 𝛽, 𝛿 > 0. We can deduce an
oscillatory time path.
oscillatory time paths
Explosive
Uniform
Damped
It is this that gives rise to the cobweb phenomena. Three possible
𝛿> 𝛽
𝛿=𝛽
𝛿< 𝛽
if
if
if
An example:
Find the time path of price given the following
Qdt =10 – 2Pt
Qst = -4 + 3Pt-1
Qdt= Qst
10 – 2Pt = -4 + 3Pt-1
10 + 4 = 2Pt + 3Pt-1
2Pt + 3Pt-1 = 14
Pt + (3/2)Pt-1 = 7
Pt+1 + (3/2)Pt = 7, the equation is equivalent to
3
𝑦𝑡+1 + 𝑎𝑦𝑡 = 𝑐
𝑦 = 𝑝, 𝑎 = 2 , 𝑐 = 7
3
7
𝑃𝑡 = 𝐴(− 2)𝑡 + 1+3/2
3
𝑃𝑡 = 𝐴(− 2)𝑡 +
3
14
(𝑃0 −
14
5
𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
5
𝑃0 = 𝐴(− 2)0 +
𝑃0 = 𝐴 +
𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
14
5
14
5
)=𝐴
𝑃𝑡 = (𝑃0 −
14
5
3
)(− 2)𝑡 +
14
5
𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
10
11