Lee Ohanian
Macroeconomic Analysis
UCLA
1 A Two Period Model
The optimal growth model was a major advance over the Solow-Swan model.
The Cass-Koopman’s model retains the production structure of the Solow-Swan
model, but it specifies an optimizing consumer that makes choices over con-
sumption and investment, rather than the exogenous savings rate assumption
of Solow-Swan model. In fact, note that there are no optimizing choices what-
soever in the Solow-Swan model!
First let’s take a look at a two period problem. There is a representative
household (or equivalently many identical households) which chooses how much
output to save, and how much to consume, to maximize their utility. We will
assume that there is one unit of labor in the economy, and that labor is supplied
inelastically, but we will relax that assumption later. We will also assume that
households are impatient in that they value future consumption less than current
consumption. We will call this “discounting”, and we will denote the household’s
“discount factor” as β, in which 0 < β < 1.
2 The Social Planning Problem and Solution
We will first derive the solution to a social planning problem. Then we will
derive the solution to the competitive equilibrium of the economy, and show
that they are identical, given the first welfare theorem, which states that the
competitive equilibrium of a competitive economy with perfectly functioning
markets is Pareto optimal. We will call the two periods "0" and "1".
The social planner maximizes utility of the household, subject to resource
constraints. Note that there are no markets of any kind. Rather, the social
planner makes all choices for the individuals in the economy. Preferences are
assumed to be:
max{ln(c0) + β ln(c1)} (1)
subject to the resource constraints:
F (k0, 1) + (1− δ)k0 = c0 + k1 (2)
F (k1, h1) + (1− δ)k1 = c1 (3)
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Note that the capital stock available for production today, k0, is exogenous.
As in the Solow model, we will assume that output can be costlessly transformed
into either consumption or investment, and that the production function is
constant returns to scale. Since the economy is over at the end of the second
period, there is no incentive to save in period 1. Therefore, all resources will be
consumed in this last period.
We set this up as a Lagrangian
L = max{ln(c0)+β ln(c1)+λ0[F (k0, 1)+(1−δ)k0−c0−k1]+λ1[F (k1, 1)+(1−δ)k1−c1]}
(4)
The first order necessary conditions choosing consumption in both periods,
and for choosing and capital for the second period are (note that Fk refers to
the partial derivative of F with respect to k is :
1
c0
− λ0 = 0 (5)
β
c1
− λ1 = 0 (6)
λ0 = λ1[Fk1 + 1− δ] (7)
(8)
We also have the two resource constraints:
F (k0, 1) + (1− δ)k0 − c0 − k1 = 0 (9)
F (k1, 1) + (1− δ)k1 − c1 = 0 (10)
We can substitute out for the Lagrange multipliers λ0 and λ1. We get the
following equations once we make those substitutions:
1
c0
= β
1
c1
[Fk1 + (1− δ)] (11)
along with the resource constraints
F (k0, 1) + (1− δ)k0 − c0 − k1 = 0 (12)
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F (k1, 1) + (1− δ)k1 − c1 = 0 (13)
We have 3 equations in 3 unknowns, which are c0, c1, and k1. We can solve
these equations as functions of the exogenous variable k0. (Note that these will
be nonlinear equations, so we will need the help of the computer to solve them).
Once we have done that, we can then assess how changes in k0 or the coeffi cients
in the model, affect the endogenous variables.
2.1 How Preferences and Technologies Affect Consumer
Choices
This section shows how different preferences and technologies affect choices. We
will work with the planner’s first order condition
1
c0
= β
1
c1
[Fk1 + (1− δ)] (14)
Let’s begin with the power utility function. This is also called constant rela-
tive risk aversion (CRRA) utility function, and constant intertemporal elasticity
of subsitution utility function.
u(c) =
c1−σ − 1
1− σ
, σ ≥ 0 (15)
There are several features of this function. One is that this utility function
is given by ln(c), if you take limits as σ → 1. (You dont need to do this, but if
you would like the practice, you use L’hopital’s rule.) Note that the "-1 " term
that appears in the utility function is only needed for this limiting case.
