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004: Macroeconomic TheoryLecture 10

Mausumi Das

Lecture Notes, DSE

August 18, 2014

Das (Lecture Notes, DSE) Macro August 18, 2014 1 / 43

2X2 System of Simultaneous Difference Equations: Linear& Autonomous

In the last class we considered the following 2× 2 simultaneoussystem of difference equations (written in matrix form):[

xt+1yt+1

]=

[a11 a12a21 a22

] [xtyt

]+

[b1b2

]i.e., Xt+1 = AXt + B.

where Xt represent the time-dependent state vector, A representsthe (time invariant) coeffi cient matrix and B represents the (timeinvariant) vector of non-homogenous terms of the system.We are going to apply the Superposition Principle to solve thissystem.


2X2 System of Linear & Autonomous SimultaneousDifference Equations: Solution (Contd.)

We have already seen that if the coeffi cient matrix was diagonal (suchthat a12 = a21 = 0), then the general solution to the above system ofsimultaneous difference equations would be given by:[

xtyt

]=

[C1 (a11)

t

C2 (a22)t

]+

[xy

]; C1, C2 arbitary constants.

where (x , y) represents the steady state of the system, defind asfollows (assuming a11, a22 6= 1):

x =b1

1− a11;

y =b2

1− a22.


A Digression: Eigen Values, Eigen Vectors &Diagonalization of a Square Matrix

In the last class we had indicated that if the coeffi cient matrix A isnon-diagonal, then it can be transformed into a diagonal matrix byapplying a procedure called ‘Diagonalization’, which usescorresponding eigen values and eigen vectors of the square matrix A.Recall that for any square matrix An×n, if there exists a scaler λ anda non-zero vector X such that

AX = λX ,

then the scaler λ is called an eigen value of the Matrix A, and thevector X is called the eigen vector associated with λ.The eigen value(s) of Matrix A can be found by solving the followingcharacteristic equation:

det [A− λI ] = 0


Eigen Values & Eigen Vectors of a Square Matrix: (Contd.)

For the 2× 2 coeffi cient matrix A ≡[a11 a12a21 a22

], the characteristic

equation is given by:

det[a11 − λ a12a21 a22 − λ

]= 0

i.e., λ2 − (a11 + a22)λ+ (a11a22 − a12a21) = 0

This is a quadratic equation in λ, which can be solved to derive twoeigen values and the corresponding eigen vectors of matrix A − whichcan then be used to transform matrix A into a diagonal (or almostdiagonal) form.There are three mutually exclusive possibilities. The eigen values (i.e.,the roots of the above quadratic equation) are

real and distinct;real and repeated;complex conjugate.


Diagonalization of a Square Matrix when the eigen valuesare real & distinct:

Let the two roots of the above quadratic equations are given by λ1and λ2 such that

λ1 and λ2 are both real numbers;λ1 6= λ2.

Let E1 ≡(e11e21

)and E2 ≡

(e12e22

)be a corresponding set of

eigen vectors such that

[A− λ1I ]E1 = 0;

[A− λ2I ]E2 = 0.

Then one can construct an invertible matrix P ≡[e11 e12e21 e22

], such

that

P−1AP =[

λ1 00 λ2

]≡ D.


2X2 System of Linear & Autonomous SimultaneousDifference Equations: SolutionCase A: the coeffi cient matrix has real and distinct eigen values

Let us now go back to our 2× 2 system difference equations given by:[xt+1yt+1

]=

[a11 a12a21 a22

] [xtyt

]+

[b1b1

]or, Xt+1 = AXt + B.

Recall that the Superposition Principle requires us to find (i) ageneral solution to the homogenous part of the system; and (ii) oneparticular solution to the entire (non-homogenous) system.Also recall that to find a particular solution to the entire system, wesimply have to identify its steady state solution (x , y) (which alsohappens to be a particular solution to the system, corresponding tothe initial condition: x0 = x and y0 = y .)


2X2 System of Linear & Autonomous SimultaneousDifference Equations: SolutionCase A: the coeffi cient matrix has real and distinct eigen values (Contd.)

The steady state of the system is given by the solution to thefollowing two equations:

x = a11x + a12y + b1;

y = a21x + a22y + b2.

Let the steady state solution be represented by:

X =[xy

].

Once the steady state solution X is identified (assuming that itexists), we then have to find the general solution to the homogenouspart, given by:

Xt+1 = AXt .



