Macroeconomic Theory Notes

ECON503: Lecture NotesMacroeconomic Theory II

by Jorge Rojas

This document has been written using LATEX based on the Lectures of ProfessorOksana Leukhina. I assume full responsibility for any mistakes or typos. I promise,I have minimized them.1

Contents

1 Market Structures, Welfare Theorems. 21.1 Arrow-Debreu Competitive Equilibrium. . . . . . . . . . . . . . . . . . . 21.2 First Welfare Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Sequential Market Structure. . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Neoclassical Growth Model 92.1 SETUP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 (SP) Sequential Problem Formulation. . . . . . . . . . . . . . . . . . . . 9

2.2.1 Direct approach (Lagrange) . . . . . . . . . . . . . . . . . . . . . 102.3 Functional Equation (Recursive approach) . . . . . . . . . . . . . . . . . 12

3 Topology 143.1 Contraction Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . 18

4 NGM REVISITED 204.1 SETUP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

4.1.1 Features of the Model . . . . . . . . . . . . . . . . . . . . . . . . 204.1.2 Households . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204.1.3 Firms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4.2 Competitive Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . 214.3 Solving for the Equilibrium Path . . . . . . . . . . . . . . . . . . . . . . 224.4 Solving NGM using the Sequential Approach . . . . . . . . . . . . . . . . 254.5 Calibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

1Without Equality in Opportunities, Freedom is the privilege of a few, and Oppression the reality ofeveryone else.


by Jorge Rojas

1 Market Structures, Welfare Theorems.

We will use a simple dynamic exchange economy with two agents to define a equilibriumconcept, in other order to discuss different market structures and the First WelfareTheorem.

Assumptions:

1. Time is discrete t = 0, 1, 2, . . .

2. two individuals that live forever. There are no other identities.

3. Preferences defined over allocations ci = {cit}∞t=0, i = 1, 2

u(ci) =∞∑t=0

βt ln(cit) ; β ∈ (0, 1)

This is a pure exchange economy. For example:

e1t =

{2 if t is even0 otherwise

and e2t =

{0 if t is even2 otherwise

Commodities here are consumption goods at t = 0, 1, 2, . . . .Information: all public, full enforcement.{pt} denote prices of these commodities.Market structure: trading takes place at a central market place in period 0. We furtherassume that agents take prices as given.No more trading takes place after this only one meeting, after that only delivery is carriedout. Another crucial assumption is that the market always clears.

1.1 Arrow-Debreu Competitive Equilibrium.

Definition 1. A Competitive Arrow-Debreu equilibrium is given by {pt}∞t=0 and(c1, c2) such that:

1. Given {pt}∞t=0, {cit}i=1,2 solves:

Max{ct}∞t=0

∑βtln cit

s.t.∑

ptcit ≤

∑pte

it

cit ≥ 0 ∀t

2. Prices have to be such that market is clear (also called feasibility constraint).

c1t + c2

t = e1t + e2

t ∀t ← there is no storage, otherwise ≤

In this model, we can actually find equation prices and allocations analytically. Settingup the Lagrangian:

L =∑

βtln cit − λi(∑

ptcit −∑

pteit

)

University of Washington Page 2


by Jorge Rojas

The FOC’s are given by:

βt

cit= λipt

βt+1

cit+1

= λipt+1

combining them, we get:cit+1

cit= β

ptpt+1

⇒ βptcit = pt+1c

it+1 (1)

Notice that equation (1) establishes the solution, so hats are needed for mathematicalformality.We know sum equation (1) across individuals i = 1, 2:

βpt(c1t + c2

t ) = pt+1(c1t+1 + c2

t+1)

For our example, we have that c1t + c2

t = c1t+1 + c2

t+1 = 2 since markets always clear. Thus,

⇒ β =pt+1

pt(2)

Normalizing p0 = 1 in (2), we get that p1 = β and by recursive iteration:

pt = βt (3)

Using equation (1) into (3), we get that:

cit = cit+1 = ci0 ∀i = 1, 2

Now, we may use the budget constraint to achieve specific values for our optimal con-sumption. For agent 1 we have:

c10

∞∑t=0

βt = 2 + 0 + 2β2 + 0 + 2β4 + . . .

c10

1

1− β= 2

∞∑t=0

β2t

= 2∞∑t=0

(β2)t

= 21

1− β2

⇒ c10 =

2

1 + β> 1

For agent 2, on the other hand, we will have that:

c20 = 2− 2

1 + β< 1

Agent type 1 gets to eat more than half of the aggregate endowment in each period.Why? Because the value of his endowment is larger since agent 1 receives 2 units of thegood in period zero, while agent 2 receives nothing in period zero. As we can see, if youget higher endowment at the beginning, it will have a net benefit for the rest of the “life”.So, we should tax the bequest from parents to children to build a better society (Jorge’scomment), otherwise, inequality will keep going on, regardless of talent.



by Jorge Rojas

Definition 2. An allocation {c1t , c

2t} is feasible if cit ≥ 0 ∀t, i and c1

t + c2t ≤ e1

t + e2t

Definition 3. An allocation {c1t , c

2t} is Pareto Optimal if:

1. it is feasible

2. there is no other feasible allocation {c1t , c

2t} such that:

i. u(cit) ≥ u(cit) for both i, and

ii. u(cit) > u(cit) for at least one agent i.

Example 1. c1t = 0 and c2

t = 2 ∀tThis allocation is efficient, since any change makes agent 2 worse off, given local nonsatiation in the utility function and “selfishness” of the agents.

Example 2. Autarky economy (no trade, closed economy)c1t 2 0 2 ...c2t 0 2 0 ...

any feasible allocation without zero is better, because given the log utility of the agents,we have that for ln(0)→ −∞. Therefore, this allocation is not Pareto Efficient.

Proposition 1. An allocation is Pareto Optimal if and only if (iff) it solves the Planner’sproblem for some α.

Max{ct}∞t=0

α∑

βtln c1t + (1− α)

∑βtln c2

t ;α ∈ [0, 1]

s.t. c1t + c2

t = e1t + e2

t ∀tcit ≥ 0 ∀t, i

Prove it as an exercise.

1.2 First Welfare Theorem.

Theorem 1. A competitive equilibrium is Pareto Efficient. In other words, let {c1t , c

2t}

be the C.E. allocation, then it is also Pareto optimal.

Proof. By contradiction.Suppose is not, i.e., ∃{c1

t , c2t} such that it is feasible, and

i. u(ci) ≥ u(ci) ∀i, and

ii. u(ci) > u(ci) for at least one i.

Without loss of generality, suppose u(c1) ≥ u(c1).

Step 1.It must be the case that {c1

t} was not affordable for 1 at C.E. prices, or else she wouldhave chosen it. Hence,

∞∑t=0

ptc1t >

∞∑t=0

pte1t

Step 2.We show that it must be the case that this {c2

t} allocation was either not affordable or



by Jorge Rojas

exactly as affordable as {c2t} for agent 2 as equilibrium prices.

Suppose not, i.e., (it means that c2t was strictly affordable)∑

ptc2t <

∑pte

2t (4)

define δ to be the difference (δ > 0).

