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Difference Equations

Gerald P. Dwyer

Clemson University

January 2016

Outline

1 Difference EquationsIntroduction

OperatorsMultipliers and Impulse Response Functions

Solutions to Difference EquationsSolution by IterationGeneral method of solution

Solve First-Order Difference Equation

Method of Undetermined CoefficientsLag Operator to Solve EquationsSecond-order difference equationSummary

Times Series – Unemployment rate

5.0

0

2

4

6

8

10

12

1/1/48 1/1/53 1/1/58 1/1/63 1/1/68 1/1/73 1/1/78 1/1/83 1/1/88 1/1/93 1/1/98 1/1/03 1/1/08 1/1/13

The Great Moderation: U.S. Unemployment RateJanuary 1948 to December 2015

Sources: BLS, NBER, Haver Analytics

Times Series – CRSP stock price index

0

20

40

60

80

100

120

140

160

30 40 50 60 70 80 90 00 10

vwcrspx


0

1,000

2,000

3,000

4,000

5,000

30 40 50 60 70 80 90 00 10

vwcrsp


4,000.0

400.0

40.0

4.0

0.430 40 50 60 70 80 90 00 10

vwcrsp

Times Series – Return on CRSP stock index

-.20

-.16

-.12

-.08

-.04

.00

.04

.08

.12

.16

30 40 50 60 70 80 90 00 10

vwretd

Autoregressions

Unemployment rate

CRSP price index

CRSP return

Terminology I

Time-series analysis makes heavy use of terms such as “stable” and“roots of the equation”, even “eigenvalues” and “eigenvectors”

What does all this mean?

A first-order linear difference equation (one lagged value of variableon left)

yt = a0 + a1yt−1 + xt , t = 1, ...,T

I The variable y at period t has the value represented by ytI The variable yt−1 is the same variable one period earlierI Time, t, runs from 1 to TI a0 and a1 are constant coefficients in the equationI xt is a forcing process that affects yt

F xt can be deterministic or stochastic

Terms “First Order” and “Linear”

Why is this equation called a first-order linear difference equation?

yt = a0 + a1yt−1 + xt , t = 1, ...,T

I Difference – can write in terms of first difference

yt − yt−1 = a0 + (a1 − 1) yt−1 + xt , t = 1, ...,T

I Linear – involves no nonlinear functions of yt

Order of Difference Equation

The “order” of a difference equation is the maximum lag included onthe right-hand side

First-order difference equation

yt = a0 + a1yt−1 + xt

Second-order difference equation

yt = a0 + a1yt−1 + a2yt−2 + xt

k ’th order difference equation

yt = a0 + a1yt−1 + a2yt−2 + ... + akyt−k + xt

Another k ’th order difference equation

yt = a0 + akyt−k + xt

Can also define in terms of number of differences can take of yt butthat is not particularly helpful and it’s the same

Deterministic and stochastic

“Deterministic” means perfectly predictable from its own past

I For example trend xt = bt with b knownI Quarterly dummy variablesI A deterministic difference equation

yt = a0 + a1yt−1 + bt, t = 1, ...,T

“Stochastic” means evolves according to a probability law, e.g.xt = εt ∼ N(0, 1)

I Another word is “random”

F Does not mean “arbitrary”

I “Random” means for example xt = εt ∼ N(0, 1)I A stochastic difference equation

yt = a0 + a1yt−1 + εt , t = 1, ...,T

εt ∼ N(

0, σ2)

Behavior of difference equations I

First-order linear difference equation with no forcing variable

yt = a0 + a1yt−1, t = 1, ...,T

At time 1, this is

y1 = a0 + a1y0

At time 2, it is

y2 = a0 + a1y1

which also equals

y2 = a0 + a1 (a0 + a1y0) = a0 + a1a0 + a21y0

At time 3,

y3 = a0 + a1(a0 + a1a0 + a21y0

)= a0 + a1a0 + a21a0 + a31y0

Behavior of difference equations II

At time 4,

y4 = a0 + a1(a0 + a1a0 + a21a0 + a31y0

)= a0 + a1a0 + a21a0 + a31a0 + a41y0

Can prove by induction

yt =t−1∑i=0

ai1a0 + at1y0, t = 1, ...,T

Well defined paths depending on value of a1 – see spreadsheet

Operators I

Two operators are often used in analyzing difference equations

I An operator is a symbol that indicates an algebraic operation to betaken

I A plus sign (+) is an example of an operator, as are minus (−), times(· and x), and division (/)

Difference operator ∆

∆yt = yt − yt−1

∆yt+h = yt+h − yt+h−1, h = ...,−1, 0.1, ...

