Lecture 4: Solving Linear Difference Equations

Spring Semester ’12-’13 Akila Weerapana

Lecture 4: Solving Linear Difference Equations

I. OVERVIEW

• In the first few lectures of this course we derived the IS-MP/AD-IA model of economic fluctuations. We showed how the IS-MP model could be represented using matrix algebra and solved to obtain the AD curve. However, we were unable to do the same for the AD- IA model because the IA curve was a difference equation, which related π _t to past values, specifically π t −1

• Solving a model with two variables x t and y _t turns out to be a lot easier than solving a model with x t and x t −1 . In today’s class and the next we will lay the mathematical groundwork needed to work with models where the key macroeconomic variables are dynamic in nature, which is a fancy way of saying that past values influence current values and current values influence future values.

• Many economic variables, known as ‘stock’ variables behave in this fashion. For example, the capital stock in a country or a firm is dependent not just on current variables like interest rates, investment, openness, corruption etc. but is also a function of how much capital we had in the last period.

• There are two basic types of models based on how they represent time: models where variables change in discrete time and models where variables change in continuous time. The economy is changing continuously so in many ways continuous time models would seem to be the most appropriate to use. On the other hand, data about the economy is only collected discretely, so if we ever want to test a model with data we would need to use a discrete time model. Rather than choosing one over the other we will discuss models in both continuous and discrete time when appropriate.

• Today’s lecture focuses on solving difference equations (discrete time). A future lecture will cover differential equations (continuous time).

II. SOLVING FIRST ORDER LINEAR DIFFERENCE EQUATIONS

• A first order linear difference equation is one that relates the value of a variable at a particular time in a a linear fashion to its value in the previous period as well as to other exogenous variables. In other words a first order linear difference equation is of the form y _t = α _t y _{t −1} + x _t where x is an exogenous variable.

• A classic example is the evolution of the capital stock in an economy K t+1 = (1 − δ)K t + I t

where I t is investment and δ is the depreciation rate. This equation says that the number of machines next period will be the undepreciated portion of this period’s capital stock plus any new machines we add to the economy.

• Another classic example is the level of government debt in the economy. D t = (1 + i)D _{t −1} + b _t

where D _t is debt, i is the nominal interest rate and b _t is the primary budget deficit this

year. So the debt level will depend on past debt, any interest payments on that debt and

the primary budget deficit (defined as the difference between government expenditures and

government revenues this period).

• The IA curve that we were interested in finding a solution for is another classic example:

π _t = π _{t −1} + λ ˆ Y _t + ϵ _t .

• An exact solution to a first order difference equation is a sequence of values {y t }, expressed as a function of α, x, time (t), and a single value of the endogenous variable: whether it be an initial condition (a value at the beginning) or a terminal condition (a value at the end), or some other arbitrary point in time.

• In other words solving a difference equation means finding a time path for the variable y t

that is consistent with the difference equation.

Method 1: The General Method

• The general method, as the name suggests, provides a systematic general framework for solving first-order difference equations of the form y t = αy t −1 + β with a known initial value y ₀ or terminal value y _T . Note that this is a more restrictive form than what we discussed above, in that α _t = α and x _t = x. This is called a first order difference equation with constant coefficients.

• I should also point out that this method can only be used when |α| ̸= 1.

• The solution to this difference equation consists of two parts: a part named the particu- lar solution, denoted y P which is any solution to the first order difference equation given above and a part named the homogenous solution, denoted y _H which solves the difference equation y _t = αy _{t −1} . The general solution is the sum of y _P and y _H .

• Let’s begin by thinking about the homogenous solution, i.e. the solution to y t ^H = αy _t ^H ₋₁ .

• This can be solved fairly easily. Since each value of y ^H is α times the previous value we can guess that y ^H _t = α ^t A (for some arbitrary constant A) would work as a solution. We can verify this easily: suppose y _t ^H = α ^t A, then y _t ^H ₋₁ = α ^{t −1} A and αy _t ^H ₋₁ = α(α ^{t −1} A) = (α ^t A), which is equivalent to y _t ^H .

• Another way we can arrive at the same solution is by iteration. y ^H _t = αy _t ^H ₋₁ = α(αy _t ^H ₋₂ ) = α ² y _t ^H ₋₂ . Subbing in again, we get y ^H _t = α ² (αy ^H _{t −3} ) = α ³ y ^H _{t −3} . Continuing the iteration, y _t ^H = α ^t y ^H ₀ . We can express this as y _H = α ^t A where A ≡ y ^H ₀ is an arbitrary constant.

• Now let’s think of the particular solution, which can be ANY solution to the equation y t = αy _{t −1} + β. Since any solution is allowed, we can, for example, find the easiest solution, one that sets all the y’s equal. If we set y = αy + β, we get the solution y ^P = _{1 −α} ^β Combining y _t = y ^P + y ^H we get the general solution of the form

y t = α ^t A + β 1 − α

• This is where knowing one additional value of y comes into play. Suppose that we knew the value of y 0 , then we can pin down the value of A. Setting t = 0 we get y 0 = A + _{1 −α} ^β ⇒ A = y 0 − _{1 −α} ^β . So the solution to the difference equation is of the form:

y _t = α ^t [

y ₀ − β 1 − α

]

+ β

1 − α

Example:

• We can illustrate the general method with a specific numerical example: Solve the difference equation y _t = 0.2y _{t −1} + 4 where y ₀ = 10 using the general method. Let’s begin by thinking about the homogenous solution y _t = 0.2y _{t −1} . The solution is of the form y ^H _t = 0.2 ^t A where A is a constant.

• The particular solution is the solution to y = 0.2y + 4 ⇒ y = 5 So y ^P = 5. Combining we get y t = 0.2 ^t A + 5.

