PROBLEM SET 1
Question 1
First we need to show existence using Weierstrass Theorem. Then we need to show that
the objective function is continuous and its domain is compact. Let
U=
T
X
β t u[f (kt ) − kt+1 ].
t=0
Since f (·) and u(·) are continuous function, then U is a continuous function. Now we
need to show that the set of sequences {kt+1 }Tt=0 ∈ S ⊆ RT++1 is compact. We first show
it is bounded. Since kt+1 ≥ 0, S is bounded below. Also, using the budget constraint
kt+1 ≤ f (kt ) − ct ≤ f (kt ) since ct ≥ 0. Therefore, S is bounded and the bounds are in
S, then it is closed. Since, U is continuous and S is compact there exist {kt+1 }Tt=0 that
maximizes U .
To prove uniqueness we assume there are two sequences that maximizes U and show
that any linear combination of those sequences yields higher value of U. Let k̃ and k̂
be sequences of capital that maximize U. Note that since the constraint set is convex,
any linear combination of those two sequences is a feasible sequence. Therefore, for any
λ ∈ [0, 1]:
U (λk̃ + (1 − λ)k̂) =
T
X
β t u[f (λk̃t + (1 − λ)k̂t ) − λk̃t+1 + (1 − λ)k̂t+1 ] ≥
t=0
T
X
β t u[λf (k̃t ) + (1 − λ)f (k̂t ) − λk̃t+1 + (1 − λ)k̂t+1 ] >
t=0
T
X
β t {λu[f (k̃t ) − k̃t+1 ] + (1 − λ)u[f (k̂t ) + k̂t+1 ]} = λU (k̃) + (1 − λ)U (k̂),
t=0
which is a contradiction.
Question 2
The problem is given by
(1)
U=
max
{kt+1 }T
t=0
T
X
β t u[f (kt ) + (1 − δ)kt − kt+1 ]
t=0
Date: Sept 6.
1
PROBLEM SET 1
2
subject to:
(i) kt+1 − (1 − δ)kt ≥ 0
(ii) kt+1 ≥ 0
(iii)
Given
k0
where constraint (i) represents the non-negative investment. The Lagrangian is
L=
T
X
t
β u[f (kt ) + (1 − δ)kt − kt+1 ] +
t=0
T
X
λt [kt+1 − (1 − δ)kt ] +
t=0
T
X
µt kt+1 .
t=0
Since we need capital to produce kt+1 will be positive for all t. If possible the social planner
would choose kT +1 = 0 but that would require kT = 0 which implies no consumption.
Therefore, kt+1 > 0 for all t, which implies µt = 0 for all t. The FONCs are:
(kt+1 )
− β t u0 [f (kt ) + (1 − δ)kt − kt+1 ] + λt − (1 − δ)λt+1 +
β t+1 u0 [f (kt+1 ) + (1 − δ)kt+1 − kt+2 ][f 0 (kt+1 ) + (1 − δ)] = 0
(kT +1 )
− β T u0 [f (kT ) + (1 − δ)kT − kT +1 ] + λT ,
plus the conditions related to λt . Note that since marginal utility is positive, λ∗T > 0,
which implies that the investment in time T is zero and that household will consume all
the production in time T . However, the remaining capital is going to be wasted next period.
Overall consumption will be lower since the non-negative constraint is more restrictive than
only kt+1 .
Question 3
The problem is
(2)
U=
max
{ct ,bt+1 }T
t=0
T
X
β t ln(ct )
t=0
subject to:
(i)
ct + bt+1 ≤ ȳ + τt + (1 + r)bt
(ii)
ct ≥ 0
(iii)
Given
(iv)
bT +1 = 0.
The FONCs for this problem are:
b0 = 0, ȳ
and {τt }Tt=0
PROBLEM SET 1
3
(i) β t /ct − λt = 0
(ii)
− λt + (1 + r)λt+1 = 0 ∀t 6= T
(iii) φT bT +1 = 0
(3)
(iv) φT − λT = 0
(v) λt ≥ 0;
ct + bt+1 = ȳ + τt + (1 + r)bt
where we used Inada conditions to get rid of the multiplier for constraint (ii) and the
fact that utility is increasing to set the budget constraint in equality. Then we have the
Euler equation, ct+1 = (1 + r)βct , and solving recursively we find the intertemporal budget
constraint
t
t
T T X
X
1
1
ct =
(ȳ + τt ) + (1 + r)b0 .
1+r
1+r
t=0
t=0
Let b0 = 0 and use the Euler equation to find ct = β t (1 + r)t c0 which yields
t
t X
t
T T T X
X
1
1
1
t
t
β (1 + r) c0 = ȳ
+
τt
1+r
1+r
1+r
t=0
t=0
T
X
T X
t=0
T X
t
1
1
+
τt
c0
β = ȳ
1+r
1+r
t=0
t=0
t=0
" T t X
t #
T X
1
1−β
1
ȳ
c0 =
+
τt ,
1 − β T +1
1+r
1+r
t
t
t=0
t=0
which implies
" T t X
t #
T X
1
−
β
1
1
c∗t = β t (1 + r)t
ȳ
+
τt .
1 − β T +1
1+r
1+r
t=0
t=0
Note that
∂c∗t /∂ ȳ
∂c∗t /∂τt
PT
=
t=0
1
1+r
t
t
> 1.
1
1+r
Thus, we have that an increase in permanent income has a larger effect on data t consumption than an increase in the transitory component.
Question 4
The Lagrangian for this problem is
L=
T
X
t=0
that yields the following FONCs:
β
t
ln(Aktα
− kt+1 ) +
T
X
t=0
λt kt+1 ,
PROBLEM SET 1
4
1
1
+ µt = 0
+ β t+1 α
Aktα − kt+1
Akt+1 − kt+2
1
(kT +1 ) − β T
+ µT = 0
α
AkT − kT +1
(µt ) kt+1 ≥ 0; µt kt+1 = 0
Since marginal utility of consumption goes to infinity when consumption goes to zero, it is
not optimal choose zero capital for any period but T + 1. Therefore µ∗t = 0 for all t 6= T .
Then we have the Euler equation:
(kt+1 )
− βt
α−1
Aαkt+1
1
,
=β α
Aktα − kt+1
Akt+1 − kt+2
which can be simplified to
kt+2
1−
= αβ
α
Akt+1
Aktα
−1 .
kt+1
α
such that
Now let’s say zt = kt /Akt−1
αβ
.
zt
= 0 since kT +1 = 0, we have
zt+1 = 1 + αβ −
Solving recursively noticing that zT +1
zt =
αβ + . . . + (αβ)T −t+1
St
αβ − (αβ)T −t+2
=
,
=
1 + αβ + . . . + (αβ)T −t+1
1 + St
1 − (αβ)T −t+2
since St = (αβ − (αβ)T −t+2 )/(1 − αβ). Then solving for capital we have:
kt =
t
koα
t−1
Y
s=0
A
αs
α
t Y
αβ − (αβ)T −s+2
s=1
1 − (αβ)T −s+2
t−s
.