ECA5102-Lecture2

ECA5102: Lecture NotesTopic 02: Ramsey-Cass-Koopmans Model

Aamir Rafique Hashmi

National U of Singapore

August 22, 2014

Hashmi (National U of Singapore) RCK Model August 22, 2014 1 / 48

Why this model?

In Solow model households do not optimally choose consumption andsavings

In this model we shall explicitly model household economic behavior

This has become one of the main workhorse models of modernmacroeconomics


Outline of the Lecture

Model assumptions

Behavior of the firm

Behavior of the household

Dynamics of the economy

Comparative dynamics

Rate of adjustment to steady state


Assumptions

A large number of identical firms

Production function:

Y (t) = F [K (t),A(t)L(t)],

Usual restrictions on the production function (same as in Solow’smodel)

Competitive input and output markets

A is given and grows at an exogenously given rate g

Firms’objective is to maximize profits


Assumptions

A large number of identical households

The size of each household grows at rate n

Each member of the household supplies one unit of labor at everypoint in time

Each household rents capital to firms

Initial capital stock per household is

Initial aggregate capital stocknumber of households

=K (0)H

No depreciation


Assumptions

The (representative) household divides income between consumptionand savings to maximize utility

Household’s utility is given by

U =∞∫

t=0e−ρtu[C (t)]

L(t)Hdt (2.1)

C (t) = Consumption of each member of the household

u = Instantaneous utility function

L(t) = Total population in the economy

ρ = Discount rate(e−ρ = 1

eρ ≈ 11+ρ

)


Assumptions

The instantaneous utility function is

u[C (t)] =C (t)1−θ − 11− θ

, θ > 0 and ρ− n− (1− θ)g > 0. (2.2)

This utility function is called CRRA (constant relative risk aversion)

Risk aversion is related to the curvature of the utility function

Arrow-Pratt coeffi cient of Absolute Risk Aversion: −u′′/u′

Coeffi cient of Relative Risk Aversion:

−cu′′/u′ = −c (−θ) c−θ−1/c−θ = θ


Assumptions

θ is the inverse of the intertemporal elasticity of substitution

If θ is small (approaching zero), large swings in consumption areacceptable to the consumers

If θ → 1, u = lnC

Proof:

limθ→1

C 1−θ−11−θ = lim

θ→1

∂C 1−θ−1∂θ

∂(1−θ)∂θ

= limθ→1

−C 1−θ ln C−1

= lnC . Q.E.D.

The first equality follows from L’Hopital’s rule.

ρ− n− (1− θ)g > 0 is needed to ensure that life-time (which isinfinite here) utility does not diverge



Firm’s problem is static so we suppress the time subscripts

The firm maximizes profits taking factor prices as given

maxL,K{F (K ,AL)−WL− rK}.

We have normalized the price of output to 1

The first order conditions for this problem are

∂F∂L = W and

∂F∂K = r .



Since F is CRS,Y = F (K ,AL) = ALf (k),

where k = KAL .

=⇒ ∂F∂K = f

′(k) and ∂F∂L = A[f (k)− f ′(k)k ].

The first order conditions become

A[f (k)− f ′(k)k ] = W (2.4) , and

f ′(k) = r . (2.3)

The wage rate per effective unit of labor is

WA≡ w = f (k)− f ′ (k) k (2.5)



The household chooses a path of consumption to maximize its utilitytaking timepaths of r and W as given

Define R(t) =t∫

τ=0r(τ)dτ

This allows for time varying interest rate

If the interest rate is constant, say at r , then R(t) = r t



The household budget constraint is

∞∫t=0e−R (t)C (t)

L(t)Hdt ≤ K (0)

H+

∞∫t=0e−R (t)W (t)

L(t)Hdt (2.6)

=⇒ K (0)H +

∞∫t=0e−R (t)[W (t)− C (t)] L(t)H dt ≥ 0 (2.7)

=⇒ lims→∞

[K (0)H +

s∫t=0e−R (t)[W (t)− C (t)] L(t)H dt

]≥ 0 (2.8)



The expression in square brackets in (2.8) is the present value ofcapital holdings of the household at time s.

To see this, note that the capital holdings of the household at time sare

K (s)H

=K (0)H

eR (s) +s∫

t=0eR (s)−R (t)[W (t)− C (t)]L(t)

Hdt. (2.9)

Taking the present value we get

K (s)H

e−R (s) =K (0)H

+s∫

t=0e−R (t)[W (t)− C (t)]L(t)

Hdt.

