Equilibrium Dynamics and Di¤erence Equations Craig Burnside Duke University September 18, 2010

1

Equilibrium Dynamics

When we studied the generic dynamic programming problem in Stokey and Lucas we obtained the following …rst-order and envelope conditions Fy (xt ; xt+1 ) + v 0 (xt+1 ) = 0

(1)

v 0 (xt ) = Fx (xt ; yt+1 )

(2)

We can combine the …rst-order and envelope conditions to get the Euler equation Fy (xt ; xt+1 ) + Fx (xt+1 ; xt+2 ) = 0

(3)

We also derived the transversality condition lim

t!1

t

Fx (xt ; xt+1 ) xt = 0:

(4)

In this section we are concerned with the following question: How can we use the equations (1) and (2), and possibly (3) or (4) to characterize the policy function h? As Stokey and Lucas point out, in general we cannot say much, as the theorem of Boldrin and Montrucchio (see Thm. 6.1 of Stokey and Lucas) states that just about any smooth h can be obtained as the policy function to some dynamic programming problem. One way to characterize the dynamics induced by h is to use …rst order approximations to the model in the neighborhood of a steady state. We will take this approach, which is fundamentally local, in subsequent sections. In this section I consider a di¤erent approach, which is to establish global stability of equilibrium dynamics around steady states. De…nition: A point x is globally stable if for all x0 2 X and fxt g1 t=0 such that xt+1 = h(xt ), limt!1 xt = x.

Thanks to Rosen Valchev for pointing out a typo in the previous version of these notes.

1

You can soon see how hard proving global stability will be when you realize that even proving the existence of a unique steady state is di¢ cult. It’s quite obvious that a necessary condition for a point to be a steady state is that (3) hold for xt = xt+1 = xt+2 = x. Stokey and Lucas also point out that this is a su¢ cient condition if F is concave. Uniqueness of the steady state, on the other hand, is more elusive, but is a necessary condition for global stability. One approach to establishing global stability is to use a Liapounov function, if one is available for the problem in question. Lemma 4.2 in Stokey and Lucas shows how these functions could be useful. Lemma: Let X

Rl be compact and let h : X ! X be continuous with x = h(x) for some

x. If there exists a continuous function L : X ! R such that (a) L(x) 0; with equality i¤ x = x, (b) L[h(x)]

L(x); with equality i¤ x = x,

then x is a globally stable solution to (1) and (2). You should consult the proof of this theorem in Stokey and Lucas. The problem, for many cases, is that it is not obvious how to form L, nor whether an L exists. Stokey and Lucas argue that one way to proceed is to check whether the following form of L satis…es the two conditions of the Lemma: L(x) = (x

x) [v 0 (x)

v 0 (x)]:

(5)

The reason this form of function is worth trying is that it automatically satis…es the Lemma’s condition (a) when v is strictly concave.1 It’s then a matter of trying to verify (b). This turns out to be much harder. Stokey and Lucas motivate it as follows. They begin by noting that for any pair of points (x; y), (x0 ; y 0 ) the strict concavity of F implies2 (x

x0 ) [Fx (x; y)

Fx (x0 y 0 )] + (y

y 0 ) [Fy (x; y)

Fy (x0 y 0 )]

0

with equality only when the two points are the same. We can therefore let (x; y) be (x; h(x)) and (x0 ; y 0 ) be (x; x) so that (x

x) fFx [x; h(x)]

Fx (x; x)g + [h(x)

x] fFy [x; h(x)]

Fy (x; x)g

The optimality conditions imply that v 0 (x) = Fx [x; h(x)] and Fy [x; h(x)] = (x

x) [v 0 (x)

v 0 (x)]

[h(x)

1

x] fv 0 [h(x)]

v 0 (x)g

0:

(6)

v 0 [h(x)], hence 0:

(7)

The strict concavity of v implies that v 0 (x) < v 0 (x) for x > x and v 0 (x) > v 0 (x) for x < x, hence (a) is satis…ed. 2 We will usually need to assume F is concave to prove that v is concave. Also, the result holds for the same reason as given in the previous footnote.

2

Now notice that L[h(x)]

L(x) = (h(x)

x) [v 0 (h(x))

v 0 (x)]

Obviously we could use (7) to show that L[h(x)]

L(x)

(x

x) [v 0 (x)

0 if

v 0 (x)]:

= 1, but it is not. So we

cannot get a generic result that will work for any concave F and v, but we are very close. If we could somehow start from (x

x) fFx [x; h(x)]

x) fFy [x; h(x)]

Fx (x; x)g + (h(x)

instead of (6), we would have the desired result that L[h(x)]

Fy (x; x)g

0

(8)

0. So on a case by

L(x)

case basis one could try to show that (8) holds in order to demonstrate that L given in (5) is a Liapounov function. In the end, I o¤er two proofs of the global stability of the dynamics of the neoclassical model. One of these is from Stokey and Lucas, while the other is based on Brock and Mirman (1972). Both proofs start by working out the steady state of the model. Steady State Here I conjecture that the model has a steady state in which ct = c and kt = k for all t. In such a steady state, the Euler equation and resource constraint are u0 [g(kt )

kt+1 ] = u0 [g(kt+1 )

kt+2 ]g 0 (kt+1 )

ct + kt+1 = g(kt ). The steady state in which kt = kt+1 = kt+2 = k is the solution to 1 =

g 0 (k)

c = g(k)

(9) (10)

k:

We can solve (9) for k once we note that g 0 (k) = f 0 (k) + 1

. If we rewrite (9) in terms

of the rate of time preference, = 1 1 we have f 0 (k) = + . Since > 0, 0 < 1, 0 00 0 0 f (k) > 0, f (k) < 0, limk!0 f (k) = 1 and limk!1 f (k) = 0, there is a unique, positive,

…nite solution to (9). Once we have solved for k , notice that we can easily obtain c from (10). We need to verify that the solution for c is positive. This requires verifying that g(k ) k = f (k ) k > 0. Let k be the unique point at which f (k) = k. That such a point exists follows from the fact that f is increasing, concave, has in…nite slope at 0, and 0 slope at 1. Obviously f 0 (k)
k0 . We want to prove that h(k1 ) > h(k0 ). Assume the contrary, so let h(k1 )

u0 [g(k0 )

h(k0 ). Then since v is strictly concave h(k0 )] = v 0 [h(k0 )]

v 0 [h(k1 )] = u0 [g(k1 )

h(k1 )]:

