Heijdra Foundations of Modern Macroeconomics (Oxford, 2002)

b t

for 0 < t < N and of rc,r. 0 < t < N then:

(PS)

e simple case with f(t)

'31 order of f (t) so that

-order time derivatives

(P6)

Iced so far and introduce means of the following

Ind v = f(t).

•

Here goes the three-step procedure:

Step 1: Set up the subsidiary equation. By taking the Laplace transform of (A.88) and noting (P6) we get:

By substituting (A.89) in (A.90) we obtain the subsidiary equation of the differential equation including its initial conditions.

We now do the easy stuff of algebraically manipulating the expression (A.91) in s-space. We notice that the quadratic on the left-hand side of (A.91) can be written as s2 + 4s + 3 = 3) so we can solve

for L{y, s} quite easily: 4

3s + 133(s + 1) + 10

ED/, s} = (s + 1)(s + 3) = (s + 1)(s + 3)

310

▪s + 3 (s + 1 )(s+ 3)

Step 3: Invert the transform to get the solution of the given problem. We have now written the (Laplace transform of the) solution in terms of known transforms. Inversion of (A.92) is thus straightforward and results in:

Of course we could have obtained this solution quite easily using the standard techniques so for this simple example the Laplace transform technique is not that

4 We show the trivial steps leading to the final result in order to demonstrate that the algebra involved in s-space is indeed trivial. In general, the work involved in step 2 of the procedure is always easier than tackling the problem directly in t-space.

683

Mathematical Appendix

useful. It has some value added but not a lot. The thing to note, however, is that the method is essentially unchanged for much more complex problems. We now study two such cases.

Assume that the differential equation (A.88) is replaced by:

where g(t) is some (piecewise continuous) forcing function which is time-dependent and has a unique Laplace transform L{g , s}. The initial conditions continue to be as given in (A.89). Using the same procedure as before we derive the solution of the subsidiary equation in terms of the Laplace transforms:

The first term on the right-hand side is the same as before (see (A.92)) and results from the initial conditions of the problem. The second term on the right-hand side represents the influence of the time-varying forcing term. Two further things must be noted about equation (A.95). First, the expression is perfectly general. A whole class of shock terms can be used in (A.95) to solve for y(t) after inversion. Second, it should be noted that all of the model's dynamic properties are contained in the quadratic function appearing in the denominator. In fact, H(s) (s+1)(s+3) is often referred to as the transfer function in the engineering literature since it transfers the shock (the "input") to the variable of interest (the "output")—see for example Boyce and DiPrima (1992, p. 312). The inverse of H(s), denoted by h(t) L-1 {H(s)}, is called the impulse response function of the system.

A.6.3 Systems of differential equations

The transform method is equally valuable for systems of differential equations. Suppose that the dynamic model is given in matrix form by:

where A is the two-by-two Jacobian matrix with typical element 84, and gi(t) are (potentially time-varying) shock terms. Note that a system like (A.96) occurs quite regularly in analytical low-dimensional macro models.

By taking the Laplace transform of (A.96), and noting property (P5) we get:

where A (s) s/ -- A is a t know from matrix _

A(s) -1 = 1 ac'.

IA(s)I

where adjA(s) is the adj the determinant of A

and:

A (s) I = (s - 311)k,

where tr(A) and 1 Al al, The quadratic equati, -

I

IA(s)I = (s - A.1 J o

where A l and A2 are th

Before going on we no two case we have:

tr(A) = + A2,

i.e. the sum of the ch and the product of ti., is often very useful fo why this is so by lo( [tr(A)] 2 > ( ) 41A tr(A) > 0 there must lo one positive (unstu, .L Let us now considt literature for which I

5 These characteristic the (system of) different; ,• positive (negative) roots for all t. See section A.6.4 t

note, however, is that problems. We now

(A.94)

ik :h is time-dependent Mons continue to be as the solution of the

(A.95)

(A.92)) and results !rm on the right-hand Two further things is perfectly general. A y(t) after inversion.

roperties are contained

tact, H(s) (s+1(s+3) is

:ture since it transfers put")—see for example I by h(t) ,C- 1 {H(s)},

iferential equations.