Define c2t =

{c2t for t = 1, 2, . . .c2

0 + δ for t = 0

By “eating” c2t , agent 2 would have done better than with {c2

t} because u(c2) ≥ u(c2),but u(c2) > u(c2) ⇒⇐ (stands for contradiction).Hence, ∑

ptc2t ≥

∑pte

2t (5)

Step 3.Suming equations (4) and (5),∑

pt(c1t + c2

t ) >∑

pt(e1t + e2

t )

and using the fact that (c1t + c2

t ) ≤ (e1t + e2

t ), we get:∑pt(e

1t + e2

t ) ≥∑

pt(c1t + c2

t ) >∑

pt(e1t + e2

t )

Transitivity of the inequality gives us the contradiction ⇒⇐. Thus, we have proven theFWT.

1.3 Sequential Market Structure.

AD: agents meet once at t = 0 and traded claims on future consumption goods.e.g. In our economy, after trade is completed agent 1 holds claims on 2/(1 + β) of agent2’s “apples” (our consumption good) in every odd periods. To get those, agent 1 tradeda promise to deliver 2β/(1 + β) apples to agent 2 in all even periods.

A much better approximation to reality is a Sequential Market (SM) structure, whereagents trade each period. It turns out that if Arrow securities (bonds) can also be traded,then equivalent allocations and prices are achieved in both market structures.An Arrow security is a claim traded at time t to deliver 1 apple at t+ 1. Notice that thisis a deterministic environment. Let qt be the price of 1 bond at time t. Spend qt applesat time t, to receive 1 apple at time (t+ 1).

Definition 4.

1 + rt =1

qt

Let ait+1 denote the number of Arrow securities (bonds) purchased in period t. Im-portant: bonds are traded after apple deliveries have been made.

The budget constraint is given by:

cit + qtait+1 = eit + ait ∀t

qt is the price of 1 bond at time t. So, equivalently we may write:

cit +ait+1

(1 + rt+1)= eit + ait ∀t



by Jorge Rojas

Note that ait+1 < 0 means that agent i is borrowing against tomorrow’s income.ait+1 < 0⇔ i holds negative bonds (you are borrowing and you have to pay back).⇔ you get qta

it+1 extra apples today in exchange for a promise to deliver ait+1 apples to

the other agent. Assume that ai0 ∀i.

Definition 5. A competitive SM equilibrium is allocations {cit, ait+1}∞t=0 ∀i, and interestrates {rt+1}∞t=0 such that:

1. For each i, given {rt+1}∞t=0, {cit, ait+1}∞t=0 solves:

Max∑

βtln cit

s.t. cit +ait+1

1 + rt+1

≤ eit + ait ∀t

ai0 = 0 ∀icit ≥ 0 ∀tait+1 ≥ −Ai ← (Ai is a “big” number).

The last constraint is necessary for existence, although it will not be binding in theequilibrium.

2. Feasibility:∑

i cit =

∑i eit ∀t Goods Market clearing

Asset market clearing:∑

i ait = 0 ∀t

P

Proposition 2. Suppose {cit, ait+1}∞t=0∧{rt+1}∞t=0 form a SME. Then ∃ an AD equilibrium{cit}∞t=0 ∧ {pt}∞t=0 such that cit = cit ∀t, i

Suppose allocations {cit}∞t=0 ∧ {pt}∞t=0 form an AD equilibrium such that pt+1

pt< 1.

Then ∃(Ai)i=1,2 and a SME {cit, ait+1}∞t=0 ∧ {rt+1}∞t=0 such that cit = cit ∀t, i

Notice that the equivalence only holds if the No Ponzi constraints do not bind, andrt+1 > 0 ∀t. Although, this last condition is for simplicity, but is not crucial for theresult.

Proof. By definitions.

Step 1: We want to show that SME allocations {cit} satisfy the AD budget constraintwhen AD prices are related to the SM interest rates as follows:Set p0 = 1 and let pt+1 = pt

1+rt+1∀t. Thus, qt = pt+1

ptby definition is the price of apples

at time (t+ 1) in terms of apples at time t.We know that SM budget constraints hold:

[BC0] : ci0 +ai1

1 + r1

= ei0 + 0

[BC1] : ci1 +ai2

1 + r2

= ei1 + ai1

From [BC1] ⇒ ai1 = ci1 +ai2

1+r2− ei1 and this into [BC0], we get:

ci0 +ci1

(1 + r1)+

ai2(1 + r1)(1 + r2)

= ei0 +ei1

(1 + r1)



by Jorge Rojas

by recursive iteration, we get:

T∑t=0

citΠtj=1

+aiT+1

ΠT+1j=1 (1 + rt)

=T∑t=0

eitΠtj=1(1 + rj)

(6)

We define Π0j=1(1 + rj) = 1. We also have that:

Πtj=1(1 + rj) =

p0

��p1

×��p1

��p2

× · · · ×��pt−1

pt=p0

pt=

1

pt(7)

recall that 1 + rt = pt−1

pt. Using (7) into (6), we get:

T∑t=0

ptcit +

aiT+1

ΠT+1j=1 (1 + rt)

=T∑t=0

pteit (8)

taking limit when T →∞, equation (8) becomes:

∞∑t=0

ptcit + lim

T→∞

aiT+1

ΠT+1j=1 (1 + rt)

=∞∑t=0

pteit (9)

We know that ait+1 ≥ −Ai for all t, and in particular for T . Then:

0 ≥ limT→∞

aiT+1

ΠT+1j=1 (1 + rj)

≥ limT→∞

−Ai

ΠT+1j=1 (1 + rj)

= 0 since rt > 0 ∀t

Hence:

limT→∞

aiT+1

ΠT+1j=1 (1 + rj)

= 0

Therefore,

∞∑t=0

ptcit =

∞∑t=0

pteit (10)

Step 2: Now, suppose {cit, pt} comprise the AD competitive equilibrium. We want toshow that {cit} satisfies every SM budget constraint if we define {rt+1} and {ait+1} asfollows:

1 + rt+1 =ptpt+1

ait+1 =∞∑τ=1

pt+τ (cit+τ − eit+τ )pt+1

← excess demand!

so, ait+1 is the value of the excess demand in terms of (t+ 1) apples. Thus,

ait+1 = cit+1 − eit+1 +cit+2 − eit+2

(1 + rt+2)+

cit+3 − eit+3

(1 + rt+2)(1 + rt+3)+ . . .

ait = cit − eit +cit+1 − eit+1

(1 + rt+1)+

cit+2 − eit+2

(1 + rt+1)(1 + rt+2)+ . . .

dividing period (t+ 1) by (1 + rt+1) and subtracting it to ait, we get:

ait −ait+1

(1 + rt+1)= cit − eit



by Jorge Rojas

This shows what we wanted.

By assumption rt > 0, then pt+1

pt≤ ξ < 1. Hence,

ait+1 > −∞∑τ=1

pt+τeit+τ

pt+1

> −∞∑τ=1

ξτ−1eit+τ︸︷︷︸Ai

Thus, ait+1 ≥ −Ai.