The parameter σ governs the degree of concavity in the utility function. For
linear utility (σ = 0) , the function is weakly concave. For σ > 0, the function
is strictly concave.
First we discuss the role of σ in risk aversion of the consumer. We will as-
sume that consumers maximize expected utility. This is an assumption regarding
how consumers deal with uncertainty. Maximizing expected utility means that
the consumer makes decisions that maximize the probability-weighted sum of
the possible utilities that could occur. For example, suppose that the consumer
has preferences over consumption, and that there are two possible outcomes
for her consumption, xH and xL, each with a 50% probability. Then expected
utility is given by:
EU(x) = .5 ∗ U(xH) + .5 ∗ U(xL) (16)
Note that the expected value of x is given by:
E(x) = .5 ∗ xH + .5 ∗ xL (17)
These two expressions will be the same if σ = 0, which is the case of a risk
neutral consumer. (For now, assume that the -1 in the utility function above is
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not present). Specifically, a risk neutral consumer is one who has no aversion to
uncertainty. Suppose that you are risk neutral and someone says “I will flip a
coin. If it is “heads”, then I will pay you $1,000,000. If it is “tails”, then I will
pay you nothing. Alternatively, I will pay you $500,000 for sure, and I won’t
flip the coin.”
A risk neutral individual is indifferent between these two options: (1) $500,000
with certainty, or (2) a 50% chance of $1,000,000/50% change of 0.
Consider which one of those options you would pick. If you are like most
people, then you would pick $500,000 with certainty, rather than the gamble of
50% chance of $1,000,000, or 50% chance of zero. This is because most people
are risk averse – they do not like uncertainty. A risk averse consumer has a
strictly concave utility function.
The parameter σ is called the coeffi cient of relative risk aversion. This is
a measure of how well a consumer tolerates risk. Specifically, risk tolerance is
related to the curvature in the utility function, which itself is related to the size
of σ. Note that for σ = 0, the utility function is linear. For σ slightly larger
than zero, we get a small departure from linearity, with a curve that has a slope
that is slightly decreasing as c increases. For large values of σ, we get a highly
curved function which has a slope that is rapidly decreasing as c increases. Note
that what is important in terms of understanding how the consumer tolerates
risk is the slope of the utility function, which is marginal utility, but also the
rate of change of the marginal utility, which measures the degree of concavity
in the utility function.
There are two widely used measures of risk aversion that use both the first
and second derivatives of the utility function. The coeffi cient of absolute risk
aversion is −u
′′(c)
u′(c)
. The coeffi cient of relative risk aversion is −u
′′(c)∗c
u′(c)
.
First, note that the linear utility function is one in which there is no risk
aversion. Since the second derivative of the function is zero, both measures of
risk aversion are zero.
For the case of σ > 0, we get that absolute risk aversion for this utility
function is σc
−σ−1
c−σ
and for relative risk aversion, we get σ. Many economists
use the the power utility function because the consumer’s tolerance of risk does
not depend on their consumption level. Thus, the consumer’s risk preference is
invariant to their level of consumption with this utility function.
The parameter σ also is important in understanding the consumer’s willing-
ness to substitute consumption over time. This elasticity measures how the ratio
of consumption between today and tomorrow changes with respect to a change
in the return to savings.
Thus, we have d(c2/c1)
dR
R
(c2/c1)
. (Note that the return to savings is defined as
R ≡ Fk2 + 1− δ). Our first order condition is:(
c1
c0
)σ
= βR (18)
Next, differentiating, we get:
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d(c1/c0)
dR
=
1
σ
(βR)
1−σ
σ β (19)
Forming the elasticity d(c1/c0)
∂R
R
(c1/c0)
, we get:
1
σ
(βR)
1−σ
σ β
R
(βR)
1
σ
=
1
σ
(20)
Note that the IES is infinite for the case of linear utility. This is because with
linear utility, the marginal utility of consumption is constant, so consumption
today and tomorrow are perfect substitutes. This means that a very small
change in the return to savings will cause the consumer to consume either zero
in period 1, (if R > 1/β) or consume everything in period 0 (R < 1/β).
