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CHAPTER 17

Exercise 17.2

1. (a) yt+1= yt+ 7 (b) yt+1= 1.3yt (c) yt+1= 3yt 9

(a) Iteration yieldsy1= y0+ 1, y2= y1+ 1 =y0+ 2, y3= y2+ 1 = y0+ 3, etc. The solution

is yt= y0+t= 10 +t.

(b) Since y1= y0, y2= y1= 2y0, y3= y2= 3y0, etc., the solution is yt = ty0= t.

(c) Iteration yields y1 = y0 , y2 = y1 = 2y0 , y3 = y2 = 3y0

2 , etc. The solution is yt = ty0 (t1+

t2 +. . .+| {z }a total of t terms

+ 1).

(a) yt+1 yt = 1, so that a = 1and c = 1. By(17.90), the solution is yt= y0+ ct= 10 + t.

The answer checks.

(b) yt+1yt= 0, so that a = , and c = 0. Assuming 6= 1,(17.80)applies, and we have

yt =y0t =t. It checks. [ Assuming = 1 instead, we find from (17.90) that yt = ,

which is a special case ofyt = t.]

(c) yt+1 yt = , so that a = , and c = . Assuming 6= 1, we find from (17.80)

that yt = y0+ 1t

1 . This is equivalent to the earlier answer, because we can

rewrite it as yt = y0t

1t

1

= y0t (1 ++2 +. . .+t1).

(a) To find yc, try the solution yt = Abt in the homogeneous equation yt+1+ 3yt = 0. The

result is Abt+1 + 3Abt = 0; i.e., b = 3. Hence yc = Abt = A(3)t. To find yp, try the

solution yt = k in the complete equation, to get k+ 3k = 4; i.e. k = 1. Hence yp = 1.

The general solution isyt= A(3)t + 1. Settingt = 0in this solution, we get y0= A + 1.

The initial condition then gives us A= 3. The definite solution is yt= 3(3)t + 1.

(b) After normalizing the equation to yt+1 (12)yt = 3, we can find yc = Ab

t = A( 12)t, and

yp= k = 6. Thus,yt = A(12)t + 6. Using the initial condition, we get A = 1. The definite

solution is yt = (12)

t + 6.

(c) After rewriting the equation as yt+1 0.2yt = 4, we can find yc = A(0.2)t, and yp = 5.

Thus yt = A(0.2)t + 5. Using the initial condition, this solution can be definitized to

yt= (0.2)t + 5.

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Exercise 17.3

1. (a) Nonoscillatory; divergent.

(b) Nonoscillatory; convergent (to zero).

(c) Oscillatory; convergent.

(d) Nonoscillatory; convergent.

2. (a) From the expression3(3)t, we have b = 3(region VII). Thus the path will oscillate

explosively around yp= 1.

(b) Withb = 12 (region III), the path will show a nonoscillatory movement from 7 toward

yp= 6.

(c) With b = 0.2 (region III again), we have another convergent, nonoscillatory path.

But this time it goes upward from an initial value of 1 toward yp= 5.

3. (a) a = 13 , c = 6, y0 = 1. By (17.80), we have yt = 8(

13)

t + 9 nonoscillatory and

convergent.

(b) a= 2, c = 9, y0= 4. By(17.80), we have yt = (2)

t + 3 oscillatory and divergent.

(c) a = 14 , c = 5, y0 = 2. By (17.80), we have yt = 2(

14)

t + 4 oscillatory and

convergent.

(d) a= 1,c = 3,y0= 5. By(17.90), we have yt = 5+3t nonoscillatory and divergent

(from a moving equilibrium 3t).

Exercise 17.4

1. Substitution of the time path(17.120)into the demand equation leads to the time path ofQdt,

which we can simply write as Qt (since Qdt= Qst by the equilibrium condition):

Qt =

Pt = P0

P

t

P

Whether Qt converges depends on the

tterm, which determines the convergence ofPt

as well. Thus Pt and Qt must be either both convergent, or both divergent.

2. The cobweb in this case will follow a specific rectangular path.

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3. (a) = 18, = 3, = 3, = 4. Thus P = 217 = 3. Since > , there is explosive

oscillation.

(b) = 22, = 3, = 2, = 1. Thus P = 244 = 6. Since < , the oscillation is

damped.

(c) = 19, = 6, = 5, = 6. Thus P = 2412 = 2. Since = , there is uniform

oscillation.

4. (a) The interpretation is that if actual pricePt1exceeds (falls short of) the expected price

Pt1, then P

t1 will be revised upward (downward) by a fraction of the discrepancy

Pt1 P

t1, to form the expected price of the next period, P

t. The adjustment

process is essentially the same as in (16.34), except that, here, time is discrete, and

the variable is price rather than the rate of inflation.

(b) If = 1, then Pt =Pt1 and the model reduces to the cobweb model (17.10). Thus

the present model includes the cobweb model as a special case.

(c) The supply function gives Pt = Qst+

, which implies that Pt1 =

Qs,t1+

. But

since Qst = Qdt= Pt, and similarly, Qs,t1= Pt1, we have

Pt =+ Pt

and Pt1=

+ Pt1

Substituting these into the adaptive expectations equation, and simplifying and shift-

ing the time subscript by one period, we obtain the equation

Pt+1

1

Pt=

(+)

which is in the form of (17.6) with a =

1

6= 1, and c= (+)

.

(d) Since a 6= 1, we can apply formula (17.80) to get

Pt = P0 +

+1

t

++

+

=

P0 P

1

t+ P

This time path is not necessarily oscillatory, but it will be if

1

is negative,

i.e., if +

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(e) If the price path is oscillatory and convergent (region V in Fig. 17.1), we must have

1 < 1

< 0, where the second inequality has to do with the presence

of oscillation, and the first, with the question of convergence. Adding ( 1), and

dividing throught by, we have1 2

<

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[These results are the same as Table 17.2 withset equal to 0.] To have a positive P, we must

have k

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It follows that

k=

+

P

=

+

P

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