∆2yt = ∆∆yt = ∆ (yt − yt−1) = ∆yt − ∆yt−1

= yt − 2yt−1 + yt−2

I Higher orders of differencing possible tooI Statisticians sometimes use ∇

Operators II

Lag operator L

L yt = yt−1

L2 yt = yt−2

Li yt = yt−i

I Statisticians sometimes use B (backshift)

Polynomial in lag operator

When more than one lag, often form polynomial in lag operator

yt = a0 + a1yt−1 + a2yt−2 + xt

yt = a0 + a1 L yt + a2 L2 yt + xt

yt − a1 L yt − a2 L2 yt = a0 + xt(1− a1 L−a2 L2

)yt = a0 + xt

a (L) yt = a0 + xt , a (L) = 1− a1 L−a2 L2

or yt = a0 + a∗ (L) yt−1 + xt , a∗ (L) =

1

∑i=1

ai Li = a1 + a2 L



yt = a0 + a1yt−1 + a2yt−2 + xt

yt = a0 + a1 L yt + a2 L2 yt + xt

yt − a1 L yt − a2 L2 yt = a0 + xt

(1− a1 L−a2 L2

)yt = a0 + xt

a (L) yt = a0 + xt , a (L) = 1− a1 L−a2 L2

or yt = a0 + a∗ (L) yt−1 + xt , a∗ (L) =

1

∑i=1

ai Li = a1 + a2 L



yt = a0 + a1yt−1 + a2yt−2 + xt

yt = a0 + a1 L yt + a2 L2 yt + xt

yt − a1 L yt − a2 L2 yt = a0 + xt(1− a1 L−a2 L2

)yt = a0 + xt

a (L) yt = a0 + xt , a (L) = 1− a1 L−a2 L2

or yt = a0 + a∗ (L) yt−1 + xt , a∗ (L) =

1

∑i=1

ai Li = a1 + a2 L

Lag Operator

Lag operator rules

L−i is well definedL−i yt = yt+i

Lag operator applied to a constant c

L c = c

Distributive (Li + Lj

)yt = yt−i + yt−j

Associative

Li Lj yt = Lj Li yt = Li+j yt = Lj+i yt = yt−(i+j) = yt−i−j

Same rules for the difference operator ∆Note ∆yt = (1− L) yt

Multipliers and Impulse response functionsThe multiplier in period t for any period t + h indicates the responseof a variable h periods in the future to an impulse in some period t

I First-order difference equation

yt = a0 + a1yt−1 + xt , t = 1,T (1)

I The h’th period multiplier simply is

∂yt+h

∂xt

F Holding constant any other forcing variables

I If you iterate equation (1) with one initial nonzero forcing value xt , youwill find

yt+h =h

∑i=0

ai1a0 + ah+11 yt−1 + ah1xt

I Therefore∂yt+h

∂xt= ah1

for a first-order difference equation

Impulse response function

The impulse response function is the set of{∂yt+h

∂xt

}, n = 0, 1, 2, ...

I For a linear first-order difference equation, this is just{∂yt+h

∂xt= ah1

}, h = 0, 1, 2, ...

Properties for a linear equation such as this one

I The size of the impulse response is independent of the magnitude of xtor yt

I The size of the impulse response is independent of the particular timeperiod t

F All that matters is a1 and h

Solution

A “solution” to a difference equation is a representation of thedifference equation in which the value depends on the coefficients, thesequence of xt values, denoted {xt} and possibly one or more initialvalues

For example, for the first-order difference equation with no forcingvariable

yt = a0 + a1yt−1 (2)