(A.96)

ement 8q, and gi(t) are (A.96) occurs quite

Lefty (PS) we get:

(A.85)

Mathematical Appendix

where A(s) s/- A is a two-by-two matrix depending on s and the elements of A. We know from matrix algebra that the inverse of this matrix, A(s) -1 , can be written as:

where adjA(s) is the adjoint matrix (see above in section A.2.4) of A(s) and I A(s)1 is the determinant of A(s). For the simple two-by-two model adjA(s) and 1 A (s)1 are:

where tr(A) andl Al are, respectively, the trace and the determinant of the matrix A. The quadratic equation in (A.100) can be factored as follows:

Before going on we note—by comparing (A.100) and (A.101)—that for the two-by- two case we have:

i.e. the sum of the characteristic roots equals the trace of the Jacobian matrix A and the product of these roots equals the determinant of this matrix. This property is often very useful for deducing the signs of these roots. It is not difficult to see why this is so by looking at (A.102). We note that the roots are real (imaginary) if [tr(A)]2 > < ) 41A1 and that they are distinct provided [tr(A)] 2 41A1. Also, if tr(A) > 0 there must be at least one positive root. Finally, if Al1 < 0 there is exactly one positive (unstable) and one negative (stable) real characteristic root. 5

Let us now consider the two cases encountered most often in the economics 'literature for which the roots are real and distinct, i.e. [tr(A)] 2 > 41A1.

5 These characteristic roots are going to show up in exponential functions, eAi t , in the solution of

Mathematical Appendix

Both roots negative (Aq, A2 < 0)

We can use (A.97), (A.98), and (A.101) to derive the following expression in Laplace transforms:

which is in the same format as (A.95), with H(s) adj A(s)/[(s — Ai)(s — A2)] acting as the transfer function. To solve the model for particular shocks it is useful to reexpress the transfer function. We note that for the two-by-two case adjA(s) has the following properties:

A l — A2

where the second result follows from the first. We can now perform a partial fractions expansion of the transfer matrix:

By using (A.106) in (A.104) we obtain the following general expression in terms of the Laplace transforms:

Suppose that the shocks are step functions and satisfy gi(t) = gi for i = K, Q and t > 0. The Laplace transform for such step functions is Ligi, ) = gils which can be substituted in (A.107). After some manipulation we obtain the following result:

where B adjA(A1)/(kt The expression is no child's play:

[K(t) ]_ [Be).,t _ Q(t) I

B

A,

Equation (A.109) transition, and long-n correctly we verify that t = 0 and the long-: for t = 0 we have that

t oo, exit 0 (sinc

I

K(oo) i =

Q(00)

[

= ad- •

which is the same sc. shock in (A.96) and it effects check out! I We could have chec terms of Laplace tra shocks). We need the

Property 6 If the

liMf(t) = lim s _

S-. ,

and the final-value the

lim f (t) = lirn sC1 too

PROOF: See Spiegel ( 1!

6 These weighting m..

B I — B adjAo' Xi + X2 ---ki

These results are used bek where n is the order of A (

g expression in Laplace

(A.104)

' f (s - X 1 )(s - A2)] acting shocks it is useful to re- 'wo case adjA(s) has the

now perform a partial

(A.106)

1 expression in terms of

(A.107)

C{gQ, s}

(t) = gi for i = K, Q and s} = gds which can be the following result:

(A.108)

gK gQ

Mathematical Appendix

where B adj A (A MA - A2) and I -B -adj A(X2)/(X1 — A2) are weighting matrices. 6 The expression is now in terms of known Laplace transforms so that inversion is

child's play:

Equation (A.109) constitutes the full solution of the problem. It yields impact, transition, and long-run results of the shock. To check that we have done things correctly we verify that we can recover from (A.109) the initial conditions by setting t = 0 and the long-run steady state by letting t -+ oo. The first result is obvious: for t = 0 we have that exit = 1 so that K(t) = K(0) and Q(t) = Q(0). Similarly, for t oo , -> 0 (since both roots are stable) and we get from (A.109):

which is the same solution we would have obtained by substituting the permanent shock in (A.96) and imposing the steady state. So at least the initial and ultimate effects check out!

We could have checked our results also by working directly with the solution in terms of Laplace transforms (i.e. (A.107) in general and (A.108) for the particular shocks). We need the following two properties to do so.

Property 6 If the indicated limits exist then the initial-value theorem says:

PROOF: See Spiegel (1965, p. 20).

6 These weighting matrices also satisfy:

BI — B adjA(0) adjA =

AV- A2 - - A1A2 AlA2

These results are used below. Note that we have used the fact that adjA(0) = adj(—A) = (— 1)n- ladj where n is the order of A (n = 2 here). See Lancaster and Tismenetsky (1985, p. 43).