Step 3: Consider a SM equilibrium allocation {cit, ait+1} and {rit+1} such that rt+1 > 0 ∀tand ait+1 > −Ai. We show that {cit} = {cit} constitute an AD equilibrium allocation forthe earlier definition for {pt}. {cit} satisfies market clearing and also maximizes utilitysubject to the AD constraint. If not, i.e., if there were another allocation chosen underAD markets, we know that it would also satisfy SM budget constraints, so it should havebeen chosen under SM too.

quot erat demonstrandum



by Jorge Rojas

2 Neoclassical Growth Model

2.1 SETUP

1. Time is discrete t = 0, 1, 2, . . .

2. Commodities

i. labour services ht

ii. capital services kt

iii. final output yt

3. Technology: yt = F (kt, ht)

i. F (·) continuous differentiable,

ii. strictly increasing in both arguments,

iii. strictly concave,

iv. homogenous of degree 1,

v. Inada Conditions

a. F (0, h) = F (k, 0) = 0

b. limk→0 Fk(k, h) =∞, limk→∞ Fk(k, h) = 0

c. limh→0 Fh(k, h) =∞, limh→∞ Fh(k, h) = 0

4. Market clearing: yt = ct + it

5. Law of Motion for capital: kt+1 = (1− δ)kt + it ; δ ∈ [0, 1]

6. Households: a large number of infinitely lived households preferences can be repre-sented by a time-separable utility function.

u({ct}∞t=0) =∞∑t=0

βtu(ct) (11)

7. Endowments

i. t = 0 each household has k0 units of capital

ii. t = 0, 1, 2, . . . each household has 1 unit of labour

2.2 (SP) Sequential Problem Formulation.

W (k0) = Max{ct,kt+1,ht}∞t=0

∑βtu(ct)

s.t. ct + kt+1 − (1− δ)kt︸︷︷︸it

= F (kt, ht)

ct, kt+1 ≥ 0, 0 ≤ ht ≤ 1

k0 ≤ k0given (but we know that k0 = k0)



by Jorge Rojas

Notice that it can be negative kt+1 < (1− δ)kt (we may disinvest).W (k0): total lifetime utility of the representative household if the Planner chooses{ct, kt+1, ht}∞t=0 to solve the above.From the utility function we get that ht = 1 since the agent does not care about leisurein this model. We also have that k0 = k0.There is no cost of adjustment, so we may define:

f(k) = F (k, 1) + (1− δ)k (12)

Therefore, we can restate the problem as:

Max{kt+1}∞t=0

∞∑t=0

βtu(f(kt)− kt+1)

0 ≤ kt+1 ≤ f(kt)← from law of motion for k

k0 given

This problem is stationary, i.e., time-invariant. Theorem 6.11 in Acemoglu establishesexistence and uniqueness of the solution. Theorem 6.12 in Acemoglu establishes thatFOC’s and TVC are necessary and sufficient to characterize the optimal plan {k∗t+1}∞t=0.

2.2.1 Direct approach (Lagrange)

We may “tackle” this problem using the Lagrange approach:

L =∞∑t=0

βtu(f(kt)− kt+1)

L = u(f(k0)− k1) + βu(f(k1)− k2) + β2u(f(k2)− k3) + . . .

FOC’s:

[kt+1] : 0 = −βtu′(f(kt)− kt+1) + βt+1u′(f(kt+1)− kt+2)f

′(kt+1)

rearranging this equation, we get:

⇒ βtu′(f(kt)− kt+1) = βt+1u

′(f(kt+1)− kt+2)f

′(kt+1) (13)

Equation (13) is the so-called Euler equation or intertemporal choice equation.

The Transversality condition (TVC) in Sequential Markets ensures No-Ponzi Game Schemes.

limt→∞

βtu′(f(kt)− kt+1)f

′(kt)kt = 0

This condition is equivalent to:limt→∞

λtkt+1 = 0

Example 3. Solve the above problem using the following functions:

1. u(c) = ln c

2. F (k, 1) = kα



by Jorge Rojas

3. δ = 1

Using the Euler equation and defining the auxiliary variable zt = kt+1

kαt, we analyse different

cases and we obtain the policy rule kt+1 = αβkαt . Then,

W (k0) =1

1− β

(αβ

1− αβln (αβ) + ln (1− αβ)

)+

α

1− αβln k0

NOTE: If there is a steady-state, then we do NOT need the TVC.

Theorem 2. If a sequence {kt+1}∞t=0 satisfies the Euler Conditions and the TransversalityConditions, then it solves the related planning problem.

Proof. We have the following functional equation for the Social Planner’s problem:

V (k(0)) = Max{k(t+1)}∞t=0

∞∑t=0

U(k(t), k(t+ 1)) (14)

subject to:

k(t+ 1) ∈ G(k(t)) ∀t ≥ 0

k(0) is given

We define:

Φ(k(t)) = {{k(s)}∞s=t : k(s+ 1) ∈ G(k(s)) for s = t, t+ 1, . . . }

In other words, Φ(k(t)) is the set of feasible choices of vectors starting from k(t).

This is one way proof (⇒). So, we want to show that if the conditions are satisfied by agiven sequence, then this sequence solves the social planner’s problem.Consider an arbitrary k(0), and let us define the sequence k ≡ (k(0), k(1), . . . ) ∈ Φ(k(0))as a feasible nonnegative sequence satisfying the Euler equations (15) and TVC (16).

∂U(k(t), k(t+ 1))

∂y+ β

∂U(k(t+ 1), k(t+ 2))

∂x= 0 (15)

and

limt→∞

βt∂U(k(t), k(t+ 1))

∂x· k(t) = 0 (16)

where x stands for the first argument and y for the second argument of the utility function.

We will show that k yields higher value than any other k ∈ Φ(k(0)). For any k ∈ Φ(k(0)),define:

∆k ≡ limT→∞

infT∑t=0

βt[U(k(t), k(t+ 1))− U(k(t), k(t+ 1))] (17)

Notice the subtle detail that the limit is actually limit inf, since there is no guaranteethat for an arbitrary k ∈ Φ(k(0)), the limit will exist.

We assume that the utility function is well-behaved. That is, U is continuous, concaveand differentiable. Since U is concave, a Taylor expansion of order 1 around (k(t), k(t+1))tells that:

U(y) = U(x) +DU(x)T (y − x) + o(||y − x||2)



by Jorge Rojas

where DU(x) is the Jacobian of U(x).We also know that U(x) > 0∀x ∈ R2 ando(||y − x||2) is also positive. Thus, we get:

∆k ≥ limT→∞

infT∑t=0

βt[DxU(k(t), k(t+1))·(k(t)−k(t))+DyU(k(t), k(t+1))·(k(t+1)−k(t+1))]

(18)for any k ∈ Φ(k(0)),where Dx = ∂

∂xand Dy = ∂

∂y.

