ECON 4310
Fall 2009
Problem set 1
ECON 4310
Fall 2008
Espen Henriksen
Problem set 3
Due next seminar.
1. Theory
Verify the form of the true value function: Consider a model economy
where the social planner chooses an infinite sequence of consumption
and next period’s capital stock {ct, kt+1}∞
t=0 in order to
max
{ct,kt+1}∞
t=0
∞
t=0
βt
u (ct)
subject to
ct + kt+1 ≤ yt, ∀t
ct, kt ≥ 0, ∀t
k0 > 0. given
Assume the following functional forms
u (ct) = ln ct, ∀t, σ > 0
yt = F (kt, 1) = γkα
t , ∀t, α ∈ (0, 1)
Reformulate this optimization problem as a dynamic programming
problem and write up the Bellman equation. Verify that the value
function solving the functional equation (i.e. the Bellman equation) is
of the following form
v (k) = a + b ln k.
Find a and b as functions of the model economy’s structural parame-
ters.
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1
(a) the investment/capital ratio2
(b) the capital/labor ratio3
(c) the capital/output ratio
(d) the factor prices
(e) capital’s and labor’s shares of national income
(f) the investment/output ratio
(g) the consumption/output ratio
2. Computations
Consider a model economy where the social planner chooses an infinite
sequence of consumption and next period’s capital stock {ct, kt+1}∞
t=0
in order to
max
{ct,kt+1}∞
t=0
∞
t=0
βt
u (ct)
subject to
ct + kt+1 ≤ f (kt) + (1 − δ) kt, ∀t
ct, kt ≥ 0, ∀t
k0 > 0. given
Assume the following functional forms
u (ct) =
c1−σ
t − 1
1 − σ
, ∀t, σ > 0
yt = γkα
t , ∀t, α ∈ (0, 1)
where α = .35, β = .98, δ = .025, σ = 2, and γ = 5.
As we have derived in class we know that we can rewrite this as a
recursive problem and that the Bellman equation is
v (kt) = max
kt+1
{u (kt, kt+1) + βv (kt+1)} .
In order to compute the stationary value function you choose to use
discrete value function iteration. The capital stock can take three discrete
values; k ∈ k(1), k(2), k(3) = {2.85, 3.00, 3.15} . That means
that v (kt) and v (kt+1) are 3×1 vectors (Figure 1) whereas u (kt, kt+1)
is a 3 × 3 matrix (Figure 2).
2
Hint: use the law of motion for capital.
3
Hint: use the intertemporal optimality condition, the Euler equation.
2
2
(a) Compute a (3 × 3) dimensional consumption matrix C (i, j) with
the value of consumption for all the (3 × 3) values of kt and kt+1.
Then compute a (3 × 3)-dimensional matrix with the utility of
consumption for all the (3 × 3) values of kt and kt+1 similar to
the one in Figure 2.
(b) Assume
v (kt+1) =


167.6
168.1
168.6

 .
Before you maximize {u (kt, kt+1) + βv (kt+1)} you need to compute
the sum of two elements u (kt, kt+1) and βv (kt+1). But since
u (kt, kt+1) is a 3×3 matrix whereas v (kt+1) is a 3×1 vector you
need to transform v (kt+1) into a 3 × 3 matrix. Note that v (kt+1)
is independent of kt. The resulting matrix should therefore be
like the matrix represented in Figure 3.
Hint: To transpose a matrix in Matlab you simply use ′.
(c) Now that you have {u (kt, kt+1) + βv (kt+1)} you can compute
v (kt);
v (kt) = max
kt+1
{u (kt, kt+1) + βv (kt+1)} .
Hint: lookup help for max, i.e. type help max.
3. Measurement
Compute annual real interest rate for the period 1979 to 2006 using
annualized 3M NIBOR rates from Norges Bank
http://www.norges-bank.no/Pages/Article____41851.aspx
and CPI series from Statistics Norway
http://statbank.ssb.no//statistikkbanken/default_fr.asp?PLanguage=1
3
3
Appendix: Figures
kt
v k
(1)
t
v k
(2)
t
v k
(3)
t
(a) v (kt)
kt+1
v k
(1)
t+1
v k
(2)
t+1
v k
(3)
t+1
(b) v (kt+1)
Figure 1: v (kt) and v (kt+1)
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kt+1
kt
u k
(1)
t , k
(1)
t+1
u k
(2)
t , k
(1)
t+1
u k
(3)
t , k
(1)
t+1
u k
(1)
t , k
(2)
t+1
u k
(2)
t , k
(2)
t+1
u k
(3)
t , k
(2)
t+1
u k
(1)
t , k
(3)
t+1
u k
(2)
t , k
(3)
t+1
u k
(3)
t , k
(3)
t+1
Figure 2: u (kt, kt+1)
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kt+1
kt
u k
(1)
t , k
(1)
t+1 + βv k
(1)
t+1
u k
(2)
t , k
(1)
t+1 + βv k
(1)
t+1
u k
(3)
t , k
(1)
t+1 + βv k
(1)
t+1
u k
(1)
t , k
(2)
t+1 + βv k
(2)
t+1
u k
(2)
t , k
(2)
t+1 + βv k
(2)
t+1
u k
(3)
t , k
(2)
t+1 + βv k
(2)
t+1
u k
(1)
t , k
(3)
t+1 + βv k
(3)
t+1
u k
(2)
t , k
(3)
t+1 + βv k
(3)
t+1
u k
(3)
t , k
(3)
t+1 + βv k
(3)
t+1
Figure 3: u (kt, kt+1) + βv (kt+1)
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