I a solution is the function

yt =t−1∑i=0

ai1a0 + at1y0, t = 1, ...,T

I which we determined above

Verify solution

It is possible to verify this is a solution by substituting it into the leftand right-hand sides of equation (2)

yt = a0 + a1yt−1 with solution yt =t−1∑i=0

ai1a0 + at1y0

t−1∑i=0

ai1a0 + at1y0 = a0 + a1

(t−2∑i=0

ai1a0 + at−11 y0

)t−1∑i=0

ai1a0 + at1y0 = a0 +

(t−1∑i=1

ai1a0 + at1y0

)t−1∑i=0

ai1a0 + at1y0 =t−1∑i=0

ai1a0 + at1y0

Therefore holds as an identity

Verify solution



ai1a0 + at1y0

t−1∑i=0

ai1a0 + at1y0 = a0 + a1

(t−2∑i=0

ai1a0 + at−11 y0

)

t−1∑i=0

ai1a0 + at1y0 = a0 +

(t−1∑i=1

ai1a0 + at1y0

)t−1∑i=0

ai1a0 + at1y0 =t−1∑i=0

ai1a0 + at1y0


Verify solution



ai1a0 + at1y0

t−1∑i=0

ai1a0 + at1y0 = a0 + a1

(t−2∑i=0

ai1a0 + at−11 y0

)t−1∑i=0

ai1a0 + at1y0 = a0 +

(t−1∑i=1

ai1a0 + at1y0

)

t−1∑i=0

ai1a0 + at1y0 =t−1∑i=0

ai1a0 + at1y0


Verify solution



ai1a0 + at1y0

t−1∑i=0

ai1a0 + at1y0 = a0 + a1

(t−2∑i=0

ai1a0 + at−11 y0

)t−1∑i=0

ai1a0 + at1y0 = a0 +

(t−1∑i=1

ai1a0 + at1y0

)t−1∑i=0

ai1a0 + at1y0 =t−1∑i=0

ai1a0 + at1y0


Methods to solve difference equations

1 Iteration

2 General method

3 Method of undetermined coefficients

Another example of solution by iteration

Equation isyt = a0 + a1yt−1 + εt , εt ∼ N(0, σ2)

I No initial condition – push back to minus infinity instead of havinginitial condition

Solve iteratively backward

yt = a0 + a1yt−1 + εt

yt−1 = a0 + a1yt−2 + εt−1Therefore yt = a0 + a1 (a0 + a1yt−2 + εt−1) + εt =a0 + a1a0 + a21yt−2 + a1εt−1 + εt

yt = a0 + a1a0 + a21a0 + a31yt−3 + a21εt−2 + a1εt−1 + εt

yt =t−1∑i=0

a0ai1 +

t−1∑i=0

ai1εt−i + at1y0

If |a1| < 1, then perhaps yt =a0

1−a1 +∞∑i=0

ai1εt−i

Cannot solve this way if a1 = 1 or |a1| > 1





yt = a0 + a1yt−1 + εt

yt−1 = a0 + a1yt−2 + εt−1

Therefore yt = a0 + a1 (a0 + a1yt−2 + εt−1) + εt =a0 + a1a0 + a21yt−2 + a1εt−1 + εt


yt =t−1∑i=0

a0ai1 +

t−1∑i=0

ai1εt−i + at1y0


1−a1 +∞∑i=0

ai1εt−i






yt = a0 + a1yt−1 + εt

yt−1 = a0 + a1yt−2 + εt−1Therefore yt = a0 + a1 (a0 + a1yt−2 + εt−1) + εt =a0 + a1a0 + a21yt−2 + a1εt−1 + εt


yt =t−1∑i=0

a0ai1 +

t−1∑i=0

ai1εt−i + at1y0


1−a1 +∞∑i=0

ai1εt−i


Solution?

If |a1| < 1, then yt =a0

1−a1 +∞∑i=0

ai1εt−i is a solution

Solution?

Have yt =t−1∑i=0

a0ai1 +

t−1∑i=0

ai1εt−i + at1y0

Want limit as push time 0 further back

Use t0 instead of 0 for initial time period which implies

yt =t−1−t0

∑i=0

a0ai1 +

t−1−t0∑i=0

ai1εt−i + at−t01 yt0

Let h = t − t0 which implies yt =h−1∑i=0

a0ai1 +

h−1∑i=0

ai1εt−i + ah1yt−h

limh→∞h−1∑i=0

a0ai1 →

∞∑i=0

a0ai1 =

a01−a1

limh→∞h−1∑i=0

ai1εt−i →∞∑i=0

ai1εt−i if εt−1 is well behaved

limh→∞ ah1 = 0 and assume limh→∞ εt−h is bounded so thatlimh→∞ ah1εt−h = 0

Yielding yt =a0

1−a1 +∞∑i=0

ai1εt as the solution

General method of solution

Similar to way differential equations are solved

1 Solve the homogenous equation (no constant term)

2 Solve for a particular solution (complete equation)

3 The general solution is the sum of the particular solution and a linearcombination of all homogeneous solutions

4 There will be an arbitrary constant which can be eliminated byimposing the initial condition on the general solution or by driving thesolution back infinitely far

Rules for solving linear difference equations with constantcoefficients

Linear difference equation – nth order

yt = a0 +n

∑i=1

aiyt−i (3)

Homogeneous form

yt =n

∑i=1

aiyt−i (4)

1 If the homogenous equation (4) has n independent solutionsy = yj (t) , then y = ∑

jyj (t) also is a solution.