Since k(0) = k(0), DxU(k(0), k(1)) · (k(0)− k(0)) = 0, we can rewrite the inequality as:

∆k ≥ limT→∞

infT∑t=0

βt[DyU(k(t), k(t+ 1)) + βDxU(k(t+ 1), k(t+ 2))] · (k(t+ 1)− k(t+ 1))

− limT→∞

inf βT+1DxU(k(T + 1), k(T + 2)) · k(T + 1)

+ limT→∞

inf βT+1DxU(k(T + 1), k(T + 2)) · k(T + 1)

Recall that lim inf(−yn + xn) ≥ − lim sup yn + lim inf xn.

∆k ≥ limT→∞

infT∑t=0

βt[DyU(k(t), k(t+ 1)) + βDxU(k(t+ 1), k(t+ 2))] · (k(t+ 1)− k(t+ 1))

− limT→∞

sup βT+1DxU(k(T + 1), k(T + 2)) · k(T + 1)

+ limT→∞

inf βT+1DxU(k(T + 1), k(T + 2)) · k(T + 1)

We know that k satisfies equations (15) and (16). Thus, the terms in the first linefrom above are equal to zero because of the Euler equations. The same is true forthe second line because of transversality condition. For the last line, we know thatDxU(k(T + 1), k(T + 2)) > 0 since U is increasing in both arguments by assumption. Inaddition, we know that k > 0 since we cannot have negative capital. Therefore, we get:

∆k ≥ limT→∞

inf βTDxU(k(T + 1), k(T + 2)) · k(T + 1) ≥ 0 (19)

Hence, ∆k ≥ 0. This means that {kt+1}∞t=0 yields higher value than any other feasiblesequence in Φ(k(0)), and is therefore optimal.

2.3 Functional Equation (Recursive approach)

The recursive formulation, also known as the “Bellman Equation” consists in writingthe problem in the form:

v(k) = Max0≤k′≤f(k)

{u(f(k)− k′) + βv(k′)} (20)

where:

(a) k: state variable

(b) k′: control variable

(c) v(k): value function



by Jorge Rojas

The solution is the function v∗(k) and the corresponding optimal policy function isk′= g(k). Thus,

v∗(k) = u(f(k)− g(k)) + βv∗(g(k))

In general, we will find conditions under which:

v(k)︸︷︷︸sol. to FE

= W (k)︸︷︷︸sol. to SP

and the optimal policy function g(k) from the FE corresponds to the optimal plan{k∗t+1}︸︷︷︸

SP

.

Example 4. We have the functions u(c) = ln c and f(k) = kα. So, we get:

v(k) = Max0≤k′≤kα

{ln(kα − k′) + βv(k′)} (21)

We conjecture that the solution has the form v(k) = A+Bln k. Thus,

A+Bln k = Max0≤k′≤kα

{ln(kα − k′) + β(A+Bln k′)}

FOC’s:

− 1

kα − k′+βB

k′= 0⇒ k

′=

βBkα

1 + βB= g(k)

at the optimal:

A+Blnk = ln

(kα − βBkα

1 + βB

)+ β

(A+Bln

(βBkα

1 + βB

))= ln

(kα

1 + βB

)+ Aβ + βBln

(βBkα

1 + βB

)= (1 + βB)ln(kα)− ln(1 + βB) + βA+ βBln

(βB

1 + βB

)= α(1 + βB)ln(k)− ln(1 + βB) + βA+ βBln

(βB

1 + βB

)equating the terms with the variable capital k:

⇒ B = α(1 + βB)⇒ B =α

1− αβ

Therefore,

g(k) =βBkα

1 + βB⇒ g(k) = αβkα and it holds!!!

In a similar fashion we obtain A.

A =1

1− β

(αβ

1− αβln(αβ) + ln(1− αβ)

)So,

v(k) =1

1− β

(αβ

1− αβln(αβ) + ln(1− αβ)

)+

α

1− αβln(k)



by Jorge Rojas

3 Topology

In macroeconomics, we study functional equations(equations that give functions as aresult, instead of numbers) of the form:

v(x) = Maxy∈Γ(x)

{F (x, y) + βv(y)} (22)

In the case of the Neoclassical Growth Models we have that:

x = k

y = k′

F (x, y) = u(f(k)− k′)Γ(x) = {k′ ∈ R|0 ≤ k ≤ f(k)}

Definition 6. A function operator is a mapping that takes functions as an input andreturns out a function as an output.

Definition 7. (S, d) is a metric space with norm d(f, g) with f, g ∈ S, if the followingconditions are satisfied:

(i) d(f, g) ≥ 0

(ii) d(f, g) = 0⇔ f = g

(iii) d(f, g) = d(g, f)

(iv) d(f, h) ≤ d(f, g) + d(g, h)

Example 5. S = R, d : R×R 7→ R, and d(x, y) = |x−y|. Then (S, d) is a metric space.

Example 6. S = `∞ = {x = {xt}∞t=0|xt ∈ R,∀t supt |xt| < ∞} space of all infinitesequences of real finite values.We define the norm d : `∞ × `∞ 7→ R as d(x, y) = sup0≤t≤∞ |xt − yt|. So, (S, d) is ametric space.

Example 7. S = C(X) = {f |f : X 7→ R, f is continuous and bounded, X ⊆ RL} andwe define the norm d : C(X) × C(X) 7→ R as d(f, g) = supx∈X |f(x) − g(x)|. Later, wewill show that (S, d) is indeed a metric space. (This follows directly from the propertiesfor the supremum operator).

Definition 8. Cauchy Sequence.A sequence {xn}∞n=0 with xn ∈ S is Cauchy if:

∀ε > 0,∃Nε ∈ N such that d(xn, xm) < ε ∀n,m ≥ Nε

Definition 9. We say that (S, d) is a complete metric space if every Cauchy sequence{fn}∞n=0 with fn ∈ S ∀n converges to some f ∈ S.

Definition 10. Convergence of a sequence.

∀{fn}∞n=0 ∃f ∈ C(X) such that ∀ε > 0 ∃Nε satisfying d(fn, f) < ε ∀n ≥ Nε

Example 8. Consider the set S = C(X) to be the set of all continuous bounded functionsdefined on a subset of RL. Let the metric d : C(X)×C(X) 7→ R be the supnorm metric.Prove that (S, d) is a complete metric space.



by Jorge Rojas

Solution:We want to prove that (S, d) is a complete metric space. Therefore, we can prove thisclaim in two stages:

(1) we prove that (S, d) is a metric space,

(2) we prove that every Cauchy sequence {fn}∞n=0 with fn ∈ S ∀n converges to somef ∈ S.

First part:To prove that (S, d) is a metric space, we need to show that for the set S = C(X) and themetric d(f, g) = supx∈X |f(x)− g(x)| the following conditions are satisfied ∀f, g, h ∈ S:

(i) d(f, g) ≥ 0

(ii) d(f, g) = 0⇔ f = g

(iii) d(f, g) = d(g, f)

(iv) d(f, h) ≤ d(f, g) + d(g, h)

We know that f ∈ S, where S is the set of all continuous bounded functions in a subsetof RL, we can see that supx∈X |f(x)| <∞. Thus,

(i) supx∈X |f(x)−g(x)| < supx∈X |f(x)|+supx∈X |g(x)| by property of the supremum.Moreover, the functions are bounded. This implies that supx∈X |f(x)−g(x)| < λ1+λ2 = λwhere λ ∈ R+ ∪ {0}, so we will have that supx∈X |f(x) − g(x)| ∈ [0, λ], and therefore,d(f, g) ≥ 0.