And since u is strictly concave this means g(k0 )

h(k0 )

g(k1 )

h(k1 ):

implying g(k0 )

g(k1 )

h(k0 )

h(k1 )

0:

But since g is strictly increasing this means k0 k1 , which is a contradiction. With this result we can proceed to …nish the proof. For any k 2 K+ , the strict concavity of v implies

h(k)] fv 0 (k)

[k

v 0 [h(k)]g

(11)

0

with equality only at k = h(k), i.e. only at the steady state.3 The …rst order condition of the model states that u0 [g(k) v 0 (k) = u0 [g(k)

h(k)] = v 0 [h(k)] and the envelope condition states that

h(k)]g 0 (k). I will use the shorthand u0 (c) to indicate u0 [g(k)

h(k)]. If we

substitute these conditions into (11) we get [k

h(k)] [u0 (c) g 0 (k)

u0 (c)= ]

0

(12)

Since u0 (c) is strictly positive everywhere, we can take it out as a common factor in (12) and preserve the inequality: [k

h(k)] [g 0 (k)

1= ]

0;

(13)

which, again, only holds with equality if k = h(k). So for k 6= k we have [k

h(k)] [g 0 (k)

3

1= ] < 0:

(14)

This looks like the Liapounov function from equation (5), but Stokey and Lucas’proof does not use a Liapounov argument.

4

Since g 0 (k) = 0

g (k)

1

at k and g is concave we know that g 0 (k)

1= > 0 for k < k and

1= < 0 for k > k . So if k < k this tells us that h(k) > k. We also know that if

k > k then h(k) < k. With this result and the fact that h(k) is increasing, Stokey and Lucas are able to draw their Figure 6.1 [equivalent to my Figure 2 below], and this guarantees the global stability of the model’s dynamics in K+ . Brock and Mirman’s Proof Take as given the regularity assumptions made in the previous section and the results that v is concave and h is increasing. Next we can show that the policy function for consumption, c(kt ) = g(kt )

h(kt ) is increasing. Pick some k0 < k1 .

Our second result means h(k0 ) < h(k1 ). This, in turn, implies that v 0 [h(k0 )] > v 0 [h(k1 )]. And, from the …rst order condition this means u0 [c(k0 )] = u0 [g(k0 )

h(x0 )] > u0 [g(k1 )

h(k1 )] = u0 [c(k1 )]

From the concavity of u this means c(k0 ) < c(k1 ). Now de…ne A(k)

u0 [c(k)] A(k)g 0 (k):

B(k)

(15) (16)

We need to plot A(k) and B(k) against k. To do this notice that B(k) > A(k) for k < k since g 0 (k) >

1

for k < k . Similarly, B(k ) = A(k ) and B(k) < A(k) for k > k . Also,

both functions are downward sloping. To see this, notice that A0 (k) = u00 [c(kt )]c0 (kt ) < 0 since u00 < 0 and c0 > 0. Also

B 0 (k) = A0 (k)g 0 (k) + A(k)g 00 (k) < 0 since A > 0, A0 < 0, g 0 > 0 and g 00 < 0. The nonnegativity constraint on k 0 given k also implies that limk!0 c(k) = 0 so that both functions asymptote towards 1 as k goes to zero. Thus we can draw A(k) and B(k) as in Figure 1.

Now consider the dynamics of the capital stock starting from an arbitrary k0 < k . Notice

that the optimality conditions require that u0 (c0 ) = u0 (c1 )[f 0 (k1 ) + 1 that A(k0 ) = B(k1 ):

5

], or, equivalently

FIGURE 1 The Transition Dynamics of the Neoclassical Model The Functions A and B

A, B

AÝk 0 Þ = BÝk 1 Þ

AÝk 1 Þ = BÝk 2 Þ AÝkÞ

BÝkÞ k0

k1

k2

k*

k

From Figure 1, this implies that k0 < k1 < k . If k0 > k we have the opposite implication that k < k1 < k0 . As we can see in the diagram, the dynamics clearly mean that kt converges into the steady state at k . It is also clear that since c0 > 0, if k0 < k then c0 < c1 < c , if k0 > k then c < c1 < c0 and that ct converges to its steady state value. Figure 2 shows the results we have obtained regarding h(k). First it is increasing. Second, it is above the 45 degree line for k < k and below it for k > k . Also we know that limk!0 h(k) must be zero since limk!0 c(k) = 0 and limk!0 g(k) = 0. A Phase Diagram Another diagram that is often used to illustrate dynamics is a phase diagram. A phase diagram for the system of di¤erence equations u0 (ct ) = u0 (ct+1 )g 0 (kt+1 )

(17)

ct + kt+1 = g(kt ).

(18)

plots ct against kt , the two variables governed by the di¤erence equations. The phase diagram illustrates what would happen to ct and kt if we began from arbitrary values of k0 and c0 (as opposed to the optimal c0 given k0 ) and used the system of di¤erence equations to solve for (k1 ; c1 ), then (k2 ; c2 ), etc. Notice that if you know ct and kt , then you can solve (18) for kt+1 and then solve (17) for ct+1 . 6

FIGURE 2 The Transition Dynamics of the Neoclassical Model The Policy Function

kt +1

45O line

h( k t ) k3 k2

k1

k0

k1

k*

k2

k

kt

Usually we begin with a diagram such as Figure 3 that demarcates regions in which ct and kt will be either increasing or decreasing. To draw this diagram, we ask the following questions: 1. When is kt+1 > kt ? From (18) we can see that kt+1 > kt whenever ct < g(kt ) 0

kt .

0

2. When is ct+1 > ct ? Notice that ct+1 > ct implies u (ct+1 ) < u (ct ). This, in turn, using (17), implies g 0 (kt+1 ) > 1 and, hence, kt+1 < k . Unfortunately, we would like a condition that characterizes ct+1

ct in terms of (kt ; ct ) not in terms of kt+1 . Notice, however, from

(18), that kt+1 < k is equivalent to ct > g(kt )

k .

