part 1: mathematical introduction christian philipp schröder · part 1: mathematical introduction...

Macroeconomic ModelsPart 1: Mathematical Introduction

Christian Philipp Schröder

Chair of Economic Policy (520a)University of Hohenheim

Winter Term 2016/2017

Introduction

Dynamic Models in One Equation

Dynamic Systems in Discrete Time

Excursus: Matrix Algebra

Discrete-Time Dynamic Systems in Matrix Notation

Roundup & Outlook

Organizational Issues

I WhoI Part 1: Christian Philipp SchröderI Part 2: Arash Molavi VasséiI Part 3: Benjamin SchmidtI If you want to contact us, you can find us at the chair’s website:

↗wipol.uni-hohenheim.deI When & where

I Typically, the lecture will be on Mondays (2pm-4pm, HS 35) and thetutorial on Tuesdays (12am-2pm, PC-Raum 3)

I This might change sometimes due to scheduling issuesI Literature for part 1

I [A] Azariadis (1993): Intertemporal Macroeconomics, BlackwellI [CW] Chiang/Wainwright (2005): Fundamental Methods of Mathematical

Economics, 4th ed., McGraw-HillI [G] Gandolfo (2010): Economic Dynamics, 4th ed., SpringerI [SB] Simon/Blume (1994): Mathematics for Economists, Norton

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https://wipol.uni-hohenheim.de

What Is the Aim of this Tutorial?

I We want to describe (macro-)economiesI We want to do so formally, i.e., in mathematical modelsI This is an intermediate/advanced course

I For instance, one of the main issues is the time dimension ofmacroeconomic events (dynamics)

I We assume you have a basic knowledge of both maths and economicsI Depending on your previous education, some of you might already be

familiar with some/a lot of the issues presented hereI The tutorial somewhat supplements the lecture

I We don’t just fill the (computational) blanks of the lectureI However, the tutorial is not completely stand-alone eitherI We present certain tools you need in monetary macroeconomics and discuss

some (more or less) simple models

4

What Are Macroeconomic Models?

Generally SpeakingI Models are sets of simultaneous equations in endogenous and exogenous

variablesI We are always in search of the solution of a given system (equilibrium)

I What happens to the equilibrium if anything changes? (Comparative statics)I What does the adjustment look like? How long does it take? (Dynamics)

I Counting equations: To solve a system, we need as many ‘meaningful’equations as endogenous variables

I There are different kinds of equationsI Equilibrium conditionsI Behavioral equationsI Budget constraintsI Definitions (identities)

I Especially in larger models, the pure amount of variables and equationscan become a source of confusion

5

Specific Example: The IS-LM ModelI Two markets shall clear simultaneously → two equilibrium conditions:

I Goods market: Y != ZI Money market: Ms != Md

I Several other equations are substituted into these conditionsI Aggregate demand (definition): Z ≡ C + I + G + (Ex − Im)I Keynesian consumption function: C = c0 + c1YaI . . .

I Which variables are endogenous, which are exogenous?I Definitely endogenous: i , YI Surely exogenous (in typical IS-LM settings): G, MsI What about taxes, for instance? Depends on whether they are lump-sum

(T ) or income taxes (T [Y ] = tY )I . . .

I Static modelI We can compare equilibria (e.g., after G ↓ or Ms ↑)I But we don’t know when these changes are coming into effect, for instance

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Counting Equations in the Closed-Economy IS-LM Model

I Behavior within theIS-LM model isdescribed by thefollowing equations:

Z = C + I + G ⟨1⟩C = Co + c1Ya ⟨2⟩

Ya = Y − T ⟨3⟩I = I(r , Y ) ⟨4⟩

Md = PL ⟨5⟩L = L(r , Y ) ⟨6⟩

I Cataloguing all variables:

no. var. endog. exog.1 Y 12 Z 23 C 34 I 45 r 56 G 17 Co 28 Ya 69 T 310 Ms 411 Md 712 P 513 L 8

I We have only sixequations for eightendogenousvariables thus far

I The model is‘closed’ by the twomarket-clearingconditions

Y != Z ⟨7⟩

Ms != Md ⟨8⟩

which eventuallyyield the IS and LMequations

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Going Forward

“In a static model [. . .] the problem is to find the values of the endogenousvariables that satisfy some specified equilibrium condition(s).

Applied to the context of optimization models, the task becomes one of findingthe values of the choice variables that maximize (or minimize) a specificobjective function—with the first-order conditions serving as the equilibriumcondition.