The right-hand side of the last equation is the same as the bracketedexpression in (2.8)



Substituting the last equation in (2.8) we get

lims→∞

[K (s)H

e−R (s)]≥ 0, (2.10)

which says that the present value of the household assets cannot benegative in the limit

Using the household budget constraint in (2.6), we have derived aboundary condition in (2.10).



Before solving the household maximization problem it is instructive towrite the objective function and constraints in terms of variables pereffective unit of labor

Recall the utility function

U =∞∫

t=0e−ρt C (t)

1−θ

1− θ

L(t)Hdt.

Define c(t) = C (t)A(t) =⇒ C (t) = c(t)A(t) = c(t)A(0)egt .

Also recall L(t) = L(0)ent . The utility can then be written as

U =∞∫

t=0e−ρt [c(t)A(0)e

gt ]1−θ

1− θ

L(0)ent

Hdt.



U =∞∫

t=0e−ρt [c (t)A(0)egt ]1−θ

1−θL(0)ent

H dt

= [A(0)]1−θL(0)H

∞∫t=0e−[ρ−n−(1−θ)g ]t c (t)1−θ

1−θ dt

= B∞∫

t=0e−βt c (t)1−θ

1−θ dt, (2.12)

.

where B = [A(0)]1−θL(0)H and β = ρ− n− (1− θ)g > 0.



To rewrite the budget constraint, note that C (t) L(t)H = c(t)A(t)L(t)H ,

W (t) L(t)H = w(t)A(t)L(t)H and K (0)H = k(0)A(0)L(0)H .

Substituting these in (2.6) we get

∞∫t=0e−R (t)c(t)

A(t)L(t)H

dt ≤ k(0)A(0)L(0)H

+∞∫

t=0e−R (t)w(t)

A(t)L(t)H

dt.

Using A(t)L(t) = A(0)L(0)e(n+g )t and dividing by A(0)L(0)H we have

∞∫t=0e−R (t)c(t)e(n+g )tdt ≤ k(0) +

∞∫t=0e−R (t)w(t)e(n+g )tdt. (2.14)



To rewrite (2.10), note that K (s) = k(s)A(s)L(s). Substituting thisin (2.10)

lims→∞

[k(s)

A(s)L(s)H

e−R (s)]≥ 0.

=⇒lims→∞

[A(0)L(0)

He−R (s)e(n+g )sk(s)

]≥ 0.

Dividing by A(0)L(0)H we have

lims→∞

[e−R (s)e(n+g )sk(s)

]≥ 0. (2.15)



The objective of the household is to choose c(t) to maximize (2.12)subject to (2.14).

The Lagrangian for the problem is

L = B∞∫

t=0e−βt c (t)1−θ

1−θ dt+

λ

[k(0) +

∞∫t=0e−R (t)w(t)e(n+g )tdt −

∞∫t=0e−R (t)c(t)e(n+g )tdt

]. (2.16)

The first order condition with respect to c(t) is

Be−βtc(t)−θ = λe−R (t)e(n+g )t . (2.17)



Take logs and use R(t) =t∫

τ=0r(τ)dτ

lnB − βt − θ ln c(t) = lnλ−t∫

τ=0r(τ)dτ + (n+ g)t. (2.18)

Differentiate with respect to time

−β− θc(t)c(t)

= −r(t) + (n+ g). (2.19)

Substitute ρ− n− (1− θ)g for β and solve for c (t)c (t)

c(t)c(t)

=r(t)− ρ− θg

θ. (2.20)



Recall C (t) = c(t)A(t). Take logs and differentiate with respect totime

C (t)C (t)

=A(t)A(t)

+c(t)c(t)

Substitute (2.20) and g = A(t)A(t) into the last equation

C (t)C (t)

= g +r(t)− ρ− θg

θ

which simplifies to

C (t)C (t)

=r(t)− ρ

θ. (2.21)



If r(t) > ρ, consumption is rising because people consume less (andsave more) now and consume more in future

If r(t) < ρ, consumption is falling because people are impatient andthey want to consume more now

If θ is low, intertemporal elasticity of substitution is high. Peopleallow greater variation in consumption for a given difference betweenr(t) and ρ

If θ → ∞, intertemporal elasticty of substitution approaches zero.People want perfect consumption smoothing.