To summarize, if ct < g(kt ) kt , kt is increasing, and if ct > g(kt ) k , ct is increasing. So our diagram is divided into four regions, A (where ct and kt both increase), B (where ct increases but kt decreases), C (where ct and kt both decrease), and D (where ct decreases but kt increases). Hence the directions of the little arrows I have drawn into the diagram. We know that if ct is chosen optimally, the path of (kt ; ct ) will lie in region A or C. Why? Because in the previous section we showed that if k0 < k , the capital stock and consumption increase monotonically to their steady state values (this only happens in region A). On the other hand if k0 > k , the capital stock and consumption decrease monotonically to their steady state values (this only happens in region C). 7

FIGURE 3 Constructing the Phase Diagram for the Neoclassical Model, Step 1 g (k ) − k *

ct

C B (k * , c * ) g (k ) − k

A D

k kt

Obviously this means that one of the paths for (kt ; ct ) (these are called phase trajectories), which are illustrated in Figure 4, corresponds to ct = g(kt ) h(kt ) and converges in towards the steady state. How did I draw in the other paths? The reason the paths in Figure 4 are not smooth is that this we are illustrating the dynamics of a di¤erence equation not a di¤erential equation, which would have smooth phase trajectories. Paths Starting in B Suppose our arbitrary choice of (k0 ; c0 ) is in B. Clearly, the path of subsequent values of (kt ; ct ) moves up and to the left (NW). But we can also show that the path will eventually violate the nonnegativity constraint on kt or make ct unde…ned. To see this, notice that (18) implies kt+1 kt = g(kt ) kt ct . The size of the decrease in the capital stock is equal to the distance ct is above the curve g(kt )

kt . Clearly that distance

increases as we move along a path that leads to the NW, so the decreases in kt become larger at each step. This implies that eventually the rule for solving for kt+1 will imply kt+1 < 0 (this happens if ct > g(kt )) violating feasibility, or kt+1 = 0 (this happens if ct = g(kt )), making it impossible to solve for the value of ct+1 .

8

FIGURE 4 The Phase Diagram for the Neoclassical Model g (k t ) ct

C

c t = g ( k t ) − h( k t ) B

(k * , c * )

D

A

kt

k

Paths Starting in D If our arbitrary choice of (k0 ; c0 ) is in D the path of subsequent values of (kt ; ct ) clearly moves down and to the right (SE). However, we can also show that any path that starts in D must converge towards kt = k, ct = 0. To see this, notice that it is obvious that any path starting in D cannot escape into regions A and B, because those regions cannot be reached by SE movement that begins inside D. But we can also show that the dynamics cannot escape into region C or the boundary between C and D. To see this, imagine a (kt ; ct ) pair in D. De…ne kt as the largest solution to ct = g(kt )

kt . Notice that

since (kt ; ct ) 2D, kt > kt . But I can also show that kt > kt+1 . To see this notice that kt+1 < kt , g(kt ) , g(kt )

c t < kt

g(kt )

kt < kt

, g(kt ) < g(kt ) which is clearly true since kt < kt . With kt < kt+1 < kt and ct+1 < ct we have the result that (kt+1 ; ct+1 ) lies inside D. The reason the dynamics must converge to (k; 0) is that they always move to the SE (so they cant stall anywhere inside D) and, because the next solution for kt+1 is always positive, the next solution for ct+1 will always be positive as well, so the dynamics cannot go across the x-axis. An interesting fact about paths that begin in D is that they violate the transversality condition of the model because they all converge to (k; 0). (Here you should refer to the notes 9

on dynamic programming where transversality conditions are discussed.) The transversality condition is lim

t!1

t 0

u (ct )g 0 (kt )kt = 0:

Notice that for any path that begins in D there is some …nite T for which k^ < kt < k for ^ = 1. For any t > T we t T , where k^ is the point where g(k) k is maximized; i.e. g 0 (k) have u0 (ct ) = u0 (ct 1 )=[ g 0 (kt )] = u0 (ct 2 )=[ =

2 0

g (kt )g 0 (kt 1 )]

= u0 (cT )=[

(t T ) 0

g (kt )g 0 (kt 1 )

g 0 (kT +1 )]

Since kt > k^ for t Hence u0 (ct ) >

^ = 1 for t T , and that g 0 (kt ) < 1 for t T . T this means g 0 (kt ) < g 0 (k) (t T ) 0 u (cT ) for t T . This means t u0 (ct ) > T u0 (cT ) for t T . Therefore t 0 u (ct )g 0 (kt )kt > T u0 (cT )g 0 (k)k^ for t T . Hence lim

t!1

t 0

u (ct )g 0 (kt )kt >

T

u0 (cT )g 0 (k)k^ 6= 0:

Paths Starting in A If we start with a point (k0 ; c0 ) 2A there are three possibilities: the subsequent values of (kt ; ct ) (i) could stay in A, (ii) move into B or (iii) move in to D. We can easily eliminate the fourth possibility that the path moves into C: since ct > g(kt ) for any point in A, this means kt+1 = g(kt )

k

ct < k , implying (kt+1 ; ct+1 ) 2C. =

One path that clearly lies in A, and stays in A, is the optimal path along which ct = g(kt ) h(kt ). We can eliminate the possibility that any other phase trajectory starting in A stays in A. Any phase trajectory that stays in A converges to (k ; c ).4 But any such path satis…es (i) the di¤erence equations (17) and (18) and (ii) the transversality condition.5 From Theorem 4.15 in Stokey and Lucas this would imply that this other path was also optimal. But we know there is a unique optimal path. Therefore only the optimal path stays in A. Paths that start out above the optimal path eventually get into set B. To see this let k1 = h(k0 ) and c0 = g(k0 )

k1 . If c0 > c0 the phase trajectory stays strictly above the

optimal path. To see why notice that c0 > c0 implies k1 < k1 . Hence, u0 (c0 ) < u0 (c0 ) and g 0 (k1 ) > g 0 (k1 ). Thus, if we solve for c1 along the phase trajectory using (17) we will necessarily need u0 (c1 ) < u0 (c1 ), where c1 is the value of c1 that would prevail along the 4

The other possibilities for paths that stay in A would be (i) that the path converges to a point in A other than (k; c) or (ii) the path cycles around in A without ever converging to any point. The …rst possibility is impossible because there is only one steady state ot the system (17)–(18). The second possibility is also not possible. We know the dynamics in A go in one direction: to the NE, and do not cycle. 5 Along such a path limt!1 u0 (Ct )[f 0 (Kt ) + 1 ]Kt = u0 (C)K= , so that limt!1 t u0 (Ct )[f 0 (Kt ) + 1 ]Kt = 0.

