THE ECONOMICS OF MONEY,BANKING, AND FINANCIAL MARKETS 175

CHAPTER 7

Stocks, Rational Expectations, and the Efficient Market Hypothesis

143

The generalized dividend valuation model requires that we compute the present value of an infinite stream of dividends, a process that could be difficult, to
say the least. Therefore, simplified models have been developed to make the calculations easier. One such model is the Gordon growth model, which assumes
constant dividend growth.

The Gordon
Growth Model

Many firms strive to increase their dividends at a constant rate each year. Equation 4
rewrites Equation 3 to reflect this constant growth in dividends:
P0
where

D0

g )1

(1

1

g )2

(1

D0

(1

ke )

D0

...

2

(1

(1

g)

(1

ke )

ke )

D0
g
ke

the most recent dividend paid
the expected constant growth rate in dividends
the required return on an investment in equity

(4)

Equation 4 has been simplified using algebra to obtain Equation 5.1
(1

D0

P0

(ke

g)

D1
(ke g)

g)

(5)

This model is useful for finding the value of stock, given a few assumptions:
1. Dividends are assumed to continue growing at a constant rate forever. Actually,
as long as they are expected to grow at a constant rate for an extended period
of time, the model should yield reasonable results. This is because errors about
distant cash flows become small when discounted to the present.
2. The growth rate is assumed to be less than the required return on equity, ke . Myron
Gordon, in his development of the model, demonstrated that this is a reasonable
assumption. In theory, if the growth rate were faster than the rate demanded by
holders of the firm s equity, in the long run the firm would grow impossibly large.

1

To generate Equation 5 from Equation 4, first multiply both sides of Equation 4 by (1
and subtract Equation 4 from the result. This yields:
(1

P0
(1

ke )

P0

g)

(1

D0

D0

(1

ke ) /(1

g)

g)

ke )

Assuming that ke is greater than g, the term on the far right will approach zero and can be dropped.
Thus, after factoring P0 out of the left-hand side:
c

P0

1
1

1d

ke
g

D0

Next, simplify by combining terms to:
P0
P0

(1

ke )
1

(1

(1

D0
(ke

g)

g
g)
g)

D0
D1

(ke

g)