Dynamics of the Economy

Recall c (t)c (t) =r (t)−ρ−θg

θ (2.20).

Since r(t) = f ′[k(t)] we can write (2.20) as

c(t)c(t)

=f ′[k(t)]− ρ− θg

θ. (2.24)

c(t) = 0 if f ′[k(t)] = ρ+ θg . Let us call the unique k(t) thatsatisfies this equation k∗.

c(t) > 0 if f ′[k(t)] > ρ+ θg i.e. k(t) < k∗.

c(t) < 0 if f ′[k(t)] < ρ+ θg i.e. k(t) > k∗.





For k note that

k(t) = f [k(t)]− c(t)︸︷︷︸actual investment

− (n+ g)k(t).︸︷︷︸replacement investment

(2.25)

k(t) = 0, if c(t) = f [k(t)]− (n+ g)k(t).

k(t) > 0, if c(t) < f [k(t)]− (n+ g)k(t).

k(t) < 0, if c(t) > f [k(t)]− (n+ g)k(t).





Let kG be the value of k corresponding to the maximum c in Figure(2.2)

Before combining the two graphs to study the dynamics, we need toknow whether k∗ is less than, equal to or greater than kG .

Claim: kG > k∗

Proof: Showing kG > k∗ is equivalent to showing thatf ′(kG ) < f ′(k∗) because f ′′(k) < 0

f ′(k∗) = ρ+ θg (from (2.24) when c(t) = 0)

f ′(kG ) = n+ g (from (2.25) when k(t) = 0)

Then f ′(kG ) < f ′(k∗) requiresn+ g < ρ+ θg =⇒ ρ− n− (1− θ)g = β > 0,

Which is true by assumption. Q.E.D.







In Figure (2.4), points above F will lead to c = 0 and k = 0

Points below F will lead tok > kG =⇒ f ′(k) < (n+ g) =⇒ e−R (s)e(n+g )s is rising and so isk(s)

=⇒ lims→∞

[e−R (s)e(n+g )sk(s)

]= ∞,

which implies that the PV of lifetime income minus the PV of lifetimeconsumption is infinity.

Hence the household can increase utility by increasing consumption ateach point in time.

Starting from F, the system follows the Saddle Path to point E.




Balanced Growth Path

At point E: c , y and k are constant

s = y−cy is also constant

K (= ALk) and Y (= ALy) grow at rate n+ g

KL ,

YL and C (= Ac) grow at rate g

Unlike Solow model, in this model k > kG is not possible on the BGP


Welfare

In the absence of externalities and distortionary taxes, a competitiveequilibrium will generate Pareto-effi cient allocations and welfare willbe maximized.


Comparative Dynamics (1)

We shall study the dynamics after an unexpected drop in discountrate ρ

As ρ drops, people value future more.

This is equivalent to an increase in the saving rate in Solow model.

The dynamic system consists of

c (t)c (t) = f ′[k (t)]−ρ−θg

θ . (2.24)

k(t) = f [k(t)]− c(t)− (n+ g)k(t). (2.25)

ρ appears in (2.24) only so only c = 0 line will be affected

As ρ drops k∗ increases





Now we add government to the model and study the dynamics after achange in government expenditures (G )

Assume that

an increase in G does not affect utility from private consumptionit does not affect future outputit is financed by lumpsum taxesthe government’s budget is always balanced



With the government in the model, (2.25) changes to

k(t) = f [k(t)]− c(t)− G (t)− (n+ g)k(t). (2.40)

And (2.14) changes to

∞∫t=0e−R (t)c(t)e(n+g )tdt

≤ k(0) +∞∫

t=0e−R (t)[w(t)− G (t)]e(n+g )tdt. (2.41)






Slope of the Saddle Path

We shall focus on the Saddle Path in the neighborhood of thesteady-state

To do so, we shall linearize the system around the steady-state

Recall the first-order Taylor approximation formula in case of onefunction

f (x) ' f (x∗) + f ′(x∗)[x − x∗].In case of two functions and two variables it extends to[f (x , y)

g(x , y)

]'[f (x∗, y ∗) + f1(x∗, y ∗)[x − x∗] + f2(x∗, y ∗)[y − y ∗]g(x∗, y ∗) + g1(x∗, y ∗)[x − x∗] + g2(x∗, y ∗)[y − y ∗]