10

optimal path. Thus c1 > c1 and the (k1 ; c1 ) pair along the phase trajectory lies strictly to the NW of (k1 ; c1 ). Since our trajectory path stays strictly above the optimal path, leaves A (shown above) and does not go into C (shown above) it must go into B. A symmetric argument implies that paths that start out below the optimal path eventually get into set D. Paths Starting in C If we start with a point (k0 ; c0 ) 2C there are three possibilities: the subsequent values of (kt ; ct ) (i) could stay in C, (ii) move into B or (iii) move in to D. Symmetric to the argument above the dynamics never move into A. Also symmetric to the argument above, the only phase trajectory that stays in C, is one that starts on the optimal path. Paths that start out above the optimal path remain above it and move into B, paths that start out below it remain below it and eventually move into D.

2

Solving First Order Scalar Di¤erence Equations

Here we will discuss …rst order deterministic and stochastic di¤erence equations. The discussion draws heavily on Sargent (1987), Chapter 9 and to some extent Chapter 11, which should be consulted by the reader for a much more thorough treatment.

2.1

Deterministic Di¤erence Equations

Consider the …rst-order di¤erence equation yt = y t

1

+ bxt + a

8t:

(19)

Here yt is treated as an unknown sequence which we would like to characterize. The sequence fxt g1 t=

1

is known and determinstic.

Di¤erence equations can be written using lag operator notation. The operator L lags a variable in the following sense. If we de…ne a new sequence fyt g in terms of the sequence fxt g and let yt = Lxt , then yt = xt 1 . So we could simply write xt xt

k

1

= Lxt . In general

k

= L xt .

In lag operator notation we rewrite (19) as (1

L)yt = bxt + a:

(20)

A solution that is immediately apparent is yt = (1

L) 1 (bxt + a):

11

(21)

One problem with this solution is that we have not yet discussed polynomials in the lag operator, such as 1 L, and how they are inverted. Naively it would seem that there are two possibilities, the backward inverse, (1

L)

1

=1+ L+

2

L2 +

(22)

;

and the forward inverse, (1

L)

1

=

1

L 1 (1 +

1

L

1

2

+

L

2

+

(23)

):

Apart from the issue of how (1 L) 1 should be de…ned, there is another problem with (21), which is that there is a more general solution: yt = (1

L) 1 (bxt + a) +

t

(24)

c;

where c is an arbitrary constant. This solution is easily veri…ed given that L

t

=

t 1

.

So there are in…nitely many solutions to (19). How do we narrow ourselves down to one solution? There are three common ways to do this: (i) requiring that the fyt g sequence is

bounded for all fxt g sequences that are bounded, (ii) treating one of the values of the fyt g sequence as a known scalar, (iii) requiring the fyt g sequence to converge to some …nite value either as t ! 1, or as t !

1.

Mapping Bounded Sequences to Bounded Sequences One way to obtain a unique solution for the sequence fyt g is to require that the solution to (19) should be bounded for any bounded sequence fxt g. Later we will see a theoretical argument for this assumption in the context of an economic model. For now we simply impose the restriction that polynomials in the lag operator should map bounded sequences to bounded sequences. Let the set of all bounded sequences be B. Clearly, L : B ! B since fLxt g 2 B for any fxt g 2 B . It is also clear that (1 L) : B ! B for any …nite value of . But the representation of (1 L) 1 given by (22) only maps from B to B when j j < 1, while the representation of (1

L)

1

given by (23) only maps from B to B when j j > 1.

Hence, when j j < 1, the requirement that fyt g 2 B for any fxt g 2 B implies that there is a unique solution for yt given by yt = (1 + L + 2 L2 + )(bxt + a) 1 X a j = b xt j + : 1 j=0 The boundedness assumption also requires that we set c = 0 in (24), since limt! ! 1 for any c 6= 0. 12

(25)

1

c

t

When j j > 1, the assumption that fyt g 2 B for any fxt g 2 B implies that there is a

unique solution for yt given by

1

yt = =

b

L 1 (1 + 1 X

(j+1)

1

L

1

xt+1+j +

j=0

We again set c = 0 in (24), since limt!1 c

t

2

+ a 1

L

2

+

)(bxt + a)

:

(26)

! 1 for any c 6= 0.

When j j = 1, fyt g is unbounded for at least some bounded fxt g sequences. Hence, in

this case, the boundedness restriction cannot be used to …nd a unique solution for yt .

Conditioning on a Known Value Suppose we know one of the elements of the sequence fyt g, say y0 . This assumption is also motivated by theory. Frequently in macroeconomics

we work with models in which a planner or economic agent maximizes a dynamic objective function given some initial level of resources. This initial level of resources at least partially determines the solution to the optimization problem. We will see an example of this below. Given a value of y0 , we may generate the solution for the rest of the fyt g sequence

recursively using (19):

yt =

yt 1

+ bxt + a if t > 0 1 yt+1 (bxt+1 + a) if t < 0: 1

We can also work through the recursions to get the solutions ( P t if t > 0 y0 + tj=01 j (bxt j + a) P t 1 (j+1) yt = t (bxt+j+1 + a) if t < 0: y0 j=0 These solutions work for all values of

(27)

(28)

and are unique given a value of y0 .

Convergence in the Limit There are some examples in which we know something about the limiting properties of the yt sequence and there is only one solution for yt that has these limiting properties. To take a simple example, suppose we have a = 0 and xt = 0 for all t, so that the general solution is simply yt = t c. If j j > 1 yet we know that limt!1 yt = 0 then the unique solution satisfying that convergence property is c = 0 and yt = 0 for all t. On the other hand if we knew limt!

1

yt = 0 we would not be able to pin down c.

Similarly if j j < 1 and we know that limt!

1

yt = 0 then the unique solution satisfying

that convergence property is c = 0 and yt = 0 for all t. On the other hand if we knew limt!1 yt = 0 we would not be able to pin down c.