Recall that A has eigen values that are real and distinct, given by λ1

and λ2, and the associated eigen vectors E1 ≡(e11e21

)and

E2 ≡(e12e22

)respectively, such that we can construct an invertible

matrix P ≡[e11 e12e21 e22

], where

P−1AP =[

λ1 00 λ2

]≡ D.

Let us now define a new set of state variables Zt such thatZt ≡ P−1Xt for all t.Then by construction,

Zt+1 ≡ P−1Xt+1 = P−1AXt = P−1APZt ≡ DZt .Das (Lecture Notes, DSE) Macro August 18, 2014 9 / 43


In other words, from the original homogenous system given byXt+1 = AXt , we have been able to construct a new homogenoussystem, Zt+1 = DZt , which has a coeffi ecient matrix that is diagonal:

D =[

λ1 00 λ2

].

We already know that the general solution to this new system (wherethe coeffi cient matrix diagonal) will be given by:

Zt ≡[z1tz2t

]=

[C1 (λ1)

t

C2 (λ2)t

]; C1, C2 arbitary constants.

From here we can easily get back the general solution to the originalhomogenous system, by inverting the relationship Zt ≡ P−1Xt ,which tells us that Xt ≡ PZt for all t.



Since we know that precise components of the P matrix (which hasbeen constructed from the eigen vectors of the A matrix), such that

P =[e11 e12e21 e22

],

we can now write the general solution to the entire(non-homogenous) system of equations, given by Xt+1 = AXt + B,as:

Xt = PZt + X ≡[e11 e12e21 e22

] [C1 (λ1)

t

C2 (λ2)t

]+

[xy

];

where C1, C2 are arbitary constants.



In other words,

xt = e11C1 (λ1)t + e12C2 (λ2)

t + x ;

yt = e21C1 (λ1)t + e22C2 (λ2)

t + y ;

where C1, C2 are arbitary constants.

Notice that given some pre-determined initial values, x0 and y0, theprecise particular solution can be derived form the above generalsolution by simultaneously solving for the arbitrary constants C1 andC2 such that

x0 = e11C1 + e12C2 + x ;

y0 = e21C1 + e22C2 + y .



We are now interested in the stability property of this system.It is easy to see that for any arbitrary values of C1 and C2 (i.e.,starting fro any initial condition)

if |λ1 | , |λ2 | < 1, then the system is globally asymptotically stable;if |λ1 | , |λ2 | > 1, then the system is unstable.

What happens if one of the eigen values is greater than unity (inabsolute value) while the other one is less then unity?

Without any loss of generality, let us assume that |λ1 | > 1 while|λ2 | < 1.From the general solution it is clear that then the first term in the RHSwill explode while the second term in the RHS will tend to zero overtime.Thus in general the system will not converge to the steady state -unless C1 = 0.



What does C1 = 0 imply?

Recall that the precise values of C1 and C2 depend on the given initialcondition: x0 and y0.

Thus C1 = 0 will be satisfied only for some specific initial values of xand y .

Thus the system will approach the steady state if and only if it startsfrom some these specific initial values; it will be diverge otherwise.

This particular characteristics of the dynamical system is called‘saddle point property’and the corresponding steady state is‘saddle point stable’.



When dynamical system is saddle point stable, we can find the precisepath which takes the economy to the steady state by setting C1 = 0in the general solution of the system.Recall that the genral solution is given by:

xt = e11C1 (λ1)t + e12C2 (λ2)

t + x ;

yt = e21C1 (λ1)t + e22C2 (λ2)

t + y .

Now setting C1 = 0 in the two time paths above, and eliminatingC2 (λ2)

t

xt − xe12

=yt − ye22

i.e., (yt − y) =e22e12(xt − x) .


2X2 System of Simultaneous Difference Equations:SolutionCase A: the coeffi cient matrix has real and distinct eigen values (Contd.)

Notice that this is the precise equation of the time path thatapproaches the steady state in the xt -yt plane.

Any (xt ,yt) combination that satisfies this equation will take theeconomy to its steady state in the long run.

Any other path will diverge away from the steady state over time.


2X2 System of Simultaneous Difference Equations: PhaseDiagramCase A: the coeffi cient matrix has real and distinct eigen values

We now want to diagramatically illustrate the stability properties ofthe 2× 2 system.For that we have to draw the corresponding phase diagram. Thistechnique is particularly useful when the system is non-linear, so thatderivation of explicit solution paths is not possible.