(ii) Two ways demonstration. Let us start with (⇐).If f = g, then supx∈X |f(x)− g(x)| = supx∈X |f(x)− f(x)| = supx∈X |0| = 0.Now, the other implication (⇒). If supx∈X |f(x) − g(x)| = 0 means that the maximumdistance between the two functions is zero, i.e., the two functions must be the same.Suppose is not, then |f(x) − g(x)| 6= 0 for at least one x ∈ X. This is a contradiction.We establish the desired result.

(iii) supx∈X |f(x)−g(x)| = supx∈X |−(f(x)−g(x))| by property of the absolute value.Then, supx∈X |f(x)− g(x)| = supx∈X |g(x)− f(x)|. Hence, d(f, g) = d(g, f).

(iv) From the properties for the absolute value, we know that |a + b| ≤ |a| + |b|,and for the properties from the supremum we have a similar triangle inequality given bysupx∈X{f(x) + g(x)} ≤ supx∈X{f(x)}+ supx∈X{g(x)}. So,

d(f, h) = supx∈X|f(x)− h(x)|

= supx∈X|(f(x)− g(x)) + (g(x)− h(x))| ← we add a zero

≤ supx∈X|f(x)− g(x)|+ sup

x∈X|g(x)− h(x)| ←triangle inequality

= d(f, g) + d(g, h)

Hence, d(f, h) ≤ d(f, g) + d(g, h). We have shown the first part.



by Jorge Rojas

Second part:We need to prove that:

∀{fn}∞n=0 ∃f ∈ C(X) such that ∀ε > 0 ∃Nε satisfying d(fn, f) < ε ∀n ≥ Nε (23)

We will proceed as it is usual in this kind of problems, i.e., we will apply the definitionsto look for convergency rather than attempting a proof by contradiction.

By assumption {fn(x)}∞n=0 is a Cauchy sequence, therefore, ∀ε > 0, ∃Aε such thatsupx∈X |fn(x) − fm(x)| < ε for all n,m > Aε. Now, recall that {fn(x)}∞n=0 is a sequenceof functions. Therefore, if we fixed x at a particular value in X, say, x, the sequenceof functions becomes a sequence of numbers given by {fn(x)}∞n=0. So, the sequence ofnumbers is a Cauchy sequence in R. We know that R is a complete metric space, andtherefore, {fn(x)}∞n=0 → f(x). By definition A.9 in Acemoglu (page 881), we can saythat {fn(x)}∞n=0 converges to the pointwise limit f(x). Applying the same concept for thewhole domain, X, we can claim that the sequence of functions {fn(x)}∞n=0 will convergeto the pointwise limit function f(x).

To prove convergency we will proceed by definition. We know that {fn(x)}∞n=0 isCauchy and we want to prove that goes to f(x). So, we need to prove there exists Nε

such that d(fn, f) < ε for all n ≥ Nε. Fixing x at x ∈ X and using the fact that thesequence is Cauchy, hence d(fn, fm) < δ, let us say, δ = ε

2for all m ≥ n ≥ Nε. We have

as follows:

|fn(x)− f(x)| = |fn(x)− fm(x) + fm(x)− f(x)| ← add zero

≤ |fn(x)− fm(x)|+ |fm(x)− f(x)| ← triangle inequality

≤ ε

2+ |fm(x)− f(x)| ← |fn(x)− fm(x)| < ε

2by Cauchy sequence in R

once again we apply the fact that {fn(x)}∞n=0 → f(x), therefore, |fm(x)− f(x)| < δ = ε2

for some m ≥ Nε(x). Thus, |fn(x)− f(x)| < ε and the fact that x is arbitrary. This willhold for all x ∈ X. Therefore, |fn(x)− f(x)| < ε for all n ≥ Nε, implying that under thenorm for this metric space we will have:

supx∈X|fn(x)− f(x)| = d(fn, f) ≤ ε⇒ {fn}∞n=0 → f

Once we have proven that this function f(x) is in C(X), we will be done. So, we needto prove that f(x) is bounded and continuous. First, we prove that f is bounded, i.e.,supx∈X |f(x)| <∞.

supx∈X|f(x)| = sup

x∈X|f(x)− fn(x) + fn(x)| ← add zero

≤ supx∈X|f(x)− fn(x)|+ sup

x∈X|fn(x)| ← triangle inequality

we know that supx∈X |fn(x)| < α since fn is bounded, and we also know that the sequenceconverges to f(x), whence supx∈X |f(x) − fn(x)| = supx∈X |fn(x) − f(x)| < ε where ε isa finite number.

supx∈X|f(x)| < ε+ α = δ

Thus, supx∈X |f(x)| < δ. This shows that f(x) is indeed bounded.



by Jorge Rojas

To show continuity we apply the definition for a continuous function. We say that afunction f(x) is continuous if and only if:

∀ε > 0,∀x ∈ X ∃δ(ε, x) > 0 such that ||x− y|| < δ ⇒ |f(x)− f(y)| < ε ∀x, y ∈ X

We proceed in a similar fashion as before. Assume that there is a n large enough suchthat d(fn, f) < ε

3. We also know that the sequence of functions fn(x) is continuous. So,

we choose a δ > 0 such that ||x− y|| < δ is compatible with |fn(x)− fn(y)| < ε3. Then,

|f(x)− f(y)| = |f(x)− fn(x) + fn(x)− fn(y) + fn(y)− f(y)| ← add zero

≤ |f(x)− fn(x)|+ |fn(x)− fn(y)|+ |fn(y)− f(y)| ← triangle inequality

<ε

3+ε

3+ε

3← by above paragraph

Hence, |f(x)− f(y)| < ε. This shows that f(x) is continuous.

Therefore, we have proven the second part. This concludes the proof. �

Example 9. This is an example of a metric space that is not complete.

S = {f |f : [1, 2] 7→ R, f is continuous and strictly decreasing}d(f, g) = sup

x∈[1,2]

|f(x)− g(x)|

We just need to provide one Cauchy sequence that does not converge in this space, andwe are done. The sequence fn(x) = 1

nxbelongs to the space, however, as n→∞ fn(x)→

0 and 0 is a continuous function, but is not strictly decreasing. Thus, our sequenceconverges to an element outside our space, i.e., it does not converge in S.

Let (S, d) be a metric space. T : S 7→ S an operator. We say that T is a Contraction Mappingif ∃ a number β ∈ (0, 1) such that:

d(TX , TY ) ≤ βd(X, Y ) ∀X, Y ∈ S

β is called the modulus of a contraction mapping.