In a dynamic model, by contrast, the problem involves instead the delineationof the time path of some variable, on the basis of a known pattern of change[. . .].” [CW:444-445]

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Introduction

Dynamic Models in One EquationContinuous Time: First-Order Differential EquationsDiscrete Time I: First-Order Difference EquationsDiscrete Time II: Second-Order Difference Equations




Roundup & Outlook

What Is a Differential Equation?

I In general, a differential equation contains both the variable y(t) and itsderivative (or differential) d y/ d t

d yd t︸︷︷︸

derivative/differential,

≡ y

+ a(t)︸︷︷︸coefficient

· y(t)︸︷︷︸primitivefunction

= b(t)︸︷︷︸(additive)

term

⟨9⟩

I Making notation less bulky:

y = Φ(y , t) ⟨10⟩

I y and hence also y are unknown, only their relationship as given by⟨9⟩/⟨10⟩ is known

I Solving ⟨9⟩/⟨10⟩ means finding a function that satisfies it

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Types of Differential Equations and Terminology

I The order pertains to the highest order of the featured derivatives:

d yd t ,

d2 yd t2 , . . .

I The degree is determined by the highest power attained by the derivatives:

d yd t ,

(d yd t

)2, . . .

I A linear difference equation is of first order and first degreeI And it does not contain products of the form y (d y/ d t)

I An autonomous differential equation is time-independentI y = Φ(y): Its coefficient and additive term are constantI This means that it does not matter when the process starts

I A zero-valued additive term makes the differential equation homogeneousI Multiplying all y arguments by the same constant would not violate ⟨9⟩

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autonomous non-autonomous

homogeneous y = ay y = a(t)

non-homogeneous y = ay + b y = a(t) + b(t)

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What We Will Be Working With

I For the time being, we limit ourselves to linear (i.e., first-order,first-degree) autonomous differential equations

I Therefore, we can simplify ⟨9⟩ to retrieve a ‘workhorse’ differentialequation

d yd t + ay = b ⟨11⟩

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The Solution of a Differential Equation Has Two Parts

I Homogeneity is the dividing line in the two-step solution processI The homogeneous ‘part’ of a given differential equation yields the

complementary function which determines dynamic stabilityI The complete (non-homogeneous) equation yields a particular

solution—the steady stateI Taken together, they form the general solution of the differential equation

I Actually, it is a three-step processI We obtain a definite solution by definitizing the general solution via an

initial condition

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The Complementary Function

I By ignoring the additive term b in ⟨11⟩, it becomes homogeneous:

d yd t + ay = 0 ⇔ 1

yd yd t = −a ⟨12⟩

I Integrating the left-hand side of ⟨12⟩ with respect to t gives∫1y

d yd t d t =

∫1y d y = ln |y | + C1

I Integrating the right-hand side of ⟨12⟩:∫−a d t = −at + C2

I Combining both sides and rearranging yields the complementary function

yc(t) = C︸︷︷︸≡ eC2−C1

· e−at ⟨13⟩

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The Particular Solution

I In a stationary state, y does not change anymore → y = 0 in ⟨11⟩ leads tothe particular solution yp :

ay = b ⇔ yp = ba ⟨14⟩

I This requires that a = 0I For a = 0, the solution changes slightly (cf. S19)

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The General and the Definite Solution

I The general solution to ⟨11⟩ (for a = 0) is the sum of the complementaryfunction ⟨13⟩ and the particular solution ⟨14⟩:

y(t) = C e−at +ba ⟨15⟩

I The definite solution can be obtained using an observed initial value y(0):

y(0) = C + ba ⇔ C = y(0) − b

a

in ⟨15⟩ ⇒ y(t) =[y(0) − b

a

]e−at +b

a ⟨16⟩

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The Special Case of a = 0

I If a = 0, the solution ⟨15⟩/⟨16⟩ is not sensibleI The workhorse differential equation ⟨11⟩ reduces to

y = b ⟨17⟩

I This can be solved very simply by integrating. The general solution reads

y(t) = bt + C

I C represents the complementary function (cf. S16 with a = 0)I ⟨17⟩ shows that y is constantly changing, so an invariant particular solution

is not possible → steady (but non-stationary) state bt

I Using an initial condition to definitize C , the definite solution becomes

y(t) = y(0) + bt

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Application: Government Debt Dynamics

I A popular measure to assess a country’s debt situation is the ratio of debtto nominal GDP:

d ≡ DPY

I Taking the total differential with respect to time yields

d dd t = d D

d t1

PY − d (π + y)

I The budget deficit is given by

d Dd t = iD + G − T

I The primary budget deficit as a fraction of nominal income is given by

b ≡ G − TPY

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I After some rearrangements, we have

d dd t = b + (r − y) d

I The solution therefore reads

d(t) = by − r +

[d(0) − b

y − r

]e(r−y)t

I Interpretation?