Recall that the system is represented by the two equations:

xt+1 = a11xt + a12yt + b1;

yt+1 = a21xt + a22yt + b2.

In the one dimensional case, we had simply plotted the xt+1 withrespect to xt . But now xt+1 is a function of two variables; so wecannot directly plot the function in the <2 space.


2X2 System of Simultaneous Difference Equations: PhaseDiagramCase A: the coeffi cient matrix has real and distinct eigen values

In order to draw the phase diagram for this two dimensional system,we plot instead the following two level curves in the xt -yt plane:

∆x ≡ xt+1 − xt = 0;∆y ≡ yt+1 − yt = 0.

Notice that the the intersection point(s) of the two level curves willimmediately identify the steady state value(s) of the system such thatif are at such a point then both x and y remain constant from thattime onwards.

Moreover, for any point (xt , yt) which does not lie on the two levelcurves, we can derive the signs of the two functions ∆x and ∆y atthat point to infer whether starting from that point, x and y wouldincrease or decrease over time.


2X2 System of Simultaneous Difference Equations: PhaseDiagramCase A: the coeffi cient matrix has real and distinct eigen values (contd.)

Notice that equation of the first level curve is given by:

∆x ≡ (a11 − 1)xt + a12yt + b1 = 0

i.e., yt =(1− a11)a12

xt −b1a12.

Since we know precise values of all the coeffi cients (namely, a11, a12,and b1), we can plot this line directly in the x-y plane.Similarly, equation of the second level curve is given by:

∆y ≡ a21xt + (a22 − 1) yt + b2 = 0

i.e., yt =a21

1− a22xt +

b21− a22

.

Once again since precise values of all the coeffi cients (namely, a21,a22, and b2) are known, we can again plot this line directly in the x-yplane.


2X2 System of Simultaneous Difference Equations: PhaseDiagram (Contd.)Case A: the coeffi cient matrix has real and distinct eigen values (contd.)

Notice that these two lines give you the precise loci of all the xtand ytvalues such that either ∆x = 0 or ∆y = 0 (or both).What happens when you are not on one of these loci? In whichdirection would the variables move?

That will depend on the signs of the two functions ∆x and ∆y at thatpoint.

In particular, if at that point ∆x > 0, that would imply that x-valuewill increase (opposite happens if ∆x < 0).Likewise, if at that point ∆y > 0, that would imply that y -value willincrease (opposite happens if ∆y < 0).



How does one find out whether ∆x ≷ 0 and whether ∆y ≷ 0 at thispoint?If one knew the precise location of any point, then one could easilyfind out the corresponding signs of ∆x and ∆y .For example, take any particular point, say (xt , yt), which does not lieon the two level curves.Since you already know the exact equation of the two functions:

∆x ≡ (a11 − 1)xt + a12yt + b1 ≡ F (x , y) (say)

∆y ≡ a21xt + (a22 − 1) yt + b2 ≡ G (x , y) (say)you should be able to substitute xt = xt and yt = yt in the RHS ofthe above two expressions to see whether ∆x ≷ 0 and whether∆y ≷ 0 at this point.



Problem is: to do this you have to know the precise values of (xt , yt).Moreover you have to repeat this process for every point in the xt -ytplane that does not lie on the two loci. And there are infinite suchpoints!

A more feasible approach is to use the two level curves ∆x = 0 and∆y = 0 as your reference point and then utilise any of the two partialderivatives of the two functions (i.e., Fx or Fy for the ∆x function;and Gx or Gy for the ∆y function) to deduce the sign of the ∆xfunction and the the ∆y function on either side of the level curve.We use an example below to show how this procedure works.


Drawing of the Phase Diagram - An Example:

AN EXAMPLE:

Let us first consider the phase line for ∆x .Recall that the ∆x function is given by:

∆x ≡ (a11 − 1)xt + a12yt + b1 ≡ F (x , y).

Hence the equation of the corresponding level curve ∆x = 0 is:

yt =(1− a11)a12

xt −b1a12.

Now suppose it is given that a11 < 1; a12 > 0; b1 < 0.

Then the slope of the level curve ∆x = 0 will be positive (in the x-yplane) and it will have a positive intercept with the y -axis (as shownin the slide below).