13

2.2

Stochastic Di¤erence Equations

Now consider the di¤erence equation: Et 1 yt = yt

1

8t;

+ bEt 1 xt + a

(29)

where Et 1 is the expectations operator conditional on information available at time t 1. Now xt is a stochastic process, and presumably the solution for yt will also be a stochastic process. Another way of writing (29) is yt = y t where zt

1

t

= yt

1

(30)

+ bEt 1 xt + a + t ;

Et 1 yt is a white noise process. It is also convenient to use the notation

= Et 1 xt and rewrite (31) as: yt = y t

1

+ bzt

1

(31)

+ a + t:

You might think that we could simply think of (31) as a version of (19) with bxt being replaced by bzt 1 + t . Thus you might expect the general solution to be L) 1 (bzt

yt = (1

1

+

t

+ a) +

t

(32)

c;

with c being an arbitrary constant and there being two possible forms of (1

L)

1

given

by (23) and (22) respectively. The solution implied by (22) and (32), which is yt =

1 X

j

(bzt

j 1

+

t j

+ a) +

t

(33)

c;

j=0

is valid. This is easily veri…ed by noting that given (33) we have Et 1 yt = yt

1

=

1 X

j=0 1 X

j

j

(bzt (bzt

j 1

j 1

+ a) + + a) +

j=1

1 X

j=1 1 X

j

j

t j

+

t

c

t j

+

t

c:

j=1

Hence Et 1 yt

yt

1

= bzt

1

+ a;

which is the original di¤erence equation. The solution implied by (23) and (32), which is yt =

1 X

(j+1)

(bzt+j +

j=0

14

t+1+j

+ a) +

t

c;

is easily seen to be invalid, since Et 1 yt has no terms involving the future values of

t

but yt

1

does. The fact that yt must be determined as of date t imposes an additional restriction. It turns out that the correct forward-looking solution is yt =

1 X

(j+1)

(bEt zt+j + a) +

t

(34)

c:

j=0

Once again we face the dilemma that there are in…nitely many solutions to (29). We could use either representation (33) or (34). Furthermore, both (33) and (34) represent an in…nite number of possible solutions because c could be any constant. Additionally, (33) represents an in…nite number of possible solutions because

t

could be any white noise process.

The question is whether and how we can narrow down the set of possible solutions. One way to do this is to implement a variant of the boundedness condition that is based on the property of covariance stationarity. In order to do this a de…nition and a couple of theorems from time series analysis will be useful. De…nition: A stochastic process fyt g is covariance stationary if E(yt ) = )(yt s )] = s for all t. 0 < 1 and E[(yt

and Var(yt ) =

Wold Decomposition Theorem. Any covariance stationary stochastic process fxt g can be

represented as xt = yt + zt where zt is linearly deterministic [i.e. var(zt jzt 1 ; zt 2 ; : : : ) = 0] and yt is purely non-deterministic with an MA(1) representation yt = (L) t , with P1 2 j=0 j < 1. Stationarity of ARMA Processes. An ARMA process (L)yt = (L) t is covariance stationP1 2 ary if and only if (L) is square-summable, i.e. j=0 j < 1 and the roots of (Z) = 0 lie strictly outside the unit circle.

We can restrict the set of possible solutions by requiring that yt be covariance stationary for any covariance stationary xt . Covariance stationarity of yt immediately requires that c = 0 no matter which of the two representations, (33) and (34), we use because otherwise the mean of the process yt would be time varying. We will assume that xt is purely nondeterministic, allowing us to write xt = (L)et with (L) being square-summable and et being white noise. Notice that zt = Et xt+1 = ~(L)et , where ~j = j+1 and, therefore, ~(L) is square-summable. Consider the case where j j < 1. If we impose c = 0, the solution (33) can be rewritten

as

yt

yt

1

= bzt

1

+

t

+ a = b~(L)et

1

+

t

+ a:

(35)

This makes it clear that the backward looking solution represents yt as an ARMA(1,1) processes which is covariance stationary because ~(L) and the root of 1 Z = 0 is Z = 15

1

, which lies outside the unit circle. With c = 0 the forward looking solution is yt = P1 (j+1) (bEt zt+j + a). Clearly the forward looking solution, (34), can only be …nite j=0 if a = 0. But even when a = 0, there will be many covariance stationary xt processes for which the forward looking solution is ill-de…ned.6 Hence, when j j < 1 the restriction that yt should be covariance stationary for any covariance stationary xt will lead us to use yt = =

yt 1 X

1 j

+ bEt 1 xt +

t

(bEt

+

j 1 xt j

+a t j

(36)

+ a)

j=0

as the solution for yt . Although the solution is covariance stationary it remains non-unique, since t could be any white noise process. As we will see in the context of a theoretical example, below, it is sometimes possible to eliminate the multiplicity because the variable yt is known to be in the time t

1 information set. In this case, the

t

terms drop out of

(36) and the solution for yt reduces to yt = y t

1

+ bEt 1 xt + a =

1 X

j

(bEt

j 1 xt j

+ a):

(37)

j=0

Now consider the case where j j > 1. It is immediately obvious that yt will not be

covariance stationary if we use the backward-looking solution as described by (35). This 1 is because the root of the autoregressive polynomial is and this lies inside the unit

circle. On the other hand consider the forward looking solution with c = 0 is yt = P ~ P1 (j+1) (bEt zt+j + a). Given that zt = ~(L)et we have Et zt+j = 1 k=0 j+k et k so j=0

we can write

yt = =

a

b

1

1 X

(j+1)

j=0

a

b

1

1

1 X

~j+k et

k

k=0

1 X 1 X ~j+k (

k

)et j :

j=0 k=0

I am missing a little proof — which I think can be based on the convolution formula— that P k the sequence cj = 1 is square summable. This would deliver that this reprek=0 j+k+1 sentation of yt is covariance stationary. Given this result, when j j > 1 the restriction that yt should be covariance stationary for any covariance stationary xt would lead us to use yt =

1 X

(j+1)

(bEt xt+j+1 + a);

(38)

j=0

6

Consider, for example, the case where P1 xt = xt 1 + t with j j < 1. This means zt = xt and that Et zt+j = j+1 xt . With a = 0 yt = b j=0 ( = )(j+1) xt . This is not …nite when j j < j j.

16

as the solution for yt . If j j = 1 neither solution can be shown to be covariance stationary for all possible covariance stationary xt processes, but it is common, in this case, to work with the backward looking solution which is simply an ARIMA(0,1,1).