Note that in an economy without any uncertainty (no random variables),
then risk issues are simply not present. People know with certainty the state of
the economy into the future. Thus, for deterministic models, the issue becomes
how willing consumers are to substitute consumption over time. Recall our first
order condition: (
c1
c0
)σ
= βR (21)
Note that as σ becomes large, the elasticity declines. This means the fol-
lowing. Suppose we ask "how big of a change in the return to investing (R)
is required to change c1
c0
by one percent?" For the log case, (σ = 1), c1
c0
moves
one-for-one with R. But if σ = 10, then R must change by a factor of 10. To
see this, take logs of the first order condition, and abstracting from β (it is a
constant), we get:
ln(
c1
c0
) =
ln(R)
10
(22)
3 Modifying the Economy to Have Many Peri-
ods
We now modify this model. First, households live for many periods. The
economy begins in period 0, which is the current period, and continues up to
period "T". We will ultimately assume that T becomes infinite. For the utlity
function we used before, this becomes:
max{ln(c0)− φh0 + β[ln (c1)− φh1] + ...βT [ln (cT )− φhT ]} (23)
Note that as T gets very large,βT approaches 0, which means that consumers
place a decreasing amount of value on goods and services into the far-off future.
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The economy has a resource constraint each period which tells us that output
is divided between consumption and investment:
F (k, h) = c+ i (24)
So, there are two tradeoffs that society faces each period. How much time
to allocate to work, versus how much time to allocate to other uses of time, and
the second tradeoff is how much output to allocate to consumption, and how
much to allocate to investment.
Our law of motion for the capital stock is the same:
kt+1 = it + (1− δ)kt (25)
We can write the economy’s total resource constraint as:
F (k, h) + (1− δ)kt = ct + kt+1 (26)
Note that the left hand side of the constraint is the total resources available
to society after production - which is the output that is produced using capital
and labor, plus the capital that is left over after depreciation. The right hand
side of the constraint is the use of those resources, which is consumption, and the
capital stock for next period. (Note that we can rewrite this as F (k, h) = ct+it.
given that it = kt+1− (1− δ)kt. However, it will be easier to use the constraint
written as F (k, h) + (1− δ)kt = ct + kt+1.
Now we have the Lagrangian as follows:
L = max{ln(c0)− φh0 + β[ln (c1)− φh1] + ...βT [ln (cT )− φhT }+ (27)
λ0[F (k0, h0) + (1− δ)k0 − c0 − k1] + λ1[F (k1, h1) + (1− δ)k1 − c1 − k2] + ..(28)
For each period, society chooses labor (ht), consumption (ct), and how much
capital to have in place for the following period (kt+1). Let’s start with the first
period, which is period 0. The first order conditions for choosing period 0
consumption, period 0 labor, and capital for period 1 are given by:
1
c0
− λ0 = 0 (29)
−φ+ λ0Fh0 = 0 (30)
−λ0 + λ1[Fk1 + (1− δ)] = 0 (31)
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Next, we can substitute out for λ0 as
1
c0
. We can also substitute out for λ1.
Specifically, go back to the Lagrangian and then differentiate with respect to
the variable c1. When we do this, we get the first order condition
β
c1
= λ1.
This means these three equations collapse into the following two equations:
φ =
Fh0
c0
(32)
1
c0
=
β
c1
[Fk1 + (1− δ)] (33)
The interpretation of the first equation is for the consumer to set the mar-
ginal cost of working, which is φ, to the marginal benefit of working, which is
Fh0
c0
. Note that this last term is how much additional output is produced by
working one more unit, which is Fh0 , multiplied by the marginal utility of con-
sumption: 1
c0
, which is how consumers value one additional unit of consumption.
The interpretation of the second equation is for the consumer to set the
marginal cost of saving one unit of output rather than consuming it - which is
just the marginal utility of consumption - to be equal to the marginal benefit of
saving one additional unit. That maginal benefit is the product of two terms.