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Introduction





Roundup & Outlook

What Is a Difference Equation?

yt = ayt−1 + bt ⟨18⟩

I Somewhat similar to differential equations, a difference equation containsinformation about the change in a variable over time

I This change is not expressed through a derivative but its discrete-timeequivalent, the difference

I ∆ is often used as the (first-)difference operator of a variable between twoadjacent periods:

∆yt ≡ yt − yt−1

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I The terminology is the same as with differential equations (cf. S12)I E.g., there are also second- and higher-order difference equations such as

yt = a1yt−1 + a2yt−2

yt = a1yt−1 + a2yt−2 + . . . + anyt−n

which will be discussed in the next section

I We only examine linear difference equations with constant coefficientsI Our ‘workhorse’ difference equation reads

yt + ayt−1 = b ⟨19⟩

I Be mindful of notation: Even with b autonomous in ⟨18⟩, ⟨18⟩ and ⟨19⟩ arenot quite the same

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The Solution of A Difference Equation

I The solution procedure for difference equations is similar to that fordifferential equations:

1. Find the complementary function yc2. Find the particular solution/steady state yp3. Definitize using the initial condition y0

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The Complementary Solution

I Consider the homogeneous version of ⟨19⟩:

yt + ayt−1 = 0 ⇔ yt = −ayt−1 ⟨20⟩

I Repeatedly substituting ⟨20⟩ into itself

yt = −ayt−1

= (−a) (−ayt−2)= (−a)2 (−ayt−3)= . . .

yields the complementary function

yc,t = (−a)t C ⟨21⟩

I C is substituted for yt−t = y0 here

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I In a stationary state, y does not change anymore → yt = yt−1 = yp in⟨19⟩ leads to the particular solution yp :

yp + ayp = b ⇔ yp = b1 + a ⟨22⟩

I This requires that a = −1I For a = −1, the solution changes slightly (cf. S29)

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The General and the Definite Solution

I The general solution to ⟨19⟩ (for a = −1) is the sum of thecomplementary function ⟨21⟩ and the particular solution ⟨22⟩:

yt = C (−a)t + b1 + a ⟨23⟩

I The definite solution can be obtained using an observed initial value y0:

y0 = C + b1 + a ⇔ C = y0 − b

1 + a

in ⟨23⟩ ⇒ yt =(

y0 − b1 + a

)(−a)t + b

1 + a ⟨24⟩

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The Special Case of a = −1

I With a = −1, the workhorse difference equation ⟨19⟩ becomes

yt − yt−1 = b ⟨25⟩

I y is constantly changingI Therefore, the solutions ⟨23⟩/⟨24⟩ are not sensible

I ⟨23⟩ is undefined because the particular solution is infinityI At best, ⟨24⟩ results in a constant solution (y0), which does not fit ⟨25⟩

I Same solution as before (cf. S19): ‘Switch behavior’ of yc and ypI Using a = −1 on S26 yields yp = C (which is now constant in t)I The particular solution is simply yp,t = bt (i.e., depending on t)

⇒ yt = bt + C

I Using an initial condition to definitize C , the definite solution becomes

yt = y0 + bt

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Application: The Three-Equation ‘Standard’ Model of Spahn (2012)

Ingredients

1. Phillips curve/supply function with adaptive expectations

pt = pt−1 + α (yt − y∗) ⟨26⟩

2. Aggregate demand

yt = gt − β (it − pt) ⟨27⟩

3. Taylor rule

it = r∗ + pt + γ (pt − p∗) ⟨28⟩

I Approximate Fisher equation (with adaptive expectations)

rt = it − pt ⟨29⟩

I No explicit shocks (yet)I α, β, γ > 0

I This rules out the ‘special case’ later on (cf. S29, ⟨30⟩, ⟨32⟩)

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Rearranging and SimplifyingI Plugging ⟨28⟩ into ⟨27⟩ yields the integrated demand curve

yt = gt − βr∗ − βγ (pt − p∗)

I Plugging this into ⟨26⟩ then produces the following difference equation ininflation as an ‘intermediate product:’

(1 + αβγ) pt − pt−1 = α (gt − βr∗ + βγp∗ − y∗) ⟨30⟩

I What about gt , r∗, p∗, and y∗?I Simplistic approach: Interpreting y as an output gap makes y∗ = 0 optimal

