Compulsory Macro Lecture 9

004: Macroeconomic TheoryLecture 9

Mausumi Das

Lecture Notes, DSE

August 14, 2014

Das (Lecture Notes, DSE) Macro August 14, 2014 1 / 18

First Order Linear Non-Autonmous Difference Equation:Solution & Stability

In last class we have seem that for a first order non-autonomouslinear difference equation of the form

xt+1 = axt + bt

can have two possible solutions:

A Backward Solution that assumes existence of a pre-determinedinitial value;A Forward Solution that assumes existence of a well-defined terminalvalue.

The Backward Solution to the above equation is given by:

xt = Cat +t−1∑i=0aibt−1−i .


Backward Solution: Stability

We also know that the backward solution will be globallyasymptotically stable if

1 |a| < 12 |bt | 5 B for all t, where B is a finite constant.

We now turn to the derivation (and stability) of the correspondingforward solution.


First Order Linear Non-Autonmous Difference Equation:Forward Solution

Consider the same non-autonomous difference equation as before:

xt+1 = axt + bt

But instead of an initial condition, suppose we are given a terminalcondition specifying that

limt→∞

xt = x ,

where x is a finite constant.

Since we have to use a terminal value, we could rewrite the differenceequation to express the current value of the variable in terms of itsfuture value as follows:

xt =1axt+1 −

1abt


Forward Solution: Existence & Stability

Now interating forward,

xt =1axt+1 −

1abt

=1a

[1axt+2 −

1abt+1

]− 1abt =

(1a

)2xt+2 −

(1a

)2bt+1 −

1abt

=

(1a

)3xt+3 −

(1a

)3bt+2 −

(1a

)2bt+1 −

1abt

.............

=

(1a

)nxt+n −

1a

n

∑i=0

(1a

)ibt+i .

Recall that the terminal value of xt is given by x .

Hence allowing n to go to ∞ (i.e., iterating infinite times forward) andreplacing the corresponding xt+n by x , we get the particular solutionfor the difference equation that satisfy the given terminal condition.


Forward Solution: Existence & Stability (Contd.)

This particular solution is given by:

xt = limn→∞

(1a

)nx − 1

a

∞

∑i=0

(1a

)ibt+i .

Notice however that for solution would indeed be consistent with thegiven terminal condition if and only if the RHS converges to a finitevalue. A suffi cient condition for that is given by

1 |a| > 12 bt 5 B for all t, where B is a finite constant.

Under these two conditions, the first term in the RHS above vanishes,while the last term converges to a finite value, say Bt .


Forward Solution: Existence & Stability (Contd.)

Thus under these conditions, the forward solution for xt is given by:

xt = −1a

∞

∑i=0

(1a

)ibt+i = Bt .

Notice that:

The forward solution is well-defined if the above conditions aresatisfied; a forward solution may not exist otherwise.When the above suffi cient conditions are satisfied, the forward solutionis also stable and converges to its terminal value x . Otherwise, even ifit exists, it explodes to infinity.The terminal value x cannot be just any arbitrary constant. It has tobe the limit point of Bt (which in turn depends on the parameter a andthe sequence of the non-autonomous terms {bt}∞

t=0).


System of Simultaneous Difference Equations:

In our analysis so far we have focussed on a single variable and therefor a single equation of motion capturing the evolution of this variable.

It is concievable that in an economy there are several variables whichchange over time simultaneously (e.g., capital stock and population)such that to analyse the dynamic behaviour of the economy, we haveto look at the evolution of all these state variables together.

If there are N = 2 such state variables, then the dynamical systemwill now be represented by N different equations of motions and wehave to analyse all these equations simultaneously.

In this course, we shall limit our analysis to a 2× 2 autonomoussystem, (i.e, 2 autonomous difference equations in 2 state variables)as described below:

xt+1 = f (xt , yt ; α)

yt+1 = g(xt , yt ; β)


2X2 System of Simultaneous Difference Equations: Linear& Autonomous

Consider the following two simultaneous difference equations:

xt+1 = a11xt + a12yt + b1yt+1 = a21xt + a22yt + b2

where a11, a12, a21, a22, b1 and b2 are given parameters of thedynamical system.We can write the above system in matrix form as:[

xt+1yt+1

]=

[a11 a12a21 a22

] [xtyt

]+

[b1b1

]or, Xt+1 = AXt + B.

where Xt represent the time-dependent state vector, A representsthe (time invariant) coeffi cient matrix and B represents the vectorof non-homogenous terms of the system.