One is Fk1+(1−δ) - which is the additional output society has from investing one
more unit. That additional output is the marginal product of capital, plus the
capital that is left over after production. Note that if we invest one additional
unit of capital, then we will have (1− δ) units left over after production. So the
total amount of extra resources we have is Fk1 + (1− δ).
Next, let’s assume that the T becomes infinity. We call this the "infinite
horizon" model. Then we have:
max
∞∑
t=0
βtu(ct, lt) (34)
subject to:
f(kt, ht) + (1− δ)kt = ct + kt+1 (35)
Note that we will have three equations that characterize the model solution
every period:
u′(lt) = u
′(ct)fht (36)
u′(ct) = βu
′(ct+1)[fkt+1 + 1− δ] (37)
f(kt, ht) + (1− δ)kt = ct + kt+1 (38)
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3.1 The Steady State
Our infinite horizon model is one in which the economy will reach a steady
state, just as in the Solow growth model. The basic reason why is because
depreciation utlimately exceeds investment, given the diminishing marginal pro-
ductivity of capital. In the steady state, the amount of time allocate to work
is constant, and the values of consumption, investment, output, and the cap-
ital stock are also constant. The steady state is the long-run position of the
economy.
Let’s denote the steady state levels of consumption, labor, output, invest-
ment, and capital as having "ss" subscripts, where ss stands for steady state.
The first order conditions for the steady state are:
u′(lss) =
1
css
∂f
∂h
(kss, hss) (39)
1
css
=
β
css
[
∂f
∂k
(kss, hss) + (1− δ)] (40)
Note that we can multiply the left hand and the right hand sides of the last
equation by css to get:
1 = β[
∂f
∂k
(kss, hss) + (1− δ)] (41)
The resource constraint for the steady state is:
F (kss, hss) + (1− δ)kss = css + kss (42)
This steady state resource constraint reduces to:
F (kss, hss) = css + δkss (43)
Note that we have three equations in three unknowns - kss, hss and css, and
we have three equations - the resource constraint, the labor first order condition,
and the investment first order condition. Once we solve these, then we know
the values for the steady state of this economy. We can also see how the steady
state values change if we change a parameter value.
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3.2 Adding Growth to the Model
Let’s now add growth to the model by specifiying exogenous, labor-augmenting
technological change in the production function. Assume first that the popula-
tion is not growing.
Yt = K
θ
t (Atht)
1−θ = Ct +Kt+1 − (1− δ)Kt (44)
At+1 = (1 + g)At, A0 = 1 (45)
First, we divide all the growing variables in the model (Yt, Ct, It,Kt,Kt+1)
by At. We get the following variables with "hats":
Ŷt =
Yt
At
, Ĉ =
Ct
At
, Ît =
It
At
, K̂t =
Kt
At
(46)
We also have:
K̂t+1(1 + g) =
Kt+1
At
=
Kt+1
At
At+1
At+1
(47)
Recall that this construction for Kt+1
At
is the same as we did in the Solow model.
This model will generate a steady state. Note that we are not going to di-
vided labor, h, by At. The reason is because the amount of time per person
devoted to market work is constrained by the time endowment - it cannot grow.
However, we need to be careful to make sure that we don’t change the mathe-
matical structure of the model when we divide by At. Let’s see how we can do
that now.
First, let’s assume that the utility fuction doesnt include leisure, and the
utility for consumption is given by C
1−σ
t
1−σ . To preserve the mathematical structure
ofour problem, note we can write the utility function as follows
max
∑
β̂
t Ĉ1−σt
1− σ
(48)
where β̂ is a transformation of β that takes into account that we divided C
by A:
β̂ = β(1 + g)1−σ (49)
Specifically, we divided consuption by (1+g)t raised to the power 1−σ, so we
need to undo that by multiplying the discount factor by the exact same term.
Mutliplying and dividing by the same factor leaves the problem unchanged.
You should verify that this multiplication/division leaves the original problem
unchanged yourself.