2 If the homogeneous equation (4) has the solution y = yj (t), theny = Ayj (t) also is a solution.

3 If y = y (t) is any particular solution whatever of equation (3) and ify = y(t;A1,A2, ...,An) is a solution of (4), then the general solutionof (3) is y = y (t) + y(t;A1,A2, ...,An)

Solve first-order difference equation

yt = a0 + a1yt−1

A possible particular solution is a constant y

I If y is the solution, then y = a0 + a1y and if a1 6= 1, y = a01−a1

Solve homogeneous difference equation (equation without constant)yht = a1y

ht−1 where the superscript h just indicates the homogeneous

solution

I Guess yt = Aλt where A is an arbitrary constant and λ is a parameterto be determined

I Substituting into the homogeneous equation, we find Aλt = a1Aλt−1

which holds if λ = a1 for any A


yt = a0 + a1yt−1I Particular solution of yt = a0 + a1yt−1 is yI Homogeneous: yht = Aλt is a solution to the homogeneous equation if

λ = a1

General solution isyt = Aat1 +

a01− a1

(5)

Value of A? At t = 0, equation (5) implies first-order autoregressionis

y0 = Aa01 +a0

1− a1= A+

a01− a1

⇒ A = y0 −a0

1− a1

and

yt =

(y0 −

a01− a1

)at1 +

a01− a1


In the general solution

yt =

(y0 −

a01− a1

)at1 +

a01− a1

I The term(y0 − a0

1−a1

)at1 represents transient behavior which

F is zero if y0 = a01−a1

F goes to zero as initial period gets farther back in the past if |a1| < 1

Solve first-order difference equationWhat if slope coefficient equals 1?

yt = a0 + yt−1

Solve homogeneous equation yht = a1yht−1

I Guess yht = Aλt

I Verify that yht = Aλt is a solution with λ = a1 = 1

y = a01−a1 can’t be a particular solution

Try y = bt, b a constant for trend growth (bar means “particular”,not “constant”)

I yt = a0 + yt−1 ⇒ bt = a0 + b (t − 1) , bt = a0 + bt − b, 0 = a0 − b

General solution isyt = A+ a0t

Value of A? y0 = A+ b · 0, A = y0 and

yt = y0 + a0t

Solve first-order difference equationWhat if slope coefficient is greater than 1?

yt = a0 + yt−1Solve homogeneous equation yht = a1y

ht−1

I Guess yht = Aλt

I Verify that yht = Aλt is a solution with λ = a1

y = a01−a1 can be a solution

I Verify that y = a01−a1 can be a solution

General solution isyt = Aat1 +

a01− a1

with the value of A set, it is

yt =

(y0 −

a01− a1

)at1 +

a01− a1

Transient behavior does not go to zeroI Becomes increasingly large as take initial period increasingly far back in

the past

Illustrate method of undetermined coefficients

Guess a solution, called challenge solution

Substitute it into difference equation and see if an identity in thecoefficients emerges

We already did this for the non-transient part with the first-orderdifference equation

yt = a0 + a1yt−1 (6)

I Guess answer isyt = y (7)

Substitute (7) into (6)

y = a0 + a1y

y = a01−a1

First-order autoregression with forcing variable I

Equation to solve is

yt = a0 + a1yt−1 + εt , εt ∼ N(0, 1), |a1| < 1 (8)

yt = a0 + a1yt−1 + εt , εt ∼ N(0, 1)

Iterate a few times

I yt = a0 + a1yt−1 + εtI yt = a0 + a1 (a0 + a1yt−2 + εt−1) + εt =

a0 + a1a0 + a21yt−2 + a1εt−1 + εtI yt = a0 + a1a0 + a21 (a0 + a1yt−3 + εt−2) + a1εt−1 + εt =

a0 + a1a0 + a21a0 + a31yt−3 + a21εt−2 + a1εt−1 + εt

First-order autoregression with forcing variable I

Pattern

I Constant term with a0 + a1a0 + a21a0 + ... + ai1a0I ai1yt−iI Innovations with εt + a1εt−1 + a21εt−2 + ... + ai1εt−i