2X2 System of Linear & Autonomous SimultaneousDifference Equations: Solution

One way to solve this system is to iterate the state vector

Xt ≡[xtyt

]backward and express it in terms of its initial value X0

(as we did in the single variable case).

This yields:

Xt = AtX0 +t−1∑i=0AiB.

Once again we can write the general solution to this 2× 2 system byreplacing X0 with some arbitrary matrix C .The only problem is that it is not so easy to comment on the stabilityproperty of the system because that would involve calculating apowered matrix and a geometric series on matrices!!


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

So we shall use the alternative method of Superposition Principle,which requires finding

a genreral solution to the homogenous part of the system (which iscalled the complementary function);then locating one particular solution to the entire (non-homogenous)system,and then adding the complementary function and the particularsolution together to arrive at the general solution to the entirenon-homogenous system.

The homogenous part of the system is given by:[xt+1yt+1

]=

[a11 a12a21 a22

] [xtyt

]



Notice that if the coeffi cient matrix was diagonal (such thata12 = a21 = 0), then finding the general solution to the abovehomogenous system would have been very very easy!!

Suppose A is a diagonal matrix given by[a11 00 a22

]. Then the

above homogenous system collapses into two independentsingle-variable difference equations, given by:

xt+1 = a11xt ;

yt+1 = a22yt .

The general solution to the above homogenous difference equationsare given respectively by:[

xtyt

]=

[C1 (a11)

t

C2 (a22)t

]; C1, C2 arbitary constants.



Having found the complementary functions, the SuperpositionPrinciple now requires identification of a particular solution to thegeneral (nonhomogenous) system.

Since we are looking only at non-autonomous systems, the equationswould always permit existence of some steady state.

So the easiest way to locate a particular solution is to identify asteady state of the correponding non-homogenous system.

Notice that when A is a diagonal the non-homogenous system is givenby:

xt+1 = a11xt + b1;

yt+1 = a22yt + b2.



We can easily calculate the corresponding steady states as follows(provided a11, a22 6= 1):

x =b1

1− a11;

y =b2

1− a22.

Thus we can now write the general solution to the non-homogenoussystem of equations (with a diagonal coeffi cient matrix) as:[

xtyt

]=

[C1 (a11)

t

C2 (a22)t

]+

[xy

]; C1, C2 arbitary constants.

The stability of the system will of course depend on the values of a11and a22.But this is only a special case where the coeffi cient matrix is diagonal.



What about a more general case when the coeffi cient matrix A isnon-diagonal?!

It turns out that a square matrix A can be transformed into adiagonal matrix by applying a procedure called ‘Diagonalization’.To apply this procedure we need a little knowledge of Matrix algebra,in particular, the concepts of eigen values and eigen vectors.


A Digression: Eigen Values & Eigen Vectors of a SquareMatrix

Consider a square matrix An×n.If there exists a scaler λ and a 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.Thus, to find an eigen value of the matrix A, we have to find anon-zero solution to the equation:

[A− λI ]X = 0,

where I in a n× n identity matrix.The above equation will have a non-zero solution if and only if thematrix [A− λI ] is singular or degenerate, i.e., its determinant is zero:

det [A− λI ] = 0


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

The above equation is called the characteristic equation of matrixA.

If A is a n× n square matrix, then the LHS of above equation will bea n-th order polynomial of λ, and its solution will give us all the eigenvalues of matrix A.

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 we can now solve to derivetwo eigen values and the corresponding eigen vectors of matrix A.


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

Using the eigen vectors, one can then construct an invertible matrixP, which can be used to to transform the original coeffi cient matrix Ainto a diagonal (or, in some cases, almost diagonal) form such that

P−1AP = D,

where D is a diagonal (or, in some cases, almost diagonal) matrix.

This way, one can transform the given system of difference equationsinto a new system, where the coeffi cient matrix of the new system isnone-other-than the diagonal matrix D (and therefore we can solvethe new system very easily).

We shall discuss this transformation mechanism in the next class.


Documents

Compulsory Macro Lecture 9