Example 10. S = [a, b] ⊆ R and d(X, Y ) = |X−Y | and TX = βX for β ∈ (0, 1). Then,

d(TX , TY ) = |TX − TY | = |βX − βY | = β|X − Y | ≤ β|X − Y | ⇒ T (X) is a contraction!

Simple and awesome ¨

Theorem 3. BLACKWELL (Sufficient conditions) TheoremLet X ⊆ RL and B(X) be the space of bounded functions f : X 7→ R with d being thesup-norm. Let T : B(X) 7→ B(X) such that:

(1) Monotonicity: If f, g ∈ B(X) such that f(x) > g(x) ∀x ∈ X, then

(Tf)(x) ≥ (Tg)(x) ∀x ∈ X

(2) Discounting: Let the function f + a for all f ∈ B(X) and a ∈ R+ be defined by(f + a)(x) = f(x) + a.

∃β ∈ (0, 1),∀f ∈ B(X), a ≥ 0 ∧ x ∈ X ⇒ [T (f + a)](x) ≤ [Tf ](x) + βa



by Jorge Rojas

Then T is a contraction mapping with modulus β.

Proof. We want to show that (w.t.s.) d(Tf , Tg) ≤ βd(f, g) ∀f, g ∈ B(X). Fix an arbitraryx ∈ X. Then f(x)− g(x) ≤ supy∈X |f(y)− g(y)|. So, f(x) ≤ g(x) + d(f, g). Then,

(Tf)(x) ≤︸︷︷︸by monot.

(Tg + d(f, g))(x) ≤︸︷︷︸by disctg

(Tg)(x) + βd(f, g)

⇒ (Tf)(x) ≤ (Tg)(x) + βd(f, g)

⇒ (Tf)(x)− (Tg)(x) ≤ βd(f, g) / supx∈X(·)

supx∈X |(Tf)(x)− (Tg)(x)| ≤ βd(f, g)

⇒ d(Tf, Tg) ≤ βd(f, g)

This completes the proof.

3.1 Contraction Mapping Theorem

Theorem 4. Contraction Mapping TheoremLet (S, d) be a complete metric space. Suppose that T : S 7→ S is a contraction mapping.Then:

(a) T has exactly one fixed-point v∗ ∈ S (existence and uniqueness)

(b) For any v0 ∈ S and any n ∈ N we have that d(T n(v0), v∗) ≤ βnd(v0, v∗), i.e., {vn}∞n=0

converges to v∗ at a geometric rate.

Computational approach to FE in NGM.

1. Pick an arbitrary function v0(k), say v0(k) = 0∀k, after defining a grid for k

2. update your function until convergence.

v1(k) = Max0≤k′≤f(k)

{u(f(k)− k′) + βv0(k′)}

until Maxk |vn+1(k)− vn(k)| < ε

Example 11. This may be an exam question.Prove that the Blackwell sufficient conditions theorem applies to the operator in the NGMdefined as:

Tv(k) = Maxk′∈Γ(k)

{u(f(k)− k′) + βv(k′)}

Proof. W.t.s. the operator:

Tv(k) = Maxk′∈Γ(k)

{u(f(k)− k′) + βv(k′)}

satisfies the 3 conditions.

1) T : B(X) 7→ B(X). Take an arbitrary element v(k) ∈ B(X). v(k) is bounded. Sincewe assume that u is bounded, we have that Tv is also bounded.



by Jorge Rojas

2) Monotonicity. Suppose v ≤ w (i.e. v(k) ≤ w(k) ∀k ∈ R+). We want to show thatTv ≤ Tw. Let gv(k) be an optimal policy corresponding to v. Then for all k ∈ (0,∞):

Tv(k) = u(f(k)− gv(k)) + βv(gv(k))

Tv(k) ≤ u(f(k)− gv(k)) + βw(gv(k)) ≤ Maxk′∈Γ(k)

{u(f(k)− k′) + βw(k′)} = Tw(k)

This implies that Tv(k) ≤ Tw(k). Thus, we have proven monotonicity.

3) Now, we proceed to prove discounting.

T (v + a)(k) = Max0≤k′≤f(k)

{u(f(k)− k′) + β(v(k′) + a)}

= Max0≤k′≤f(k)

{u(f(k)− k′) + βv(k′)}+ βa

= Tv(k) + βa

so it holds with equality and that is enough since we ask for ≤.

This completes the proof.

Lemma 1. Let (S, d) be a metric space and T : S 7→ S be a contraction mapping. ThenT is continuous.

Proof. Take an arbitrary s0 ∈ S and an arbitrary small ε > 0. We want to show that∃δ(s0, ε) such that whenever d(s, s0) < δ(ε, s0) we also have d(Ts, Ts0) < ε. Just chooseδ = ε. Then:

d(Ts, Ts0) ≤︸︷︷︸by T contract.

βd(s, s0) < βδ(ε, s0) = βε < ε

This shows the definition for continuity of a function, and hence T is continuous.

Now, we will proceed to prove that Contraction Mapping Theorem (Theorem 4).

Proof. This is an sketch of the proof.Choose a candidate for a fixed-point.

v∗ = limn→∞

vn

First, show that this limit is well-defined and it is in S. Do this by showing that vn isa Cauchy sequence and completeness of the metric space (this establishes existence anduniqueness of the limit).To show that this limit is indeed a fixed-point, we want to show that: Tv∗ = v∗.

Tv∗ = T ( limn→∞

vn) =︸︷︷︸by cont.of T

limn→∞

Tvn = limn→∞

vn+1 = v∗



by Jorge Rojas

4 NGM REVISITED

4.1 SETUP

4.1.1 Features of the Model

a. Includes leisure-labour trade-off, technological progress, and population growth

b. Time is discrete t = 0, 1, 2, . . .

c. One representative household

d. Nt: the number of identical members in the HH at t (or population size)

e. Population dynamics Nt+1 = Nt(1 + n) where n > 0 is known parameter

f. Initial population size N0

g. Each HH member is given 1 unit of time, trade-off between leisure (1− ht) and workht

h. Market wage is wt

i. Each household is given k0 units of capital in the initial period

j. HH’s rent their capital to the firm at rt

k. Upper case variables → aggregate variables,lower case variables → per-capita variables.

4.1.2 Households

The HH’s are described by:

a. Preference: infinite sequence of consumption, leisure and HH’s size

b. Utility function:∑∞

t=0 βtNtU(ct, 1− ht), where β ∈ (0, 1)

c. Assumption: β(1 + n) < 1

d. U(ct, 1 − ht): increasing in each argument, strictly concave, Inada conditions hold.These guarantee the existence and uniqueness of a solution

e. Taking the sequence {pt, wt, rt}∞t=0 and initial capital endowment per member (k0) asgiven, the representative HH solves:

Max{ct,kt+1,ht}∞t=0

∞∑t=0

βtNtU(ct, 1− ht)

s.t.∞∑t=0

pt[ct + (1 + n)kt+1] =∞∑t=0

pt[wtht + (rt + 1− δ)kt]

ct, kt > 0, 0 < ht < 1∀t



by Jorge Rojas

4.1.3 Firms

Firms are described by:

a. One representative firm

b. Behaviour is competitive

c. Production function F (Kt, Ht, t)

d. F (Kt, Ht, t) satisfies:

(i) homogeneous of degree 1 in Kt, Ht

(ii) F (0, H, t) = F (K, 0, t) = 0

(iii) increasing in Kt, Ht

(iv) strictly concave in Kt, Ht

(v) Inada conditions hold

e. The representative firm seeks to maximise profits taking rt and wt as given.