3

Solving Deterministic Models

The basic principles of solving for the equilibrium dynamics of both deterministic and stochastic models are easily illustrated using the determinstic neoclassical growth model. After reviewing the optimality conditions from the model, I describe how to linearize the …rst order conditions to generate a system of linear di¤erence equations that will then need to be solved. I then demonstrate how three di¤erent solution methods can be applied to these equations: (i) the Blanchard and Kahn (1980) approach, (ii) the shooting method and (iii) the method of undetermined coe¢ cients. When we solved the social planner’s problem for the neoclassical model we obtained the following optimality conditions: u0 (ct ) =

u0 (ct+1 )g 0 (kt+1 ),

ct = g(kt )

3.1

kt+1 ,

t

t

0

(39) (40)

0:

Linear Approximation

Given standard assumptions about u and f (and therefore g) equations (39) and (40) are nonlinear and we cannot determine the form of the policy function h, which yields the optimal value of kt+1 as a function of kt .7 A standard approach to working with equations like (39) and (40) is to approximate them with linear equations. The …rst step is to …nd the steady state of the model. This can be done by removing time susbcripts from the variables in (39) and (40) and solving for c and k: 1 =

g 0 (k)

c = g(k)

(41) (42)

k

Given a functional form for f , (41) can be solved for k, and, given this solution, (42) can be solved for c (we did this above). In the next step we …rst totally di¤erentiate (39) and (40) in ct , ct+1 , kt and kt+1 at their steady state values c, c, k, k: u00 (c)dct = u00 (c)g 0 (k)dct+1 + u0 (c)g 00 (k)dkt+1 ; 7

A notable exception is the case where f (k) = Ak , u(c) = ln c and

17

= 1.

dct = g 0 (k)dkt

dkt+1 :

We can simplify the …rst equation since g 0 (k) = 1 in the steady state. We can also rewrite both equations in terms of c^t = dct =c and k^t = dkt =k:8 [u00 (c)c] c^t = [u00 (c)c] c^t+1 + [ u0 (c)g 00 (k)k] k^t+1 ; [c] c^t = [g 0 (k)k] k^t

[k] k^t+1 :

We then divide the …rst equation by u0 (c) and the second equation by k: u00 (c)c u00 (c)c c ^ = c^t+1 + [ g 00 (k)k] k^t+1 ; t u0 (c) u0 (c) hci c^t = [g 0 (k)]k^t k^t+1 : k If we de…ne the coe¢ cient of relative risk aversion at the steady state as and a parameter

g 00 (k)k > 0 and note that g 0 (k) =

1

u00 (c)c=u0 (c),

, then the two equations can

be written more compactly as: c^t = c^t+1 + ( = )k^t+1 ; (43) c c^t = 1 k^t k^t+1 : (44) k In vector form, these two linear equations can be expressed as a …rst-order di¤erence equation: = 1 or

k^t+1 c^t+1

k^t+1 c^t+1

1 0 =

1= =(

=

0 1

k^t c^t

1 (c=k)

c=k ) 1 + c=( k)

k^t c^t

:

(45)

Write the di¤erence equation (45) as x t = M xt 1 :

(46)

We can use (46) to approximately solve the neoclassical model using one of three methods: (i) the Blanchard and Kahn approach, (ii) the shooting method and (iii) the method of undetermined coe¢ cients.. Each of these methods relies on speci…c properties of M , and things we know about the neoclassical model. Although we will not do so in these notes, it can be shown that the neoclassical growth model has a unique solution.9 That is, given an initial value of k0 , there are unique choices for fct ; kt+1 g1 t=0 that solve the social planner’s problem. Also, the optimal path has the property that limt!1 (ct ; kt ) = (c; k). In terms of the linearized model, this means limt!1 xt = 0. 8

Notice that the notation x ^t = dxt =x indicates the percentage deviation of xt from its steady state value, x, because the total derivative was taken in the neighborhood of x. 9 See, for example, chapters 1–4 of Stokey and Lucas (1989).

18

3.2

Blanchard and Kahn’s Method

Assuming that it is possible to diagonalize M , we begin by letting M = V V

1

, where V is

a matrix whose columns are the eigenvectors of M normalized to have unit length, and is a diagonal matrix with the corresponding eigenvalues arranged along the diagonal.10 De…ne x~t = V

1

1

xt . Premultiplying (46) by V

we get V

1

1

xt = V

V V

1

xt 1 , or (47)

x~t = x~t 1 : Thus we get a pair of scalar di¤erence equations: x~1t =

and

~1t 1 1x

x~2t =

(48)

~2t 1 : 2x

Above, we saw that the general solutions to the two di¤erence equations are x~1t = c1

t 1

and x~2t = c2

t 2;

(49)

with c1 and c2 being arbitrary constants. The next step is to pin down the values of c1 and c2 . To do so we will rely on things we know about the optimal paths of the elements of xt rather than x~t . First, we know the value of k^0 because k0 is given. Second, we know that the correct solution has the property that limt!1 xt = 0. Since xt = V x~t we can write x1t = v11 c1

t 1

+ v12 c2

t 2

x2t = v21 c1

t 1

+ v22 c2

t 2:

(50)

The fact that k^0 (equivalently, x10 ) is known puts one restriction on the constants c1 and c2 . They must jointly satisfy k^0 = x10 = v11 c1 + v12 c2 . With this one restriction, the solutions for x1t and x2t become: x1t = cv11

t 1

+ k^0

cv11

t 2

x2t = cv21

t 1

+ k^0

cv11

v22 v12

(51) t 2

(52)

We now check whether the fact that limt!1 xt = 0 allows us to pin down c. Without loss of generality, we will assume that j 1 j j 2 j. There are three cases to consider. 1. If j 1 j 1 and j 2 j 1 there is no solution for xt such that limt!1 xt = 0. Even if you set c = 0 to eliminate the cv explosive terms k^0 t and (v22 =v12 )k^0 2

t 1

terms in (51) and (52) you would be left with the t ^ 2 . These would vanish in the limit only if k0 = 0 and,

therefore, xt = 0 for all t; that is, if the system was at the steady state at time 0. On the 10

An n n matrix is diagonalizable if and only if it has n linearly independent eigenvectors. The case where M cannot be diagonalized is discussed brie‡y below.