Drawing of the Phase Diagram - An Example: (Contd.)

Now to draw the directional arrows when you are not on the ∆x = 0line, let us look at the two partial derivatives: Fx = (a11 − 1) < 0;Fy = a12 > 0.We can use any of these two partial derivatives (as long their signsare unambiguous) to draw the directional arrows - as shown in thenext slide.



Suppose we use the partial derivative Fx . In this example, Fx < 0.

The negative sign of this partial derivative implies, keeping y -valueunchanged, if we keep increasing x , then the functional value ∆x willdecrease.

Now fix the y at any arbitrary level on the y -axis, and draw ahorizontal line from there (as shown by the dotted line in the abovefigure).

Along this line, as you move from left to right, x is increasing andtherefore ∆x-value is falling (since Fx ≡ ∂∆x

∂x < 0).

Notice that this horizontal line intersects the level curve at somepoint. Call it point A (as shown in the diagram above).

By definition, at the point ∆x-value is exactly equal to zero. Use thispoint as your reference point and compare it with any other point onthe horizontal line.



Take any point to the left of A, say a point like B in the diagram.The corresponding ∆x-value would be higher here.Since ∆x = 0 at A, a higher value of ∆x must imply that ∆x > 0 atB.

This is true about any such point to the left of the level curve. So wedraw a horizontal arrow in the rightward direction here to signifythat x (which is measured on the horizontal axis ) is increasing here.Now Take any point to the right of A, say a point like C in thediagram. The corresponding ∆x-value would be lower here.Since ∆x = 0 at A, a lower value of ∆x must imply that ∆x < 0 at C .This is true about any such point to the right of the level curve. Sowe draw a horizontal arrow in the leftward direction here to signifythat x (which is measured on the horizontal axis ) is decreasing here.



Notice that we have used the sign of the partial derivative Fx todeduce the direction of the arrows. We could have used the otherpartial derivative Fy to deduce the same.

The only difference is: if you want to use Fy then you have to keepthe x-value unchanged, while increasing the y . In other words, youhave to now fix the x at any arbitrary level on the x-axis, and draw avertical line from there to see whether the functional value ∆x willincrease or decrease along such a vertical line. (Since Fy > 0 in thisexample ∆x will increase as you move upward along such a verticalline.)

Now once again identify the point where this vertical line intersectsthe level curve and use this point as your reference point and compareit with any other point on the vertical line.

The resulting direction of arrows should be exactly the same as wehave drawn above. (Verify this yourself)


Drawing of the Phase Diagram - An Example (Contd.):

Let us now consider the phase line for ∆y .Recall that the ∆y function is given by:

∆y = a21xt + (a22 − 1) yt + b2 ≡ G (x , y).

Hence the equation of the corresponding level curve ∆y = 0 is:

yt =a21

1− a22xt +

b21− a22

..

Now suppose it is given that a22 > 1; a21 > 0; b2 < 0.

Then the slope of the level curve ∆y = 0 will be negative (in the x-yplane) and it will have a positive intercept with the y -axis (as shownin the slide below).


Drawing of the Phase Diagram - An Example (Contd.):

Once again to draw the directional arrows when you are not on the∆y = 0 line, we have to look at the two partial derivatives:Gx = a21 > 0; Gy = (a22 − 1) > 0. We can use either of the twopartial derivatives.



Suppose we use the partial derivative Gy . In this example, Gy > 0.

The positive sign of this partial derivative implies, keeping x-valueunchanged, if we keep increasing y , then the functional value ∆y willincrease.

Now fix the x at any arbitrary level on the x-axis, and draw a verticalline from there (as shown by the dotted line in the above figure).

Along this line, as you move from south to north, y is increasing andtherefore ∆y -value is increasing too (since Gy ≡ ∂∆y

∂y > 0).

Notice that this vertical line intersects the level curve at some point.Call it point A′ (as shown in the diagram above).

By definition, at the point ∆y -value is exactly equal to zero. Use thispoint as your reference point and compare it with any other point onthe vertical line.



Take any point below A′, say a point like B ′ in the diagram. Thecorresponding ∆y -value would be lower here.Since ∆y = 0 at A′, a lower value of ∆y must imply that ∆y < 0 atB ′.

This is true about any such point below the level curve. So we draw avertical arrow in the downward direction here to signify that y(which is measured on the vertical axis ) is decreasing here.Now Take any point above A′, say a point like C ′ in the diagram.The corresponding ∆y -value would be higher here.Since ∆y = 0 at A′, a higher value of ∆y must imply that ∆y > 0 atC ′.