Implications as lag goes to infinity

I As i goes to infinity, term ai1yt−i goes to zeroI As i goes to infinity, sum of constants goes to a0

1−a1I As i goes to infinity, sum of innovations goes to ∑∞

i=0 ai1εt−i

Guess solution

yt = b∗0 +∞

∑i=0

bi εt−i (9)

Verify by substituting yt = b∗0 + ∑∞i=0 bi εt−i into

yt = a0 + a1yt−1 + εt

First-order autoregression with forcing variable II

Substitute yt = b∗0 + ∑∞i=0 bi εt−i into yt = a0 + a1yt−1 + εt

b∗0 +∞

∑i=0

bi εt−i = a0 + a1

(b∗0 +

∞

∑i=1

bi−1εt−i

)+ εt

Equating constants and coefficients of innovations, we see

b∗0 = a0 + a1b∗0

b∗0 =a0

1− a1bi = a1bi−1, i = 1, ..., ∞

First-order autoregression with forcing variable III

which implies

b∗0 =a0

1− a1b0 = 1

b1 = a1b0 = a1

b2 = a1b1 = a21

bi = a1bi−1 = ai1

b0 = 1 is a normalization

Therefore

yt =a0

1− a1+

∞

∑i=0

ai1εt−i

In terms of lag operator for slope coefficient less than one I

|a1| < 1

Equation is

yt = a0 + a1yt−1

(1− a1 L) yt = a0

Solution with period 0 is

yt =

(y0 −

a01− a1

)at1 +

a01− a1

If period 0 is infinitely far back in the past, transient behavior is

unimportant because lim(y0 − a0

1−a1

)at1 = 0, and solution is

yt =a0

1− a1

In terms of lag operator for slope coefficient less than oneII

Can solve for this in terms of the lag operator by writing

yt =a0

(1− a1 L)

I noting that

yt =∞

∑i=0

ai1 Li a0 =a0

(1− a1)

Might seem like we’re shutting down variation over time but moregenerally have a forcing variable xt and

(1− a1 L) yt = a0 + xt

In terms of lag operator for slope coefficient less than oneIII

I Equation can be written

yt =a0

(1− a1 L)+

1

(1− a1 L)xt

yt =a0

(1− a1)+

∞

∑i=0

ai1xt

Root of equation characterizes behavior I

Root of difference equation is important for determining behavior

A first-order difference equation is a polynomial of order 1 in in thelag operator

(1− a1 L) yt = a0

z transform

Solve for root of polynomial in lag operator

(1− a1λ) = 0

Root is

λ = a−11

|λ| > 1 implies the equation is stable in the sense of returning to itsmean in response to transient variation

Root of equation greater or less than one?

Important digression: Root less than one or greater than one forstability?

I Will see both characterizationsI A root greater than one is necessary for stability – commonly for

time-series statisticiansI A root less than one is necessary for stability – sometimes for

econometricians

They both can be correct

(1− a1λ) = 0(λf − a1

)= 0

λ > 1 and λf < 1 are the same thing: λ−1 = λf

Possible strategy for solving difference equations I

Solve for roots in polynomial in lag operator or forward operator

We are interested in solving for roots to characterize behavior ofseries, not learning to be difference-equation wizards