(this is an economic deliberation, not a mathematical one)I No ‘special-case solution’ here → solution will yield pt = pt−1 → ⟨26⟩

implies yt = y∗ = 0 in equilibriumI Thus, the starred versions of ⟨27⟩ and ⟨29⟩ imply

y∗ = g∗ − βr∗ ⇔ r∗ =g∗ − y∗

β⟨31⟩

I For simplicity, we assume gt = g∗ ∀t (focus on monetary policy)I As a long-term policy choice of the central bank, we keep p∗ in the model

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Getting a Single Equation and Solving ItI Using all of the above, ⟨30⟩ becomes

pt − 11 + αβγ

pt−1 = αβγ

1 + αβγp∗ ⟨32⟩

I The (definite) solution of ⟨32⟩ is

pt = p∗ + (p0 − p∗)(

11 + αβγ

)t

I Equilibrium inflation—the particular solution p∗—is equal to the centralbank’s inflation target

I Stability depends on the coefficient (1 + αβγ)−1

I α, β > 0 are observable, but typically not directly controllableI The assumption that γ > 0 is in fact an important result in monetary

macroeconomics: the Taylor principle

32

Introduction





Roundup & Outlook


I Our new workhorse will be the second-order difference equation

yt + a1yt−1 + a2yt−2 = b ⟨33⟩

I The solution procedure is basically the same as beforeI Since finding the complementary function will be more complicated, we

start with the particular solution:

1. Find the particular solution/steady state yp2. Find the complementary function yc3. Definitize using the initial condition y0

34


I In a stationary state, y does not change anymore →yt = yt−1 = yt−2 = yp in ⟨33⟩ leads to the particular solution yp :

yp + a1yp + a2yp = b ⇔ yp = b1 + a1 + a2

⟨34⟩

I This requires that a1 + a2 = −1I For a1 + a2 = 1, the solution changes slightly (along the lines of S29)

35

The Complementary Solution

The Characteristic EquationI Consider the homogeneous version of ⟨33⟩:

yt + a1yt−1 + a2yt−2 = 0 ⟨35⟩

I With first-order difference equations, the solution is of the form

yc,t = Cλt ⟨36⟩

I λ is a yet-unknown coefficient here

36

I Plugging ⟨36⟩ into ⟨35⟩,

Cλt + a1Cλt−1 + a2Cλt−2 = 0, ⟨37⟩

and rearranging yields a characteristic equation

λ2 + a1λ + a2 = 0 ⟨38⟩

I Solving ⟨38⟩ yields two (characteristic) roots:

λ1, λ2 =−a1 ±

√a2

1 − 4a2

2 ⟨39⟩

I The roots depend on the coefficients of ⟨33⟩I They determine the stability of the equation

37

I What do we do with two solutions?I Both λi are associated with specific arbitrary constants Ci and

complementary solutions yc,i,t (in two instances of ⟨37⟩)I With second-order difference equations, we need both arbitrary constantsI The reason is that every time we difference, we lose a constant termI By implication, we must definitize twice using two initial conditionsI Therefore, we cannot simply ‘pick’ one yc,i,t over the other at randomI Instead, we combine them: It can be shown that (yc,1,t + yc,2,t) works fine

⇒ yc = yc,1,t + yc,2,t = C1λt1 + C2λt

2 ⟨40⟩

I Unfortunately, we’re not done yetI Cf. ⟨39⟩: Depending on a1 and a2, three cases can arise

1. a21 > 4a2: distinct real roots (λ1 = λ2)

2. a21 = 4a2: repeated real roots (λ1 = λ2)

3. a21 < 4a2: complex roots

I We discuss cases 1 and 2 in turn but ignore case 3

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Case 1: Distinct Real RootsI If a2

1 > 4a2 in ⟨39⟩, the two characteristic roots λ1, λ2 are distinctI The complementary solution is exactly ⟨40⟩:

yc = yc,1,t + yc,2,t = C1λt1 + C2λt

2 ⟨41⟩

I From here, we can proceed to definitizing the general solutionI Two initial conditions, e.g., y0 and y1, are necessary

39

Case 2: Repeated Real RootsI If a2

1 = 4a2 in ⟨39⟩, the characteristic roots are equal: λ1 = λ2 = λ

⇒ yc,1,t + yc,2,t = C1λt + C2λt ≡ C3λt

I Like this, we could definitize only one constant later on

I The trick is to find another ‘second half’ for the complementary solutionthat also satisfies ⟨33⟩: C4tλt

⇒ yc = C3λt + C4tλt

I Like this, a definite solution for yt can be found

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part 1: mathematical introduction christian philipp schröder · part 1: mathematical introduction...

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