Max{Kt,Ht}

F (Kt, Ht, t)− rtKt − wtHt

4.2 Competitive Equilibrium

The Competitive Equilibrium (C.E.) is given by the sequences {c∗t , hS∗t , kS∗t+1, N∗t+1, K

D∗t , HD∗

t }∞t=0

and {r∗t , w∗t }∞t=0 such that:

i. Given {rt, wt}∞t=0, the sequence {c∗t , hS∗t , kS∗t+1}∞t=0 solves the HH’s problem

ii. Given {rt, wt}∞t=0, the sequence {KD∗t , HD∗

t }∞t=0 solves the firm problem

iii. All markets clear for each t

- labor: hS∗t N∗t = HD∗

t

- capital: kS∗t N∗t = KD∗

t

- goods: c∗tN∗t +KS∗

t+1 − (1− δ)KS∗t = F (KD∗

t , HD∗t , t)

iv. Population: N∗t+1 = N∗t (1 + n)

To solve the HH’s problem we combine the direct approach (i.e. we write the La-grangian) with the AD formulation:

L =∞∑t=0

βtNtU(ct, 1− ht)− λ∞∑t=0

pt[ct + kt+1(1 + n)− wtht − (rt + 1− δ)kt]

The FOC’s are given by:

∂L∂ct

⇒ βtU1(ct, 1− ht)Nt = λpt

∂L∂ht

⇒ −βtU2(ct, 1− ht)Nt = −λptwt

∂L∂kt+1

⇒ λpt(1 + n) = λpt+1(rt+1 + 1− δ)



by Jorge Rojas

After some algebraic manipulation, we get:

[H1]: U2(ct, 1− ht) = −wtU1(ct, 1− ht)[H2]: U1(ct, 1− ht) = βU1(ct+1, 1− ht+1)(rt+1 + 1− δ)

Now, we solve the firm’s problem:

Max{Kt,Ht}

F (Kt, Ht, t)− rtKt − wtHt

FOC’s:

[F1]: rt = F1(Kt, Ht, t) = F1(kt, ht, t)← HOD 1(Euler’s theorem)

[F2]: wt = F2(Kt, Ht, t) = F2(kt, ht, t)

From the Goods market clearing condition in aggregated terms:

Ct +Kt+1 = F (Kt, Ht, t) + (1− δ)Kt

⇒ [M]: ct + kt+1(1 + n) = F (kt, ht, t) + (1− δ)ktFinally, Transversality condition is:

[TVC]: limt→∞

βtU1(ct, 1− ht)(rt + 1− δ)Nt−1

λkt = 0

From theorem (2) in section (2.2.1), we know that if we find a “candidate” a paththat satisfies conditions [H1], [H2], [F1], [F2], [M], and [TVC], then we have found a C.E.path.

4.3 Solving for the Equilibrium Path

Let us assume the following functional forms:

U(ct, 1− ht) = ln(ct) + ηln(1− ht)F (kt, ht, t) = Akαt ((1 + g)tht)

1−α

Substituting them into the system of equations that defines the equilibrium, we get:

[H1]:η

1− ht=wtct

[H2]:1

ct=

β

ct+1

(rt+1 + 1− δ)

[F1]: rt = Aα(1 + g)t(1−α)

(htkt

)1−α

[F2]: wt = A(1− α)(1 + g)t(1−α)

(ktht

)α[M]: ct + kt+1(1 + n) = Akαt ((1 + g)tht)

1−α + (1− δ)kt

We conjecture that the equations solution exhibit balanced growth path behaviour in thelong run, i.e., all variables grow at constant rates, although not necessarily the same.

Let these long run growth rates of c, h, k, r, w, and y be denoted by 1 + γc, 1 + γh, 1 +γk, 1 + γr, 1 + γw, and 1 + γy.



by Jorge Rojas

Proposition 3. It must be the case that:

1 + γc = 1 + γk = 1 + γw = 1 + γy = 1 + g and γr = γh = 0

To prove this proposition we just need to work with the set of equations [H1]-[M], andfind the relationships for xt+1

xtthat is the growth rate of the variable xt.

Let us define the variables:

ct =ct

(1 + g)t, kt =

kt(1 + g)t

, wt =wt

(1 + g)t, ht = ht, rt = rt

So, we get:

[H1] :η

1− ht=wt��

��(1 + g)t

ct��(1 + g)t

=wtct

[H2] :ct+1(1 + g)�t+1

βct��(1 + g)t

=ct+1(1 + g)

βct= rt+1 + 1− δ

[F1] : rt = Aα

(ht

kt

)1−α

[F2] : wt = A(1− α)

(kt

ht

)α

[M] : ct + kt(1 + g)(1 + n) = Akαt h1−αt + (1− δ)kt

Proposition 4. kt cannot grow unboundedly.

Proof. By contradiction.Suppose kt →∞. We know from the utility function that ht < 1, so by [F1] we have thatrt → 0 as kt →∞. From [H2], we have that:

ct+1

ct→t→∞

(1− δ)β1 + g

< 1

taking [M] and dividing it by kt(1 + g)(1 + n), we get:

��

��:0ct

(1 + g)(1 + n)kt+kt+1

kt=��

��

��:0

A

(1 + g)(1 + n)

(ht

kt

)1−α

+(1− δ)

(1 + g)(1 + n)

Therefore,kt+1

kt< 1

This implies that kt is decreasing through time, so kt cannot go to∞, and hence kt cannotgrow unboundedly.

Notice that not growing unboundedly is not the same as converging to a Steady State(SS).

We can find the analytical solution for our economy by solving the system [H1]-[M] interms of css, wss, rss, kss, hss.



by Jorge Rojas

Proposition 5. Characterizing the equilibrium.

1) ∃ a unique equilibrium for this model (matter of faith, we won’t prove it)

2) FOC’s and TVC are sufficient conditions for being an equilibrium. We proved it intheorem (2) section (2.2.1)

Notice that our variables will not exhibit “jumps” because of consumption smoothing.

Lemma 2. If k0 = kss, then kt = kBGPt = (1 + g)tkss

Lemma 3. If k0 6= kss, then kt → kBGPt

Figure 1: Converging to the BGP.

Figure (1) shows that there are different transition paths to the BGP for differentinitial conditions. However, every transition path converges to the BGP.

Figure 2: Gross Domestic Product for the US between 1890 and 2010.



by Jorge Rojas

Assumption 1. From figure (2), we can see that for the post-war data (i.e. ignoring1940-1950) the slope of log-GDP is quite stable. Thus, in order to identify the param-eters of the model, we assume that the post-war data can be well-represented by BGPequilibrium.

