对外经贸高级宏观经济学09~10期末试卷+答案

c ˙ 1 = (r − ρ ). c θ 3 1 0<ε <1 lim r = Aba1/ε − δ lim r = ∞,
k→∞
k→0
Aba1/ε > δ + ρ ,
c ˙ 1 = (Aba1/ε − δ − ρ ). c θ Aba1/ε < δ + ρ , 0 4 1
ε <0
lim r = Aba1/ε − δ
α ln(kT ) 1 − αβ
V1 (kT −1 ) V2 (kT −2 ) Vn (kT −n ) 10 10 10 10
cT −1 cT −2 cT −n Bellman
v0 (kT )
Bellman
v1 (kT −1 )
T −1
T
v1 (kT −1 ) = max = ln(cT −1 ) + β
T v2 (kT −2 )
v2 (kT −2 ) = max ln(cT −2 ) + β [ln(1 − αβ ) +
c∈Γ(k)
αβ α ln(αβ ) + ln(kT −1 )] , 1 − αβ 1 − αβ
α kT −1 = αβ kT −2
α cT −2 = (1 − αβ )kT −2
2009 — 2010
1
CES Y = F (K , L) = A[a(bK )ε + (1 − a)((1 − b)L)ε ]1/ε . 1 2 3 4 n Euler 0<ε <1 10 5 1 1 0 10 1/(1 − ε ) 5
δ
CHale Waihona Puke BaiduRA
θ
ε <0
1
MPKL =
∂ F /∂ L (1 − a)(1 − b)ε L ε −1 = ( ) , ∂ F /∂ K abε K
−
∂ (L/K )/(L/K ) MPKL/(L/K ) 1 =− =− . ∂ MPKL/MPKL ∂ MPKL/∂ (L/K ) ε −1
1
2 2 y = F (K , L)/L = A[a(bk)ε + (1 − a)(1 − b)ε ]1/ε ,
r = dy/dk − δ = Aabε [abε + (1 − a)(1 − b)ε k−ε ](1−ε )/ε − δ .
k→∞
lim r = −δ
k→0
Aba1/ε > δ + ρ , 0 Aba1/ε < δ + ρ ,
3 2 ln(ct )
β ∈ (0, 1)
V (k0 ) = max
t
V
β t u(ct ) ∑ c ∈Γ(k )
t
∞
t =0
s.t . kt +1 = ktα − ct
V0 (kT ) = 1 2 3 4 T −1 T −2 T −n V (kt )
v2 (kT −2 ) = (1 + β )[ln(1 − αβ ) +
αβ α ln(αβ )] + ln(kT −2 ), 1 − αβ 1 − αβ
4 n vn (kT −n ) = T n
αβ α 1 − β n+1 [ln(1 − αβ ) + ln(αβ )] + ln(kT −n ), 1−β 1 − αβ 1 − αβ
∞
1
5 t c ¯|odd = 1 β c ¯|even = . 1+r 1+β
1 0 c ¯|even 2 3 c ¯|even
0 1 0
0
β /(1 + β ) β (1 + r)/(1 + β ) = 1/(1 + β ) c ¯|even
c∈Γ(k)
α ln(kT ) , 1 − αβ
α kT = kT −1 − cT −1
Bellman
α cT −1 = (1 − αβ )kT −1 α kT = αβ kT −1 ,
Bellman v1 (kT −1 ) = ln(1 − αβ ) + T −2
αβ α ln(αβ ) + ln(kT −1 ). 1 − αβ 1 − αβ
ct = (1 − αβ )ktα [0, ∞)
kt +1 = αβ ktα .
3 0 0 r 1 t 1 t
yt
β
1 2 3 10
β (1 + r) = 1 t t t t
0 t t 0 Euler 1 10 0 10
Euler
t t c ¯|even = r 1+r
0
j=0
∑ (1 + r)2 j = 1 + β .
T −n = t 1 αβ α [ln(1 − αβ ) + ln(αβ )] + ln(kt ). 1−β 1 − αβ 1 − αβ
v(kt ) = lim vn (kT −n ) =
n→∞
Bellman Bellman v(kt ) = max ln(ct ) +
c∈Γ(k)
β αβ αβ [ln(1 − αβ ) + ln(αβ )] + ln(kt +1 ) . 1−β 1 − αβ 1 − αβ