Difference Equations

• Differential Equations are for continuous functions and their derivatives, $\frac{dy}{dt}$
• what about discretely increasing (step-wise) functions?

Now, consider the function,

$$I_{2024}=f(Y_{2023})$$

Here, investment, $I$ is the dependent variable in 2024, and we have a lagged (2023) in dependent variable, income $Y$.

What is a difference equation?

A difference equation expresses a relationship between the dependent variable and a lagged independent variable, which change at discrete intervals of time.

$$\Delta y=y_t-y_{t-k}$$

where $k$ is the lag.

• when we take time as a continuous variable then we need to use a Differential Equations, but when we use time as a discrete variable, we need to use a Difference Equations.

Standard form of Difference Equations

$$y_t=b{y}_{t-1}+a$$

• it is also called a First Order Linear Difference Equation

General Formula for Definite Solution

$$y_t=\left(y_0-\frac{a}{1-b}\right)b^t+\frac{a}{1-b};b\ne 1$$

• we call it definite because $y_0$ will be given.

If $b=1$,

$$y_t=y_0+\frac{a}{1-b}$$

By solving difference equation, our objective is to find the time path. The solution should be a function of time and must not contain any Difference Expression

Problems

Solve the following Difference Equations

1. Given the difference equation,

$$y_t=-7y_{t-1}+16;y_0=5\text{\hspace{1em}(1)}$$

2. Find the time path for the national income $y_t$ when consumption $c_t$, investments $I_t$ and $y_0$ are given as follows:¹

$$\begin{array}{rl}{c}_t& =60+0.8y_{t-1}\\ I_t& =50\\ y_0& =1300\end{array}\text{\hspace{1em}(2)}$$

3. Find the

$$\begin{array}{rl}{c}_t& =0.9y_t-1+90\\ I_t& =100\\ y_0& =1000\end{array}\text{\hspace{1em}(2)}$$

Solution 1

From the equation we can infer that $a=16$, $b=-7$

Applying these and $y_0=5$ into the solution for the time path we get,

$y_t=\left(5+\frac{16}{8}\right)(-7{)}^t+\frac{16}{8}$ $⟹\boxed{y_t=(-1{)}^t(7{)}^{t+1}+2}$

Solution 2

From the equation, $y=c+i+g$ we get,

$60+0.8y_{t-1}=y_t-I_t$

$60+0.8y_{t-1}=y_t-50$

$y_t=110+0.8y_{t-1}$

So, we find the parameters as $a=110$, $b=0.8$

Hence, the time path is $y_t=\left(1300-\frac{110}{0.2}\right)0.8^t+\frac{110}{0.2}$ $y_t=750(0.8{)}^t+550$

1300 - 110/0.2
110/0.2

Drawing the time path

There are four cases:

1. $b>1$

• grows exponentially

2. $b=0$

• Shrinks to zero at once.

3. $0<b<1$

• dampening and converging

Transclude of Difference-Equations-2024-09-23-09.59.13.excalidraw

4. $-1<b<0$

• dampening and oscillating → converting to $0$

Transclude of Difference-Equations-2024-09-23-10.05.39.excalidraw

After seeing 7 situations, there are four observations

If	Then
$\Vert b\Vert >1$	Time path explodes
$\Vert b\Vert <1$	Time path converges
$b>0$	Time path doesn’t oscillate
$b<0$	Time path oscillates

Footnotes

1. $y=c+i+g$ ↩