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Difference Equations - Lecture Notes | MATH 205, Study notes of Calculus

Wellesley CollegeCalculus

Material Type: Notes; Class: Multivariable Calculus; Subject: Mathematics; University: Wellesley College; Term: Unknown 1989;

Typology: Study notes

Pre 2010

Uploaded on 08/17/2009

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Fall Semester ’07-’08
Akila Weerapana
LECTURE 13: DIFFERENCE EQUATIONS
I. INTRODUCTION
•We are now entering the “dynamics” portion of this class, where we focus on models whose
endogenous variables are changing over time. The material is going to be unfamiliar to most
of you, since your math preparation was limited to Math 205 and Math 206 for the most part.
•The first two lectures cover theory and applications in the area of difference equations. Dif-
ference equations are the first step in our journey towards being able to solve dynamic opti-
mization problems. A difference equation links the value of an endogenous variable in a given
time period to its value in other time periods as well as to other exogenous variables.
•The solution to many dynamic optimization problems relate the value of variables to their
past values. So the basic description of the behavior of a variable is different: it is no longer
merely a function of other contemporaneous exogenous variables but is also a function of time-
lagged values of itself, and in some cases a function of time itself. 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.
•Next week, we switch to the study of differential equations: which are distinguished from
difference equations by the fact that the changes in variables happen continuously instead of
discretely as in the difference equations case.
II. SOLVING FIRST ORDER LINEAR DIFFERENCE EQUATIONS
•Afirst 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
yt=αyt−1+f(xt) where xis a vector of exogenous variables.
•A solution to a first order difference equation is a sequence of values {yt}, expressed as a
function of time and of the exogenous variables. In other words we need to find a time path
for the variable ythat is consistent with the difference equation.
•An exact solution to a differential equation also requires that we have 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). An example of a first order difference equation will be of the
form yt=αyt−1with initial condition y0= 10
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Download Difference Equations - Lecture Notes | MATH 205 and more Study notes Calculus in PDF only on Docsity!

Fall Semester ’07-’ Akila Weerapana

LECTURE 13: DIFFERENCE EQUATIONS

I. INTRODUCTION

  • We are now entering the “dynamics” portion of this class, where we focus on models whose endogenous variables are changing over time. The material is going to be unfamiliar to most of you, since your math preparation was limited to Math 205 and Math 206 for the most part.
  • The first two lectures cover theory and applications in the area of difference equations. Dif- ference equations are the first step in our journey towards being able to solve dynamic opti- mization problems. A difference equation links the value of an endogenous variable in a given time period to its value in other time periods as well as to other exogenous variables.
  • The solution to many dynamic optimization problems relate the value of variables to their past values. So the basic description of the behavior of a variable is different: it is no longer merely a function of other contemporaneous exogenous variables but is also a function of time- lagged values of itself, and in some cases a function of time itself. 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.
  • Next week, we switch to the study of differential equations: which are distinguished from difference equations by the fact that the changes in variables happen continuously instead of discretely as in the difference equations case.

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 yt = αyt− 1 + f (xt) where x is a vector of exogenous variables.
  • A solution to a first order difference equation is a sequence of values {yt}, expressed as a function of time and of the exogenous variables. In other words we need to find a time path for the variable y that is consistent with the difference equation.
  • An exact solution to a differential equation also requires that we have 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). An example of a first order difference equation will be of the form yt = αyt− 1 with initial condition y 0 = 10

Method 1: The General Method

  • The general method, as the name suggests, provides a systematic general framework for solving first-order difference equations. Suppose you are given a general first order difference equation of the form yt = αyt− 1 + β.
  • The solution to this difference equation consists of two parts: a part named the particu- lar solution, denoted yP which is any solution to the first order difference equation given above and a part named the homogenous solution, denoted yH which solves the difference equation yt = αyt− 1. The general solution is the sum of yP and yH.
  • Let’s begin by thinking about the homogenous solution, i.e. the solution to yt = αyt− 1. Since at each stage y is α times the previous value, the solution to this would be of the form yH = αtA where A is a constant. The ‘A’ term is important because we have to consider all possible solutions to yt = αyt− 1.
  • Now let’s think of the particular solution, which can be ANY solution to the equation yt = αyt− 1 + β. Since any solution is allowed, we can, for example, find a solution that sets all the y’s equal. If we set y = αy + β, we get the solution yP = (^1) −βα Combining yt = yP + yH we get the general solution of the form

yt = αtA + β 1 − α

  • Since we know the value of y 0 we can pin down the value of A. Setting t = 0 we get y 0 = A + (^1) −βα ⇒ A = y 0 − (^1) −βα. So the solution to the difference equation is of the form:

yt = αt

[

y 0 − β 1 − α

]