19

other hand, if you assumed c = k^0 =v11 to eliminate the second terms in each solution, then you would be left with the explosive cv t1 terms. 2. If j 1 j < 1 and j 2 j < 1 there are arbitrarily many solutions for xt such that limt!1 xt =

0. Clearly limt!1 xt = 0 regardless of the value of c. So we are left with an in…nite number of solutions given by (51) and (52). 3. If j 1 j < 1 and j 2 j 1 there is a unique solution.11 We must set c = k^0 =v11 so that the t2 terms drops out of the solutions (51) and (52). This implies x1t = k^0 t1 and x2t = (v21 =v11 )k^0 t . This completely characterizes the paths of k^t and c^t and tell us that 1

c^t = (v21 =v11 )k^t along these paths. There is an alternative approach to solving the model. The fact that c = k^0 =v11 is equivalent to c2 = 0. If we substitute this into (49) we …nd x~2t = 0. The de…nition xt tells us that x~2t = v 21 x1t + v 22 x2t , where v ij is the ij element of V 1 . Knowing that x~2t = 0 implies that x2t = (v 21 =v 22 )x1t or c^t = (v 21 =v 22 )k^t . It is easy to verify that x~t = V

1

v21 =v11 =

(v 21 =v 22 ) so that the two ways of proceeding get to the same answer.

Appendix Proof that 0
c^t if c^t > (k=c)k^t . Thus, we can de…ne the regions A, B, C, D in Figure 5, in which c^t and k^t are both increasing (A), c^t is increasing but k^t is decreasing (B), c^t and k^t are both decreasing (C), and c^t is decreasing while k^t is decreasing (D). The optimal path, c^t = (v21 =v11 )k^t , lies entirely within regions A and C. The steady state (0; 0) is a saddle point, with the optimal solution being a saddle path. All other paths are divergent. It is now clear how the phase diagram can be used to design the shooting algorithm. Given k^0 start with a guess for c^0 . Then simulate (45) forward. If the path enters region B stop and revise the guess for c^0 downward. If the path enters region D stop, and revise the guess for c^0 upward. Keep doing this until the path gets within a tolerance limit of (0; 0). 12

See my notes on the neoclassical growth model for a discussion of the uniqueness of the model’s solution and for a complete discussion of the phase diagram. 13 The reader should verify that for the neoclassical model, v21 =v11 > 0. 14 The reader should verify that for the neoclassical model, v22 =v12 < 0.

21

FIGURE 5 The Phase Diagram for the Linearized Neoclassical Model cˆt = β −1 (k / c ) kˆt

cˆt

cˆt = (v21 / v11 )kˆt B C

cˆt = ( β −1 − 1)(k / c)kˆt kˆt

(0,0)

cˆt = (v22 / v12 )kˆt

D

A

While the shooting method is obviously feasible, there is really no reason to use it once we have linearized the model. The reason is that the exact solution to the linearized model is as easy to characterize as the phase diagram using the Blanchard and Kahn approach. A few more words about the phase diagram are in order. We obtained Figure 5 because the neoclassical model is one for which 0
1, or (ii) a bounded solution for every value of c1 , if j j < 1.

3.4

Method of Undetermined Coe¢ cients

Another approach to solving the model is to guess that the optimal solution is of the form c^t = k^t for some scalar . As we have seen, above, this conjecture is correct. If the conjecture is substituted into (45) we have 1 or

1= =(

k^t+1 = 1

c=k ) 1 + c=( k)

1

1

k^t+1 =

c=k ) + [1 + c=( k)]

=(

k^t

k^t :

Dividing the second equation through by : 1 1

1

k^t+1 =

c=k ) + 1 + c=( k)

=(

k^t :

(53)

For the two equations implicit in (53) to both be valid requires that 1

This means

c=k =

=(

) + 1 + c=( k):

is a solution to the quadratic equation 2

There are two solutions,

1

c=k + [1 1

and

2.

+ c=( k)]

=(

) = 0:

(54)

Suppose we choose one of these solutions, say

1.

We can verify if it is the correct solution by going back to one of the two equations, say 23

k^t+1 = (

^

1

1 c=k)kt .

Notice that for any k^0 , we will get limt!1 k^t = 0 if and only if

j 1 1 c=kj < 1. We could verify whether we had the correct solution by checking whether this inequality was true. If it was not we would use the solution corresponding to 2 . There’s another, more straightforward, way of thinking about solving for . Notice that we could also guess a solution of the form 1 k^t + 2 c^t = 0 for all t, and try to solve for 1 and 2.

We can write this guess as 0

0

xt = 0, for all t, where

0

0

=

1

. If we premultiply

2

(46) by we get xt+1 = M xt . By assumption xt+1 = 0 so this means 0 M xt = 0. We also have 0 xt = 0. For both of these equalities to hold for all possible xt , the elements of the vector

0

0

0

must be proportional to the elements of the vector 15

given by some scalar .

I.e. there is some

0

M , with the proportion being 0

6= 0 for which

0

=

0

M . If we transpose

both sides of the equation we get M = . But now we can see that …nding of …nding the eigenvalues and eigenvectors of M 0 . There are two eigenvectors of M 0 ,

1

and

2,

is a matter

so which one should we pick? Assume, as we 1

did above, that M can be diagonalized as M = V V

. Notice that since M 0 = (V

M 0 has the same eigenvalues as M , and its eigenvectors are the columns of (V (V 1 )0 = ( 1 2 ). Since xt = M xt 1 , xt = M t x0 = V xt =

1

1

V

t 1

v11 v12 v21 v22 0 1 x0 ) 0 1 x0 )

for our solution, we have

0 1 x0

v11 v21

xt =

) V 0,

1 0

) . I.e.

x0 or

t 1( t 1(

= If we choose

t

1 0

v12 v22

0 1 0 2

0

0

t 2

+ v12 + v22

t 2( t 2(

0 2 x0 ) 0 2 x0 )

x0 :

= 0 so that t 2( t 2(

0 2 x0 ) 0 2 x0 )

:

Notice that this means limt!1 xt = 0 if and only if j 2 j < 1. Similarly, if we choose our solution, we have

0 2 x0

2

for

= 0 so that

xt =

v11 v21

t 1( t 1(

0 1 x0 ) 0 1 x0 )

:

In this case limt!1 xt = 0 if and only if j 1 j < 1. Without loss of generality let j 1 j

j 2 j.