Can rely on Fundamental theorem of algebra

I Every polynomial of order p has exactly p roots

Second-order difference equation I

Second-order difference equation is

yt = a0 + a1yt−1 + a2yt−1

Can be written

(yt − a1yt−1 − a2yt−2) = a0(1− a1 L−a2 L2

)yt = a0

There are two roots λi of this quadratic equation such that(1− a1λi − a2λ2

i

)= 0

Solve for λ1 and λ2

If |λ1| > 1 and |λ2| > 1, then the difference equation is stable

If |λ1| = 1 and |λ2| > 1, then the difference equation has one unitroot

Second-order difference equation II

If |λ1| = 1 and |λ2| = 1, then the difference equation has two unitroots

If |λi | < 1 for either or both roots, then the difference equation isexplosive

Second-order difference equation with inverse roots(equally correct) I

Equation is

yt = a0 + a1yt−1 + a2yt−1

Can be written

(yt+2 − a1yt+1 − a2yt) = a0(F2−a1 F−a2

)yt = a0

Every polynomial can be factored to yield(F−λf

1

) (F−λf

2

)yt = a0

Solve for λf1 and λf

2

If |λf1| < 1 and |λf

2| < 1, then the difference equation is stable

Second-order difference equation with inverse roots(equally correct) II

If |λf1| = 1 and |λf

2| < 1, then the difference equation has one unitroot

If |λf1| = 1 and |λf

2| = 1, then the difference equation has two unitroots

If |λfi | > 1, for either or both roots, then the difference equation is

explosive

Solving second-order difference equation I

We are interested in solving for roots to characterize behavior of timeseries, not to learn to be difference-equation wizards

Solve for roots of polynomial in lag operator or forward operator


I Every polynomial of order p has exactly p roots

General method

Equation isyt = a0 + a1yt−1 + a2yt−1


I Every polynomial of order p has exactly p rootsI That means two roots here

Particular solution

The particular solution is

y = a0 + a1y − a2y

andy =

a01− a1 + a2

if a1 + a2 6= 1

If a1 + a2 = 1, there is at least one unit root

Suppose there is no unit root

Homogeneous solution to second-order difference equationI

Homogeneous solution – guess yht = Aαt

I Note: Using α here because the book does – same as λf above

Aαt = a1Aαt−1 + a2Aαt−2

α2 − a1α− a2 = 0

This characteristic equation has two roots given by the quadraticformula

roots of aα2 + bα + c = 0 given by α1, α1 =−b±

√b2 − 4ac

2a

α1, α2 =a1 ±

√a21 + 4a2

2

The roots can be real or imaginary

Two real roots I

Characteristic equation

α1, α1 =a1 ±

√a21 + 4a2

2

If a21 + 4a2 > 0, there are two real roots

I a21 + 4a2 is called the “discriminant”I Absolute value of roots less than one for convergent behavior

Discriminant is zero I


α1, α1 =a1 ±

√a21 + 4a2

2

If a21 + 4a2 = 0, it might seem there is only one root a1/2 but thereis another one

I The other root is t (a1/2) – a trend termI Probability of getting this with data is basically zero and I will move on

Discriminant is negative I


α1, α1 =a1 ±

√a21 + 4a2

2

If a21 + 4a2 < 0, there are two complex conjugate roots

α1, α1 =a1 ± i

√− (a21 + 4a2)

2

I where i =√−1

I Obviously must appear as a pair with

α1 =a1 + i

√−(a21 + 4a2

)2

α2 =a1 − i

√−(a21 + 4a2

)2

Discriminant is negative II

I Oscillatory behaviorI Can be stable or unstableI How can we tell? The unit circle

Argand diagram with a circle super-imposed

Unit Circle

Why does unit circle work?

Can characterize circle in terms of radius, here one, and radius for acircle from origin equals

r2 = re2 + im2

where r is the radius, re is the real part (or just x-axis) and im is theimaginary part (or just y -axis), all real numbers

Can determine location of a complex root by squaring root

Square of a complex number is the number re + i · im times itscomplex conjugate re − i · imSquare of re + i · im is re2 + im2

Absolute value of re2 + im2 is(re2 + im2

)1/2– distance from zero to

the location of the number on the circle with this radius

If roots are calculated as in textbook, (λf ) are inside unit circle, stable

If calculated as statisticians calculate roots (λ), roots outside the unitcircle for stability

Complex roots and oscillations

Complex roots can be associated with oscillations – recurrent cycles

Can be seen through relationship of complex numbers and Arganddiagram

Simple way: spreadsheet

Summary

Difference equations can be stable or unstable

I Unstable means a trajectory that divergesI Stable means a trajectory that converges to a mean, to trend growth,

oscillations, some long-run path that could be repeated forever

The roots of a difference equation summarize the behavior

I As typically calculated by economists, roots with an absolute value lessthan one are stable

Second-order difference equations have two roots

I The roots can be complex numbersI Even so, roots with an absolute value less than one are stability

Unit roots mean the equation does not converge to a constant mean

Unit roots do not mean the equation is “unstable” in the sense ofdiverging from trend behavior

I yt = a0 + yt−1 always has the value given by yt − yt−1 = a0I Stable in terms of growth, just not level

january 2016 - jerry dwyerjerrydwyer.com/pdf/clemson/topic2differenceequations.pdf · solutions to...

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