687

Mathematical Appendix

Applying Property (P7) directly to (A.108) we obtain:

= [B + I - B] KM 1= [K(0)

Q(0) Q(0)

Similarly, applying Property (P8) to (A.108) we get:

Roots alternate in sign (t,1 < 0 < A2)

A situation which occurs quite regularly in dynamic macro models is one in which the Jacobian matrix A in (A.96) has one negative (stable) root and one positive (unstable) root. The way to check for such saddle-point stability is either by means of (A.102) or (A.103). From (A.102) we observe that if I A I < 0 then we have distinct and real roots for sure since .1(trA) 2 - 4IA I > 0. Also, since I A l < Xi < 0 it must be the case that A. 1 < 0 < A2. Of course we also see this directly from (A.103).

The beauty of the Laplace transform technique is now that (A.104) is still appropriate and just needs to be solved differently. Let us motivate the alternative solution method heuristically by writing (A.104) as follows:

[

In a two-by-tb non-predetern condition (and (the value of capital, assets, variable (e.g. a It is clear from Note that th that the denoi can still obtail

s} is if the i.e. if:

. adj A (,2 )

All except one such that (A.1 in one unkn since A(A.2) is to compute Q

[ 0

Q(0) =

We next use

7 Of course, e In general, i

dent and the rar A of rank n — 1

K(0)

Q(0)

—A2S "5 — A2)

In a two-by-two saddle-point stable system there is one predetermined and one non-predetermined (or "jumping") variable so we need to supply only one initial condition (and not two as before). Let us assume that K is the predetermined variable (the value of which is determined in the past, e.g. a stock of human or physical capital, assets, etc.) so that K(0) is given. But then Q is the non-predetermined variable (e.g. a (shadow) price) so we must somehow figure out its initial condition. ?

It is clear from (A.111) how we should do this.

Note that the instability originates from the unstable root A2. For s = A2 we have that the denominator on the right-hand side of (A.111) is zero. The only way we can still obtain bounded (and thus economically sensible) solutions for L{K, s} and L{Q, s} is if the numerator on the right-hand side of (A.111) is also zero for s = A2, i.e. if:

[ K(0)

Q(0)

All except one of the variables appearing in (A.112) are determined so Q(0) must be such that (A.112) holds. At first view it appears as if (A.112) represents two equations in one unknown but that is not the case. A theorem from matrix algebra says that, since A(A2) is of rank 1 so is adjA(A 2 ).8 So, in fact, we can use either row of (A.112) to compute Q(0):

0 = [ A2 — 822 812 [ K(0) 4- L{gi A2)

0 821 A.2 — 8 11 Q(0) + LtgQ, A21

[

Q(0) = —rfgQ, A21 — ( Az — 822) [K(0) + Ligx, A211 812

• models is one in which root and one positive bility is either by means 0 then we have distinct e < XiX2 < 0 it must

ctly from (A.103).

t (A.104) is still approthe alternative solution

(A.111)

We next use (A.105), (A.111), and (A.112) to get:

(A.115)

Of course, economic theory suggests which variables are predetermined and which ones are not. 8 In general, if the n-square matrix A has distinct eigenvalues its eigenvectors are linearly independent and the rank of A(Xi) A.;/ — S is n — 1 (Ayres, 1974, p. 150). Furthermore, for any n-square matrix

A of rank n — 1 we have that adjA is of rank 1 (Ayres, 1974, p. 50).

Mathematical Appendix

where we have used (A.112) in the last step. Note that in (A.115) all effects of the unstable root have been incorporated and only the stable dynamics remains (represented by the term involving s — Xi).

Suppose again that the shocks satisfy gi(t) = gi for i = K, Q and t > 0 so that s} = s and:

s} — LIgi, A.21 — I s — I A.2 = gi

s — A2 s— A2 SA2

By using these results in (A.115) we obtain the full solution of the saddle-point stable model:

where we have used (A.105) and the result in footnote 6, and where Q(0) is obtained by substituting the shock terms in either (A.113) or (A.114). By inverting (A.116) we obtain the solution in the time dimension.

[

The key point to note is that the stable root determines the speed of transition between the respective impact and long-run results.

A.6.4 Hysteretic models

We now consider a special class of models that have the hysteresis property. With hysteresis we mean a system whose steady state is not given, but can wander about and depends on the past path of the economy. Mathematically, this property implies that the Jacobian matrix of a continuous-time system has, apart from some "regular" (non-zero) eigenvalues, a zero eigenvalue. 9 Hysteretic systems are important in macroeconomics because they allow us to depart from the rigid framework of equilibrium, ahistorical, economics. Put differently: history matters in such systems.

9 Note that in a discrete-time setting a model displays hysteresis if it contains a unit root. Amable et al. (1994) argue that it is inappropriate to equate zero-root (or unit-root) dynamics with "true" hysteresis. Strong hysteresis is a much more general concept in their view and they suggest that zero-root dynamics at best captures some aspects of this concept.