Now, let’s look at the resource constraint for the stationary problem. We
have
Ŷt = K̂
θ
t h
1−θ
t = Ĉt + K̂t+1(1 + g)− (1− δ)K̂t (50)
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Note that this model now will converge to a steady state, since we have
removed technological change. The first order condition now becomes:
u′(Ĉt)(1 + g) = β̂u
′(Ĉt+1)[θ
(
K̂t+1
ht+1
)1−θ
+ (1− δ)] (51)
Note that the only differences in this equation compared to the model with-
out any technological change is (1) that the term (1+g) appears, and this occurs
because K̂t+1 is scaled by (1+g), and (2), we have β̂ instead of β . In the steady
state, we have:
1 + g
β̂
= [θ
(
K̂ss
hss
)1−θ
+ (1− δ)] (52)
To conserve on notation, from now on I will only use the "hatted" variables
on the parameter β. This is because the model with growth is the same as the
model without growth, with the only exception that the β is slightly modified,
and we have (1 + g) appears in one of the first order conditions. And all we
need to do to recover our growing economy is just mutliply all of the "hatted"
variables by At.
3.3 Choosing Parameter Values in the Model
In this model, there are several parameters, which are the rate at which the
household discounts the future, the depreciation rate, the income shares paid
to capital and labor, the parameter values that govern the utility function, and
the growth rate of the technology. Choosing parameter values means requiring
that the equations of the model hit particular target values. Sometimes, this
involves one parameter in one linear equation. For example, for the depreciation
rate, economists typically use an average value of 8 percent, with values ranging
down to 6 percent, and up to 10 percent, depending on the time period and
the country. The economy-wide depreciation rate is somewhat higher today
than in the past, because a greater fraction of the capital stock is comprised of
computers, which tend to depreciate quickly.
It is also straightforward to pick the values for the income shares in the
production function. As we discussed previously that capital’s share of income
is about 1/3, and labor’s share of income is about 2/3. This means we have
k1/3h2/3. For technology, the average growth rate, g, depends on the country
and time period. For the U.S., we often use two percent, or .02.
For the utility function, many economists use a log specification over con-
sumption, ln(c). For leisure, it is also common to use the log specification, so
the function looks like
ln(C) + φ ln(1− h), (53)
where the total time endowment is normalized to one unit of time, and h is
hours worked. This means that leisure is equal to 1 − h. For the parameter φ,
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we choose a value for that parameter so in the steady state, households work a
fraction of their time that is equal to what people report in time use surveys.
In particular, surveys report individual time allocation between working in the
market and non-market activities. For an average household, this means that h
is about 25 percent of their time, with values ranging from 0.2 to 0.35, depending
on the time period and the country. So we pick the value of φ as follows. We
have the first order condition:
φ
1− h
=
w
c
(54)
We pick a value of φ so that this equation in the steady state is satisfied
for the value of h that you are specifying. Finally, we need to pick a value of
discounting for the household. This will govern the rate of return to investing.
To see this, we have in the steady state of the model:
1 + g = β̂[
1
3
(
h
k
)2/3
+ 1− δ] (55)
where we define R to be the total return between period t and period t+1 from
investing one unit of capital in period t. We can measure an average value of
this using stock market data, and this is about 6 percent per year. Rewriting
this equation, we get:
1 + g
β̂
= R (56)
Since g is around .02, this implies β is about 0.96, assuming that the time
period is one year.
3.4 Functional FormAssumptions for Steady State Growth
There are some restrictions we can place on the utility and production functions
to insure steady state growth, or balanced growth. These two names are inter-
changeable. In particular, suppose we want the model to be consistent with the
following empirical observations:
(1) The amount of time devoted to market economic activity is roughly con-
stant over time. Note that in the United States, the amount of time devoted to
market work has not changed very much over many decades, despite significant
economic growth.
(2) The shares of income paid to labor and capital are constant over time.
We previously discussed that these shares have not changed very much.
(3) The return to capital is constant over time. This return, measured as
the return to capital (stock market return), fluctuates, but it does not have a
positive nor a negative trend.
(4) Consumption, output, investment, and the real wage all grow at roughly
the same rate, and grow at the rate of technological change, which is g.
