The multiplier as the sum of a geometric series

The gentlemenly C level of understanding

Consider a geometric series d + ad + a2d + º+ at-1d whose sum as t Æ is [d/( (1 - a))] The multiplier is obtained by suming a geometric series

The B level of understanding

where do the terms come from: Household get their income at the end of the period and spend it in the next period. Each period the firms maintain the DI level of investments and that is the first term in each period. The other terms are b times the income received at the end of the previous period. Thus in period 4 the 2nd through the 4th terms come from b times (DI + b DI + b2DI) and so on.

Dynamic Problems

Yt = Ct + It
(1)
Ct = a + bYt-1
(2)

3. Given model above with a = 200 b = 0.75 and I0 = 150:

a) Suppose the economy is in equilibrium initially. What is the equilibrium Y?

Solution

Solve using the solution to the static model:

Y = C + I
C = a + bY
that is the formulae: k = 1/(1-b) and Y = k(a + I)

k = 1/[1 - 0.75] = 4 Y = k[a + I] = 4[350] = 1400

b) Suppose I increases from 150 to 170 in period 1. Find Y0, Y1, Y2, and Y3.

Solution

Use a table. On the quiz and tests I will not give you the format for the table because there are several ways you could set up the table and I do not want arguments with the grader. This is a sample table with an extra column relating what you are doing to the geometric series.

t a + It bYt-1 Yt Yt is
1 370 1050 1420 Y0 + DI
2 370 1065 1435 Y0 + DI + bDI
3 370 1076.25 1446.25 Y0 + DI + bDI + b2DI

c) What is the final equilibrium Y?

Solution

Use the same aproach as the initial equilibrium.

Y = k[a + I] = 4[370] = 1480

Note Yt = Y0 + DI + bDI + b2DI + º+ bt-1DI thus Y = Y0 + kDI = 1400 + 4 ·20 = 1480

To solve dynamic problems use the following worksheet.

 


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On 18 Feb 2000, 11:11.