Since our criterion for choosing the correct solution is that limt!1 xt = 0 this means: if j 1 j

1 and j 2 j

1 neither solution will have the desired property

Notice that if we de…ne a vector = M 0 , one of the equations says 1 k^0 + 2 c^0 = 0, while the other says that 1 k^0 + 2 c^0 = 0. So one equation implies that c^0 = ( 1 = 2 )k^0 while the other implies that c^0 = ( 1 = 2 )k^0 . So 1 = 2 = 1 = 2 or, in other words, 1 = 1 and 2 = 2 for some . 15

24

if j 1 j < 1 and j 2 j < 1 then either solution will work and the choice will be ambiguous. if j 1 j < 1 and j 2 j

1 then clearly we would need to use

2

as our solution.

As we saw before, for the neoclassical model the last condition holds. So there is a unique solution, corresponding to the eigenvector 2 of M 0 that corresponds to the larger eigenvalue. As we noted above, the eigenvectors of M 0 are the columns of (V 1 )0 . Hence they are the rows of V 1 . This means that 2 = ( v 21 v 22 )0 . So v 21 k^t + v 22 c^t = 0, or c^t = (v 21 =v 22 )k^t . This is the same as the solution we obtained using the Blanchard and Kahn approach.

4

Linear Approximation in the Stochastic Model

Now imagine that we have the stochastic model where output is given by zt f (kt ) and zt is some stationary stochastic process. Now the optimality conditions are: u0 (ct ) =

Et fu0 (ct+1 ) [zt+1 f 0 (kt+1 ) + 1

ct = zt f (kt ) + (1

)kt

(55)

]g

(56)

kt+1 :

It is straightforward to show that the linearized representation of these equations is k^t+1

c^t = Et c^t+1 +

Z

Et z^t+1

y c c^t = z^t + 1 k^t k^t+1 ; k k where , , c and k are de…ned as before, Z = f 0 (k)z and y = zf (k). In vector form, these two linear equations can be expressed as a …rst-order di¤erence equation: = 1

1 0

or Et

Et k^t+1 c^t+1

k^t+1 c^t+1

0

= 1

=

1

k^t c^t

1 (c=k)

1

c k

1+

c k

k^t c^t

Z=

+

0

0

+

Et z^t+1 +

Z

For the moment it is probably easiest to write this system as Et xt+1 = M xt + where t

=

0 Z

Et z^t+1 +

25

t

y k

y k

Et z^t+1 +

z^t :

y k

y k

0 y=k

z^t :

z^t

Notice that the matrix M is unchanged. Once again, we could diagonalize the system and write it as Et x~t+1 = x~t + z~t where x~t = V

1

1

xt and z~t = V

t.

In this case we end up with two scalar stochastic

di¤erence equations Et x~1t+1 =

~1t 1x

+ z~1t

Et x~2t+1 =

~2t 2x

+ z~2t :

We would like to …nd solutions for x~1t and x~2t that are (i) consistent with k^t+1 being determined at time t and (iii) imply that xt is covariance stationary. There is a very direct way of …nding the solution which is due to King, Plosser and Rebelo (2002). They suggest solving the second equation forward using (38): x~2t =

1 X

(j+1) 2

1 X

Et z~2t+j =

j=0

(j+1) 2

( v 21 v 22 )Et

t+j

j=0

Then they suggest using the fact that x~2t = v 21 x1t +v 22 x2t to solve for x2t =

(v 22 ) 1 v 21 x1t +

(v 22 ) 1 x~2t . Therefore x2t =

22

1 21

22

(v ) v x1t

1

(v )

1 X

(j+1) 2

( v 21 v 22 )Et

t+j

j=0

Then take the original di¤erence equation, which states that x1t+1 = m11 x1t + m12 x2t +

1t

and rewrite it as x1t+1 =

22

m11

1 21

m12 (v ) v

x1t +

22

m12 (v )

1t

1

1 X

(j+1) 2

( v 21 v 22 )Et

t+j

j=0

=

1 x1t

+

22

1t

m12 (v )

1

1 X

(j+1) 2

( v 21 v 22 )Et

t+j

j=0

The King, Plosser Rebelo code typically works with the assumption that Et z^t+j = which case t

y k

=

z^t = z^t

y k

Z

and Et Hence x2t =

v 21 x1t v 22

t+j

j

= 2 2

26

z^t :

(

v 21 v 22

1 ) z^t

j

zt in

x1t+1 =

1 x1t

+ ( 1 0 )

1

m12

(

2

v 21 v 22

1 )

z^t

It’s important to keep in mind that the King, Plosser, Rebelo approach only works in speci…c circumstances. Recall that we …rst solved for x~2t forward and then inverted the equation x~2t = v 21 x1t + v 22 x2t to solve for the non-predetermined variables, x2t , in terms of the predetermined variables x1t and the variable just solved for, x~2t . The invertibility of the latter mapping depends on the dimension of x2 and x~2 being the same. I.e. the number of non-predetermined variables must be the same as the number of explosive roots. If number of explosive roots is smaller, the system of equations will have a non-unique solution. If the number of explosive roots is bigger, there will be no covariance stationary solution. The ‡ipside of this analysis is that the number of predetermined variables, which we solved for in the second step, was equal to the number of stable roots. If there are more stable roots than predetermined variables, the model will have multiple solutions. If there are less stable roots than predetermined variables, there will be no covariance stationary solution. There is a way of getting these results using the same logic as was used to obtain the bounded solutions in the deterministic model, except here we use the covariance stationarity property and the fact that kt is predetermined. This could be a good homework question. References Blanchard, Olivier J. and Charles M. Kahn (1980) “The Solution of Linear Di¤erence Models under Rational Expectations,”Econometrica, 48(5), 1305–12. Christiano, Lawrence J. (2002) “Solving Dynamic Equilibrium Models by a Method of Undetermined Coe¢ cients,”Computational Economics, 20, 21–55. King, Robert G., Charles I. Plosser and Sergio T. Rebelo (2002) “Production, Growth and Business Cycles: Technical Appendix,”Computational Economics, 20, 87–116. Mirman, Leonard J. and William A. Brock (1972) “Optimal Economic Growth and Uncertainty: The Discounted Case,”Journal of Economic Theory, 4, 479–513. Sargent, Thomas J. (1987) Macroeconomic Theory. New York: Academic Press. Stokey, Nancy L. and Robert E. Lucas, Jr. (1989) Recursive Methods in Economic Dynamics. Cambridge, Mass.: Harvard University Press.

27