In the remainder studied above can ei restrict attention to literature, namely tv negative, respecth

Non-positive roots ()

Suppose that the ma so that the system Ii Clearly, since 1.6.1 = the lon - ults

. owever, the for A2 = 0, i.e. the gi

where B adj A assume that there i! t > O. Ina non-hy as the system will el determined by the 1 In stark contrast. effects. In order to d

(A.118):

Equation (A.119) a determined) initial we derive from (A.l

lim

sL{K, s}

s-÷o[ sL{Q, s}

where we have second line. Equati

in (A.115) all effects of ', 1e dynamics remains

K, Q and t > 0 so that

`'-)n of the saddle-point

gQ: 1

gx -A1

SQ s( s - xi))

Id where Q(0) is obtained 4). By inverting (A.116)

(A.117)

the speed of transition

In the remainder of this section we show that the Laplace transform methods studied above can easily be applied in low-dimensional hysteretic models also. We restrict attention to the two cases encountered most frequently in the economics literature, namely two-dimensional models with both roots non-positive and nonnegative, respectively.

Non-positive roots (A1 < 0 = A2)

Suppose that the matrix A in (A.96) satisfies lAl = X1X2 = 0 and tr(A) = A l + A2 < 0 so that the system has a zero root and is hysteretic, i.e. Al = tr(A) < 0 and A2 = 0. Clearly, since I A l = 0, the inverse matrix A -1 does not exist and we cannot compute the long-run results of a shock by imposing the steady state in (A.96) and inverting A. However, the derivations leading from (A.104) to (A.107) are all still valid even for A2 = 0, i.e. the general solution in Laplace transforms is:

where B adj A(A4)/A1 and I - B -adj A(0)/A1 are weighting matrices. Now assume that there is a temporary shock, i.e. gi(t) = gie-W for i = K, Q, > 0, and t > 0. In a non-hysteretic model such a temporary shock has no effect in the long run as the system will eventually just return to its initial steady state which is uniquely determined by the long-run values of the shock terms.

In stark contrast, in a hysteretic model, a temporary shock does have permanent effects. In order to demonstrate this result we first substitute L{gi, s) = gi/(s + into (A.118):

[

ins a unit root. Amable et al. Mimics with "true" hysteresis. west that zero-root dynamics

where we have used the fact that adj A(0) = -adj A in going from the first to the second line. Equation (A.120) shows that the hysteretic system does not return to

Mathematical Appendix

its initial state following the temporary shock. It is not unstable, however, because it does settle down in a new "steady state" (for which K(oo) = ( * (00) = 0) but the position of this new steady state depends on the entire path of the shock terms, i.e. in our example on K andQ. The ultimate steady state is thus "path dependent" which explains why another term for hysteresis is path dependency.

Intermezzo

Pegging the, nominal interest rate. Giavazzi and Wyplosz (1985, p. 355) give a simple example of a hysteretic system. Consider the following simple macroeconomic model:

where m, y, y, and p are, respectively, the money supply, actual output, full employment output, and the price level (all in logarithms), r and i are the real and nominal interest rate, respectively, and y represents the exogenous elements of aggregate demand. The monetary authority uses monetary policy to peg the nominal interest rate (at i(t) io) so the LM curve residually deter-

mines the money supply. By combining the Fisher relation with the IS curve we obtain km = wo[y(t) (t)] + io. By differentiating this expression and

the AS curve—keeping the other exogenous variables constant—we obtain the system in the required format:

where the Jacobian matrix has characteristic roots A i = –0 and X2 = 0 and it is assumed that both p and y are predetermined variables (so that dp(0) dy(0) = 0). Now consider the effects of a temporary boost in aggregate demand, i.e. dy(t) e-4Dt for 0 and t > 0. Using the methods developed in this subsection we derive:

Despite the fact that the shock is purely transitory it has a permanent effect on the price level.

Non- negative roots O.]

We now assume that . that A l = 0 and ;4, 2 = A.6.3 is relevant. The Al = 0 in (A.115):

[L{K, s) ]_[

L{Q, s}

Let us once again assu

Ligi, sl = (s + t,

LIgi, s) – .C{g„ A- s– X2

Equation (A.121) can I d

s,C{K, s} L{Q, s}

where Q(0) follows fro (P8) in (A.123) we cIL

As in the outright stab the ultimate long-rui

Intermezzo

Current account di of a small open eca

1° In going from the fir, adj A(0), and recall that a