This is true about any such point above the level curve. So we draw avertical arrow in the upward direction here to signify that y (whichis measured on the vertical axis ) is increasing here.



Once again we could have used the other partial derivative Gx todeduce the same.

The resulting direction of arrows should be exactly the same as wehave drawn above. (Again verify this yourself).



Combining the two pictures, we get the complete phase diagram forthis particular example as:

Under the given parameter specifications, the phase diagramcharacterises the steady state as a saddle point.


2X2 System of Simultaneous Difference Equations: PhaseDiagram (Contd.)

Under alternative parametric specifications, we may get a steady statewhich is globally asymptotically stable:


2X2 System of Simultaneous Difference Equations: PhaseDiagram (Contd.)

Or a steady state which is unstable:


2X2 System of Linear & Autonomous SimultaneousDifference Equations:Other Possibilities:

Case B: Suppose the coeffi cient matrix has real and repeated eigenvalues, given by λ.

In this case the steady state solution will be stable or unstabledepending on whether |λ| < 1 or > 1.

Case C: Suppose the coeffi cient matrix has complex conjugate eigenvalues, given by α± iβ.

In this case the solution paths for xt and yt will exhibit cyclicalbehaviour;The cycles will be damped (asymptotically stable), uniform (stable) orexplosive (unstable) depending on whether |α| < 1, = 1, or > 1.


Dynamical Systems in Continous Time: DifferentialEquations

Differential equations are very similar to difference equations exceptthat now time is measured in continuous units.A first order linear, autonomous differential equation is given by:

dxdt= axt + b.

We can integrate this equation directly to get the general solution as:

xt = C expat +x ,

where C is a arbitrary constant and x ≡(−ba

)is the corresponding

steady state value.Notice that the steady state still represent a ‘rest point’where thevalue of the state variable does not change anymore. However, sincetime is now continuous, a steady state of a difference equation is now

defined as x :dxdt= 0.


Differential Equations (Contd.):

Note that the solution to the above differential equation looks verysimilar to its counterpart in discrete time, i.e.,

xt+1 = axt + b,

whose solution was given by

xt = C (a)t + x .

The only dissimilarity lies in terms of the first term which isexponetial function for a differential equation and a power functionfor a difference equation. (Of course, the steady state valuesthemselves are different).

Thus the stability of the continous time system would depend onwhether a < 0 or > 0, instead of whether |a| < 1 or > 1.



In particular, a first order linear, autonomous differential equation ofthe form:

dxdt= axt + b,

which has a general solution given by:

xt = C expat +x ,

will be global asymptotically stable if a < 0; unstable otherwise.



It can be shown that the solution to a 2× 2 first order linearautomonous system of differential equations given by

dxdt

= a11xt + a12yt + b1;

dydt

= a21xt + a22yt + b2,

would also look very similar to the solutions that we derived for theanalogous system of difference equations.

Notice that once again the coeffi cient matrix will given by

A =[a11 a12a21 a22

].

Thus the nature of the solution will once again depend on the eigenvalues and the corresponding eigen vectors of the coeffi cient matrix A.



In particular, suppose the coeffi cient matrix has real and distinctroots, given by λ1 and λ2, with the associated eigen vectors

E1 ≡(e11e21

)and E2 ≡

(e12e22

)respectively.

Then we can apply an analogous method to arrive at the generalsolution to the above system, which is given by:

xt = e11C1 expλ1t +e12C2 expλ2t +x ;

yt = e21C1 expλ1t +e22C2 expλ2t +y ;

where (x , y) represent the steady state solution to this system of

differential equations, which are obtained by settingdxdt= 0 and

dydt= 0.



The stability property of the above system will also be analogous.

In particular, if the coeffi ceint matrix has real and distinct eigenvalues λ1 and λ2, then the system will be

globaly asymptotically stable if λ1,λ2 < 0;unstable if λ1,λ2 > 0;saddle point stable if one eigen value, say λ1 > 0, while the other one,λ2 < 0.



Because the method of solving system of differential equations is verysimilar to that of difference equations, we shall not discuss themseparately.


lecture 10 mausumi dasecondse.org/wp-content/uploads/2014/08/compulsory-macro...eigen values &...

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