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Previously, we saw that the Cobb-Douglas production function delivered
constant shares of income paid to capital and labor, so we will retain the Cobb-
Douglas production function. This takes care of requirement (2).
In terms of requirement (3), the Cobb-Douglas function also works. Specifi-
cally, the return to capital is given by the marginal product of capital, and the
adjustment for depreciation:
Rt+1 =
(
At+1ht+1
Kt+1
)1−θ
+ 1− δ (57)
Note that δ is constant. Note that if h is constant along the steady state
growth path, then
(
At+1ht+1
Kt+1
)1−θ
will be constant, assuming that K and A grow
at the same rate. We will return to this below.
In terms of requirement (1), we need a utility function such that the income
and substitution effects of a wage change cancel each other out. Specifically, as
an economy grows, wages rise. On the one hand, higher wages will lead to a
substitution effect in which people will want to work more. On the other hand,
higher wages will lead to an income effect in which people will want to consume
more leisure. We need a utility function so that these two effects cancel each
other out. Specifically, for our leisure first order condition, we have:
u′(l)
u′(c)
= w (58)
Since the right hand side of the equation will grow at rate g, we also need
the marginal rate of substitution u
′(l)
u′(c)
to grow at the exact same rate, and we
also need leisure to be constant, independent of the value for w.
It turns out that this will hold for the following functional forms ln(c)+v(l),
where v is an increasing and concave function and l is leisure. We also can use a
slightly different formulation, in which households receive disutility from work-
ing (rather than utility from leisure), which is given by ln(c)− µ(h) in which µ
is an increasing and convex function, and h is hours worked. Note that we can
move back and forth between leisure and hours worked by specifiying a time
endowment equation that states h+ l = TE, where TE stands for "time endow-
ment", which is the amout of time that a consumer has. Often, we normalize
the time endowment to be 1.
Another function that will work is given by the Cobb-Douglas function:
(csl1−s)1−σ
1− σ
(59)
Note that in the separable cases with log utility, the change in the mar-
ginal utility of consumption exactly offsets the growth in the real wage. In the
Cobb-Douglas case, the marginal utilities of consumption and leisure interact
to produce the same effect in the first order condition:
cs(1− s)l−s
scs−1l1−s
= w (60)
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or
1− s
s
c
l
= w (61)
Thus, the Cobb-Douglas case also works. The specific restriction that we
need is that the elasticity of substitution between consumption and leisure is 1.
That is, we need:
d ln( c
l
)
d ln(w)
= 1 (62)
The utility functions we note above have this property.
To guarantee that the return to capital is constant over time, we need prefer-
ences that have a constant intertemporal elasticity of substitution. Recall that
we discussed this above. Specifically, we need d(c2/c1)
dR
R
(c2/c1)
to be constant. The
preference specfications above that deliver a constant amount of time devoted
to market work are also consistent with this requirement. For example, consider
the log utility case. In that case, we have:
ct+1
ct
= βRt+1 (63)
If consumption grows at a constant rate, then R (the return to capital) will
also be constant. Note that these empirical requirements of our model: labor
is constant along the growth path, the return to capital is constant, and all the
real variables grow at a constant rate, substantially limit the functional forms
for utility and production that we can consider.
3.5 Solving the Model off the Steady State - An Exact
Solution
Solving the model when it is not in the steady state is challenging. This is
because it is generally diffi cult to find a closed form solution. There are two
cases that we know how to solve the model exactly. Other cases, we need to use
numerical methods to construct an approximate solution.