We usually have parameters given to us. We can solve for empirical moments alongthe BGP.

kBGPt

yBGPt

=kss��

��(1 + g)t

A(kBGPt )α(��(1 + g)thBGPt ))1−α

kBGPt

yBGPt

=1

A

(kss

hss

)1−α

Similarly, we get:

xBGPt

yBGPt

=1

A[(1 + n)(1 + g)− (1− δ)]

(kss

hss

)1−α

We solve for the other moments as in homework 2, question 4.

4.4 Solving NGM using the Sequential Approach

Consider the Neoclassical Growth Model with technical change, but no utility from leisure.Time is discrete t = 0, 1, 2, . . . Population dynamics is given by Nt+1 = Nt(1 + n), N0 isknown. Each household member is endowed with 1 unit of productive time that can useto earn the market wage wt per unit of time. The representative household’s preferencesare described by

∑∞t=0 β

tNt logct. Each household member is also endowed with k0 unitsof capital. Every period, household members rent their capital to the firm at the rate rt.Capital depreciates at rate δ. There is a competitive production sector, with technologygiven by F (Kt, Ht, t) = Kθ

t ((1 + g)tHt)1−θ, where K and H stand for capital and labour

inputs.

(a) Define the competitive equilibrium in this economy, clearly stating the representativehousehold’s and firm’s maximization problems and market clearing conditions.

(b) Derive conditions that fully characterize the competitive equilibrium. Derive bal-anced growth rates of ct, wt, kt, yt, and rt.

Solution part (a).The representative firm takes the sequences for the interest rate and the wage, {rt, wt}∞t=0,as given. The firm’s problem is to maximise profits subject to its technology. This iswritten as follows:

Max{Yt,Kt,Ht}∞t=0

∞∑t=0

Yt − rtKt − wtHt

s.t.

Yt = F (Kt, Ht, t) = Kθt ((1 + g)tHt)

1−θ ∀t ≥ 0

Yt, Kt, Ht ≥ 0

Plugging the technology into the objective function, we get the equivalent maximisationproblem:

Max{Kt,Ht}∞t=0

Π =∞∑t=0

Kθt ((1 + g)tHt)

1−θ − rtKt − wtHt (24)



by Jorge Rojas

We assume that there are interior solutions, so we can ignore the non-negativity cons-traints.

The representative household takes as given the sequences for the interest rate {rt}∞t=0

and the wage {wt}∞t=0, as well as the endowed capital k0. The household wants to maximisehis/her lifetime utility subject to the budget constraint given by:

ct + it = wt + rtkt

and

it = kt+1 − (1− δ)kt⇒ ct + kt+1 − (1− δ)kt = wt + rtkt

ct + kt+1 = wt + rtkt + (1− δ)ktct + kt+1 = wt + (rt + 1− δ)kt

So, for the whole economy this is translated into:

ctNt + kt+1Nt+1 = wthtNt + (rt + 1− δ)ktNt

Therefore, the maximisation problem that the representative household solves is givenby:

Max{ct,kt+1,ht}∞t=0

∞∑t=0

βtNt logct

s.t.

ct + (1 + n)kt+1 = wtht + (rt + 1− δ)kt ∀tct, kt ≥ 0, 0 ≤ ht ≤ 1 ∀t

There are some facts, that are worthwhile mentioning. First, there is no leisure in theutility function. Hence, ht = 1 for all t. The other fact is that the utility function satisfiesthe Inada conditions, so the non-negativity constraint for consumption will not bebinding. Finally, kt at time 0 is given by k0, and capital cannot be negative. Although,we allow for “disinvestment”. Thus, we can rewrite the problem for the household as:

Max{ct,kt+1}∞t=0

∞∑t=0

βtNt logct (25)

s.t.

ct + (1 + n)kt+1 = wt + (rt + 1− δ)kt ∀tct, kt+1 > 0, ∀tk0 given

So, the maximization problems for the firm and the household are given by (24) and (25),respectively.

The competitive equilibrium, in general, is given by the sequences:

{ct, kSt+1, hSt , Nt+1, K

Dt , H

Dt , rt, wt}∞t=0

where D is for demand and S is for supply, such that:

(i) given {rt, wt}∞t=0, the sequences {ct, kSt+1, hSt }∞t=0 solve the household problem ∀t,



by Jorge Rojas

(ii) given {rt, wt}∞t=0, the sequences {KDt , H

Dt }∞t=0 solve the firm’s problem ∀t,

(iii) Population dynamics is given by Nt+1 = Nt(1 + n).

(iv) all markets clear ∀t:

Labor market: hSt Nt = HDt

Capital market: kSt Nt = KDt

Goods market: ctNt + KSt+1 = F (KD

t , HDt , t) + (1− δ)KS

t

ctNt + KSt+1 − (1− δ)KS

t = F (KDt , H

Dt , t)

ctNt + ISt = F (KDt , H

Dt , t)

The last equation above states that the total output of the economy is equal to thetotal consumption by the households plus the total investment from the householdsin the economy.

Solution part (b).Let us start with the representative firm. The FOC’s for this problem are given by:

∂Π

∂Kt

= θ((1 + g)tHt)1−θKθ−1

t − rt = 0

⇒ θ

((1 + g)tHt

Kt

)1−θ

− rt = 0

⇒ rt = θ

((1 + g)t

(Ht

Kt

))1−θ

⇒ rt = θ

((1 + g)t

(1

kt

))1−θ

⇒ rt = θ(1 + g)t(1−θ)kθ−1t

∂Π

∂Ht

= (1− θ)(1 + g)tKθt ((1 + g)tHt)

−θ − wt = 0

⇒ (1− θ)(1 + g)t(Kt

Ht

)θ(1 + g)−θt − wt = 0

⇒ wt = (1− θ)(1 + g)t(1−θ)kθt

For the household we will have three FOC’s (ct, kt+1, ht). Let us first write theLagrangian:

L =∞∑t=0

βtNt logct − λt (wt + (rt + 1− δ)kt − ct − (1 + n)kt+1)

Thus,

∂L∂ct

: βtNt

ct− λt = 0

⇒ λt =βtNt

ct∂L∂kt

: −λt(rt + 1− δ) + λt−1(1 + n) = 0

⇒ λt(rt + 1− δ) = λt−1(1 + n)



by Jorge Rojas

Combining these two equations, we get that:

βtNt

ct(rt + 1− δ) =

βt−1Nt−1

ct−1

(1 + n) / · 1

Nt−1

β��(1 + n)

��(1 + n)

(rt + 1− δ) =ctct−1

Therefore, the condition is:ctct−1

= β (rt + 1− δ)

We already know that ht = 1, since there is no utility from leisure. The other conditionsare identical to [H1] to [M] in section (4.3) with the appropriate conversions.

4.5 Calibration

We define an economy by choosing specific functional forms and assigning values to theparameters of the model. Then, we solve for the Balanced Growth Path (BGP). If we setthe model correctly, then the data counterparts should approach the BGP good enough.In order to properly define the economy, we go in reverse: Start with the feature ofthe data we want to match, then select the parameter values that would give us those“features”.


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