]



The dynamic system that we want to linearize consists of

c(t) = f ′[k (t)]−ρ−θgθ c(t). (2.24)

k(t) = f [k(t)]− c(t)− (n+ g)k(t). (2.25)

Compared to the general case in the previous slide the variables ofinterest are c and k (instead of x and y) and the relevant functionsare c(c, k) and k(c, k) (instead of f (x , y) and g(x , y))

We take first-order Taylor approximation around k = k∗ and c = c∗

c ' [c ] c=c ∗k=k ∗

+[

∂c∂c

]c=c ∗

k=k ∗[c − c∗] +

[∂c∂k

]c=c ∗

k=k ∗[k − k∗], (2.26)

k '[k]c=c ∗

k=k ∗+[

∂k∂c

]c=c ∗

k=k ∗[c − c∗] +

[∂k∂k

]c=c ∗

k=k ∗[k − k∗]. (2.27)



To simplify the notation, note that in the steady-state, c = k = 0.

Define c = c − c∗, k = k − k∗,◦c = c and

◦k = k

◦c '

[∂c∂c

]c=c ∗

k=k ∗c +

[∂c∂k

]c=c ∗

k=k ∗k (2.28)

◦k '

[∂k∂c

]c=c ∗

k=k ∗c +

[∂k∂k

]c=c ∗

k=k ∗k (2.29)

There are four partial derivatives that we need to find and evaluate atthe steady state values of c and k



Recall

c(t) =f ′[k(t)]− ρ− θg

θc(t) (2.24).

[∂c∂c

]c=c ∗

k=k ∗

=

[f ′[k(t)]− ρ− θg

θ

]c=c ∗

k=k ∗

= 0,

because f ′(k∗) = ρ+ θg .[∂c∂k

]c=c ∗

k=k ∗

=

[c (t)

f ′′[k(t)]θ

]c=c ∗

k=k ∗

= c∗f ′′[k∗]

θ.



Recallk(t) = f [k(t)]− c(t)− (n+ g)k(t). (2.25).

[∂k∂c

]c=c ∗

k=k ∗

= −1.

[∂k∂k

]c=c ∗

k=k ∗= [f ′[k(t)]− (n+ g)] c=c ∗

k=k ∗

= f ′[k∗]− (n+ g)= ρ+ θg − (n+ g)≡ β



Substituting these derivatives in (2.28) and (2.29)

◦c ' c ∗f ′′[k ∗]

θ k (2.30)◦k ' βk − c (2.31)

Dividing (2.30) by c and (2.31) by k

◦cc '

c ∗f ′′[k ∗]θ

kc (2.32)

◦kk' β− c

k(2.33)



For simplicity, we assume◦cc =

◦kk= µ. (2.32) becomes

µ =c∗f ′′[k∗]

θ

kc.

=⇒ck=c∗f ′′[k∗]

θ

1µ. (2.34)

Substitute (2.34) in (2.33)

µ ' β− c∗f ′′[k∗]

θ

1µ. (2.35)

=⇒µ2 − βµ+

c∗f ′′[k∗]θ

= 0. (2.36)



The two roots of (2.36) are

µ1, µ2 =β±

√β2 − 4 c ∗f ′′[k ∗]θ

2.

Since c ∗f ′′[k ∗]θ < 0, one root

(β−√···

2

)is less than zero (the other is

greater than zero and implies a divergent path). Let us call this rootµ1.

To be on the linearized saddle path, at time zero c must be equal to

c∗ +c∗f ′′[k∗]

θ

1µ1(k − k∗).︸︷︷︸

(c−c ∗) from (2.34)

From there,

k(t) = k∗ + eµ1t [k(0)− k∗] = k∗ + eµ1 [k(t − 1)− k∗]c(t) = c∗ + eµ1t [c(0)− c∗] = c∗ + eµ1 [c(t − 1)− c∗]


Speed of Adjustment

Assume f (k) = kα =⇒ f ′′(k) = α(α− 1)k∗(α−2)

From (2.24) k∗ =(

αρ+θg

) 11−α

From (2.25) c∗ = k∗α − (n+ g)k∗

Parameters: α = 13 , ρ = 0.04, n = 0.02, g = 0.01 and θ = 1

Using the above functional form and parameter values we getµ1 = −5.4%

It means each period both c and k cover 5.4% of their currentdistance from c∗ and k∗


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