One case of an exact solution is the case in which the production function
and the utility function are both log-linear (Cobb-Douglas), and the depreciation
rate is 100 percent. Let’s solve this now. We will leave out leisure out of the
utility function.
max
∑
βt
c1−σt
1− σ
(64)
subject to:
Akαt = ct + it (65)
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it = kt+1 (66)
The first order condition is
1
ct
=
β
ct+1
[αAkα−1t+1 ] (67)
There is one more restriction on this model, which is called the transver-
sality condition, and this is often abreviated as "TVC". We will not go into
this in detail. The importance of the transversality condition can be seen by
substituting out for ct and ct+1 using the resource constraint. If we do this, we
get:
1
Akαt − kt+1
=
β
Akαt+1 − kt+2
[αAkα−1t+1 ] (68)
Note that with this substitution, we have a difference equation in the variable
k. A difference equation is an equation in which a variable appears at different
dates. This equation is called a second-order difference equation because k
appears at dates t, t+1, and t+2. (Note that a first order difference equation is
one in which k appears at just two points in time, etc.)
It turns out that a second order difference equation has two possible paths,
one that is stable, and one that is explosive. This is because there are two roots
to this difference equation. The transversality condition is a restriction that
rules out the explosive path. It is given by:
lim
t→∞
βtu′(ct)kt+1 = 0 (69)
In this model, the solution is that consumption and investment are constant
fractions of output, as in the Solow model:
ct = (1− s)yt, it = syt (70)
To see this, plug the candidate solutions into the first order condition. We
get:
kαt+1
kαt
= βαAkα−1t+1 (71)
Re-arranging, we get:
kt+1 = βαAk
α
t (72)
Thus, fraction βa of output is saved, and fraction (1− βα) is consumed.
Note that if we have δ < 1, then we no longer have an exact solution to
the model. This is because the first order condition is one that has non-linear
components and linear components:
1
ct
=
β
ct+1
[αAkα−1t+1 + 1− δ] (73)
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3.6 Solving the Model offthe Steady State - More General
Cases
For more general cases - specifically, the case in which depreciation is not 100
percent, we need to use numerical methods to approximate the solution of the
model. This is because the model is nonlinear, and because the equations are
connected through time.
However, the saddle path provides an idea for how to solve the model off
of the steady state. For example, we know from the saddle path that there is
a single value of date 0 consumption that will take the economy to the steady
state. Therefore, we can use the following algorithm to solve for the sequences
of optimal consumption and investment to the steady state.
This is called a shooting algorithm. We proceed as follows. Consider log
utility and no leisure for simplicity. Specifiy the initial capital stock, k0. Guess
c0. Next, use the resource constraint to find k1 : k1 = f(k0) − c0 + (1 − δ)k0.
Next, go to our first order condition to solve for c1 :
c1
c0
= β[f ′(k1) + 1− δ] (74)
Given c1, we can use next period’s resource constraint to solve for k2 : k2 =
f(k1)− c1+(1− δ)k1. Then just keep doing this for another 50 periods or more.
If you happened to have guessd the correct (saddle-path) c0, then the economy
will converge to the steady state. Suppose that you guess c0 that is too low.
What will happen? In this case, k1 will be too high relative to its saddle path
value. This means that f ′(k1) will be too low relative to its saddle path value,
and c1 will be too low as well. In this case, the capital stock will grow beyond
the steady state value. See if you can find out to which case this corresponds
to in Figure 2.5.
Finally, suppose that your guess for c0 is too high relative to its saddle path
value. In this case, k1 will be too low, f ′(k1) will be too high, and as a result,
c1 will be too high relative to its saddle path value. In this case, the capital
stock will converge to zero. Look at figure 2.5 to see if you can identify which
case this corresponds to.
If your guess for c0 is wrong. You will know whether it is too low (capital
ultimately exceeds the steady state), or whether it is too high (capital converges
to zero). If it is too low, just raise your guess for c0. If it is too high, just lower
your guess for c0. Do this until you get the right c0. This algorithm is very fast
and accurate.
4 Takeaway
The optimal growth model specifies a utility-maximzing consumer with an infi-
nite planning horizon. The consumer chooses how to allocate their time between
leisure and production, and how to allocate their income between consumption
and savings. In the absence of technological change, the model posesses a steady
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state, just as in the Solow model. With technological change, we use utility func-
tion and production functions that are consistent with steady state growth, in
which labor is constant, and that output, consumption, the wage rate, invest-
ment, and the capital stock all grow at the same rate, which is the rate of
technological change.
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