β 1 − α ≡ αty 0 + β

[

1 − αt 1 − α

]

  • The general method only works for |α| 6 = 1. If α = 1 or α = −1 you have to use a different technique

Example:

  • We can illustrate the general method with a specific numerical example: Solve the difference equation yt = 0. 2 yt− 1 + 4 where y 0 = 10 using the general method. Let’s begin by thinking about the homogenous solution yt = 0. 2 yt− 1. The solution is of the form yt = 0. 2 tA where A is a constant.
  • The particular solution is the solution to y = 0. 2 y + 4 ⇒ y = 5 So yP = 5. Combining we get yt = 0. 2 tA + 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:

yt = 0. 2 t(5) + 5

  • The steady state value is a value at which the endogenous variable exhibits no dynamic adjustment. In other words, given the difference equation yt = αyt− 1 + β , a steady state value y∗^ exhibits the property that y t∗ = y t∗− 1. A difference equation can have zero, one or more (in the case of non-linear difference equations) steady state values.
  • In the above difference equation, the steady state value is therefore y∗^ = αy∗^ +β ⇒ y∗^ = (^1) −βα.
  • You can verify this by showing that if yt = (^1) −βα then yt+1 = (^1) −βα.

Stability

  • A difference equation is said to be stable if lim t→∞ yt is a finite value. If lim t→∞ yt 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 yt = 0. 2 yt− 1 + 4 is stable because we can show using the solution yt = 0. 2 t(5) + 5 that limt→∞ yt = 5
  • For the general form yt = αyt− 1 +β, where α 6 = 1, which has the solution yt = αty 0 +β

[

1 −αt 1 −α

]

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

lim t→∞ yt = lim t→∞

[

αty 0 + β

[

1 − αt 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

tlim→∞ yt^ = lim t→∞

[

αty 0 + β

[

1 − αt 1 − α

]]

is undefined, approaching either ∞ or −∞

  • If α = 1, the solution is yt = 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

yt = y 0 if t is an even number yt = −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 unlessy 0 = β/2 Note, also, that this difference equation has a steady state at y∗^ = β/2 even though it is not stable.

IV. PHASE DIAGRAMS

  • Sometimes, we can analyze the dynamics of the first order difference equation without solving it explicitly using either the iterative method or the general method. We can do this using a diagram known as the ‘Phase Diagram’.
  • The phase diagram is a simple graph with yt and yt− 1 on the two axis. The graph shows two curves. The first plots the difference equation, the second is a 45 degree line representing the steady state (recall that the steady state is where yt = yt − 1).
  • Consider the simple difference equation we solved for earlier: yt = 0. 2 yt− 1 + 4 where y 0 = 10. The phase diagram would look as follows. -

6

yt

yt− 1

yt = 0. 2 yt− 1 + 4

yt = yt− 1

6

 ?

  • The curves intersect at yt, yt− 1 = (5, 5), the steady state of the model. Think of the horizontal axis as representing yesterday’s value and the vertical axis today’s value of y.
  • We can then use the phase diagram to trace out the stability or instability of the difference equation. If yesterday’s value was the starting value of y 0 = 10, then the difference equation tells us that today’s value will be 6. So we move to the point (10, 6) on the line representing the difference equation. This period’s yt=6 becomes next period’s yt− 1 so we then move to the mirror point (6, 6) on the 45 degree line. We repeat this process until we converge to a value of (5, 5).
  • Perhaps another example will help clarify the situation. Suppose the difference equation had instead been yt = − 0. 2 yt− 1 + 4, with a starting value of y 0 = 10. In this case the intersection will be at 10/3.

  • The phase diagram would look as follows.

6

 ` yt yt− 1 yt = − 0. 2 yt− 1 + 4 yt = yt− 1 10 3 10 3 6  6 -