Exercices : Série 1

y0
yTT
g
r g = 2 ⇔r= 0.02
y0(1 + r)T=yT
y0(1 + r)T= 2y0
(1 + r)T= 2
T ln(1 + r) = ln 2
ln (1 + r)
r r
rT ≈ln 2
rT ≈0.07
T≈0.07
r=70 g.
y0erT = 2y0
erT = 2
rT =ln 2
T=ln 2
r
T≈0.07
r=70 g.

Yt=F(Kt, Lt) = K0.5
tL0.5 t
λYt=F(λKt, λLt), λ > 0
λF (Kt, Lt) = (λKt)0.5(λLt)0.5
λF (Kt, Lt) = λ0.5+0.5K0.5
tL0.5 t
λF (Kt, Lt) = λK0.5
tL0.5 t.
→
λY > F (λK, λL), λ > 0
λY < F (λK, λL), λ > 0
K
P MKt=∂F (Kt, Lt)
∂Kt
= 0.5K−0.5
tL0.5 t
∂2F(Kt, Lt)
∂K2
t
=∂P MKt
∂Kt
= (−0.5)0.5K−1.5
tL0.5 t<0.
L
P MLt=∂F (Kt, Lt)
∂Lt
= 0.5K0.5
tL−0.5 t
∂2F(Kt, Lt)
∂L2
t
=∂P MLt
∂Lt
= (−0.5)0.5K0.5
tL−1.5 t<0.
→

L
F(Kt, Lt)
Lt
=K0.5
tL0.5 t
Lt
=K0.5
tL−0.5 t=Kt
Lt0.5
f(kt) = yt=k0.5
t
n= 0 g= 0
δ= 5% ≡0.05 sA= 10% ≡0.1
sB= 20% ≡0.2A B
kt+1 −kt=1
(1 + n)[skα
t−(n+δ)kt]
k(kt+1 =kt)
sk∗α−(n+δ)k∗= 0
⇒k∗=s
n+δ1
1−α
.
k∗
A=0.1
0.051
1−0.5
= 4.
k∗
B=0.2
0.051
1−0.5
= 16.
y∗
A=k∗α
A= 40.5= 2.
y∗
B=k∗α B= 160.5= 4.
c∗
A= (1 −sA)y∗
A= (1 −0.1)2 = 1.8
c∗
B= (1 −sB)y∗ B= (1 −0.2)4 = 3.2.

(n+g+δ)˜ k(n0+g+δ)˜ k
˜
k∗→˜ k0∗
˜y∗→˜y0∗
∗
∗
∗
∗
ln
  ′   + croissance transitoire (n'=0)
ln
     transitoire
ln
 0 
croissancetransitoire
t0
Ytg+n
g+n0

n+g n n0
Yt=F(Kt, AtLt) = Kα
t(AtLt)1−α
α= 0.3δ= 4% ≡0.04 n+g= 3% ≡0.03
K∗
t/Y∗
t= 2.5
LtAtLtk∗
t/y∗
t= 2.5˜
k∗
/˜y∗= 2.5
s˜y∗= (n+g+δ)˜ k∗
s= (n+g+δ)˜
k∗
˜y∗.
sES = (0.03 + 0.04)2.5 = 0.175.
P MKt=∂F (Kt, AtLt)
∂Kt
=αKα−1
t(AtLt)1−α
Kt
P MK∗
ES =αKα−1
t(AtLt)1−αKt
Kt
=αYt
Kt
= 0.3·2.5−1= 0.12.
˜
k∗˜c∗
max ˜c∗= (1 −s)f(˜ k∗)
t.q. sf(˜ k∗) = (n+g+δ)˜ k∗.
˜c∗=f(˜ k∗)−(n+g+δ)˜ k∗.
∂˜c∗
∂˜ k∗=f0(˜
k∗)−(n+g+δ)=0
α˜ k∗(α−1) = (n+g+δ)