• Since we know the value of y 0 we can pin down the value of Aby setting t = 0 to get y ₀ = A + 5 = 10 ⇒ A = 5 . So the solution to the difference equation is of the form:

y _t = 0.2 ^t (5) + 5

Method 2: Solving by Repeated Iteration

• Note that the general method only works for |α| ̸= 1. If α = 1 you have to use a different technique

• A difference equation can be solved fairly easily through repeated iteration either forward (if you have a terminal value) or backward (if you have an initial value). Note that this solution method does NOT require that α _t = α and x _t = x.

• Consider the simple example y t = αy t −1 + β with y 0 given. We can rewrite this as y _t = = α(αy _{t −2} + β) + β = α ² y _{t −2} + αβ + β

= α ² (αy _{t −3} + β) + αβ + β .. . = α ^t y ₀ + β(1 + α + · · · + α ^{t −1} ) y _t = α ^t y ₀ + β

( 1 − α ^t 1 − α

)

• Note that a slight rearrangement gives us the equation we had before y _t = α ^t y ₀ + β

( 1 − α ^t 1 − α

)

⇒ y t = α ^t y ₀ + β

( 1

1 − α )

− β ( α ^t

1 − α )

⇒ y t = α ^t (

y 0 − β 1 − α

) +

( β

1 − α )

• A big advantage of the repeated iteration method is that we can also handle the special cases α = 1 and α = −1 using this technique.

• In the case of α = 1, the difference equation y t = y _{t −1} + β with y ₀ given can be solved as y t = (y t −2 + β) + β = y t −2 + 2β

= (y t −3 + β) + 2β = y t −3 + 3β .. . = y ₀ + tβ

y _t = y ₀ + βt

• In the case of α = −1, the difference equation y t = −y t −1 + β with y 0 given can be solved as y t = −(−y t −2 + β) + β = y t −2

= ( −y t −3 + β) = −y t −3 + β

= −(−y t −4 + β) + β = y _{t −4} .. .

y t = y 0 if t is even y _t = β − y 0 if t is odd

Verifying Your Solution Using EXCEL

• For numerical difference equations, you can use Excel to verify your solution. First create a column of a hundred or so values of t from 0 upwards using drag and drop.

• Let the next column be the difference equation. Enter the value of y 0 into a cell and then program the cell below that to be equal to α times the prior cell plus β. Drag and paste for a few hundred rows and you have a set of y values consistent with the difference equation.

• In the third column, program in the formula: y t = α ^t (

y 0 − _{1 −α} ^β ) +

( β 1 −α

)

in one cell pointing to the entry with the t for that row, and then drag down. You now have the set of y values consistent with the solution. This column should match up to the second column’s entries if your solution is correct.

• If your EXCEL mojo is low, come see me and I can show you how to do this pretty quickly.

IV. DYNAMICS OF FIRST ORDER LINEAR DIFFERENCE EQUATIONS

Steady State Value

• Having solved for the time path, we can examine the dynamic behavior of a variable (i.e. its behavior over time). We will examine the stability of the difference equation as well as its steady state value.

• The steady state value is a value at which the endogenous variable exhibits no dynamic adjustment. In other words, given the difference equation y _t = αy _{t −1} + β, a steady state value y ^∗ exhibits the property that if y t −1 = y ^∗ then y t = y ^∗ for any t. A difference equation can have zero, one or (in the case of non-linear difference equations) more than one steady state values.

• In the simple linear first order difference equation where |α| ̸= 1, the steady state value is therefore y ^∗ = αy ^∗ + β ⇒ y ^∗ = _{1 −α} ^β .

• You can verify this by showing that if y t −1 = _{1 −α} ^β then y _t = _{1 −α} ^β . Stability

• A difference equation is said to be stable if lim

t →∞ y t exists. If lim

t →∞ y t does not exist then the

difference equation is said to be unstable.

• A stable difference equation will converge to a steady state value. For example, the difference equation y _t = 0.2y _{t −1} + 4 is stable because we can show using the solution y _t = 0.2 ^t (5) + 5 that lim _{t →∞} y _t = 5

• For the general form y t = αy _{t −1} +β, where |α| ̸= 1, which has the solution y t = α ^t y ₀ +β [ 1 −α

1 −α

] , it will only be stable if |α| < 1 in which case the steady state will be

t lim →∞ y _t = lim

t →∞

[

α ^t y ₀ + β

[ 1 − α ^t 1 − α

]]

= β

1 − α

• The equation will be unstable if α < −1 or α > 1. Why? Well if α < −1 or α > 1 then

t lim →∞ y _t = lim

t →∞

[

α ^t y ₀ + β

[ 1 − α ^t 1 − α

]]

is undefined, approaching either ∞ or −∞

• If α = 1, the solution is y t = y 0 + βt. This diverges unless β = 0. So except in the extremely uninteresting case where all the ys are the same (hence not even a difference equation per se), the difference equation with α = 1 diverges.

• If α = −1, the solution is

y t = y 0 if t is an even number y t = −y 0 + β if t is an odd number

• We can see that this sequence will oscillate between this two values as t gets larger. So the sequence will not converge unless y 0 = β/2.

• Finally, it is important to keep in mind that having a steady state and being stable are not synonymous. Having a steady state means that if the variable ever reaches that value, then it will not deviate. But being stable means that over time the variable will reach a steady state. It is possible to have a steady state that is never reached unless the economy happens to start there.

• For example, we showed above that the difference equation y t = −y t+1 + β has a steady state

at y ^∗ = β/2 but that it will not be stable unless y 0 = β/2. This is a good example of an

unstable equation with a steady state.