Shocks on the Determination of Current
Account of Two-Period Small Open Economy
KS LIAN∗
December 17, 2003
1
The Model
max log C1 + β log C2
Q2
C2
= (1 + r0 ) B0∗ + Q1 +
s.t C1 +
1 + r1
1 + r1
∗
r1 = r
1.1
Lagrange function
L = log C1 + β log C2 + λ (1 +
1.2
r0 ) B0∗
C2
Q2
− C1 −
+ Q1 +
1 + r1
1 + r1
First Order Condition
1
=λ
C1
β
λ
=
C2
1 + r1
C2
Q2
C1 +
= (1 + r0 ) B0∗ + Q1 +
1 + r1
1 + r1
∗
graduate student,Department of International Trade,NCCU, [email protected]
1
1.3
the collection of equations
C2 = β (1 + r1 ) C1
Hp budget constraint ª)
C1∗
1
Q2
=
(1 + r0 ) B0∗ + Q1 +
1+β
1 + r1
C2∗
Q2
β (1 + r1 )
(1 + r0 ) B0∗ + Q1 +
=
1+β
1 + r1
(1)
T B1 = Q1 − C1 − I1 =
CA1 = T B1 + r0 B0∗ =
(2)
1
Q2
β
Q1 −
(1 + r0 ) B0∗ +
1+β
1+β
1 + r1
(3)
β
1 − βr0 ∗
1
1
Q1 −
B0 −
Q2
1+β
1+β
1 + r1 1 + β
(4)
1
β (1 + r1 )
Q2 −
[(1 + r0 ) B0∗ + Q1 ]
1+β
1+β
(5)
T B2 = Q2 − C2 − I2 =
To derive CA2
CA2 = T B2 + r1 B1∗
By C1 + B1∗ − B0∗ = r0 B0∗ + Q1
⇒ B1∗ = (1 + r0 ) B0∗ + Q1 − C1 =
β
β
1
1
Q1 −
(1 + r0 ) B0∗ −
Q2
1+β
1+β
1 + r1 1 + β
hence, we have
1
Q
1+β 2
1
1
Q2
− 1+r
1 1+β
CA2 =
−
β(1+r1 )
1+β
[(1 + r0 ) B0∗ + Q1 ] +
β
Q
1+β 1
−
β
1+β
(1 + r0 ) B0∗
=⇒
CA2 =
β
1
β
β
Q2 −
Q1 −
(1 + r0 ) B0∗
1 + β 1 + r1
1+β
1+β
2
(6)
1.4
current account and trade balance
Initially, β = β0 , r1 = r∗ , Q1 = Q01 , Q2 = Q02
=⇒
β0
1
Q02
=
Q01 −
(1 + r0 ) B0∗ +
1 + β0
1 + β0
1 + r∗
"
T B10
#
(7)
T B20 =
i
1
β0 (1 + r∗ ) h
Q02 −
(1 + r0 ) B0∗ + Q01
1 + β0
1 + β0
(8)
CA01 =
1 − β0 r0 ∗
1
1
β0
Q01 −
B0 −
Q0
1 + β0
1 + β0
1 + r∗ 1 + β0 2
(9)
CA02 =
β0
1
β0
β0
Q0 −
Q01 −
(1 + r0 ) B0∗
∗ 2
1 + β0 1 + r1
1 + β0
1 + β0
3
(10)
2
temporary output shock
Other things being equal,let Q21 = Q01 − ∆, Q22 = Q02 ,plug into (7)∼(10)
β0 0
1
Q02
β0
∗
=
Q1 − ∆ −
(1 + r0 ) B0 +
= T B10 −
∆ < T B10
∗
1 + β0
1 + β0
1+r
1 + β0
"
T B12
#
T B22 =
i
β0 (1 + r∗ )
1
β0 (1 + r∗ ) h
Q02 −
(1 + r0 ) B0∗ + Q01 − ∆ = T B20 +
∆ > T B22
1 + β0
1 + β0
1 + β0
CA21 =
β0 0
β0
1 − β0 r0 ∗
1
1
0
0
−
=
CA
Q1 − ∆ −
B0 −
Q
∆ < CA01
1
2
∗
1 + β0
1 + β0
1 + r 1 + β0
1 + β0
CA22 =
β0
1
β0 0
β0
0
Q
−
Q
−
∆
−
(1 + r0 ) B0∗ > CA02
2
1
∗
1 + β0 1 + r1
1 + β0
1 + β0
4
3
permanent output shock
Other things being equal, let Q21 = Q01 − ∆, Q22 = Q02 − ∆, plug into(7)∼(10)
T B13 = T B10 −
T B23
=
T B20
β0
r∗
∆ < T B10
1 + β0 1 + r∗
1 − β0 (1 + r∗ )
−
∆
1 + β0
CA31 = CA01 −
1
1
[β0 (1 + r∗ ) − 1] ∆
∗
1 + β0 1 + r
CA32 = CA02 −
1
1
[1 − β0 (1 + r∗ )] ∆
∗
1 + β0 1 + r
5
4
IRP shock
Other things being equal, let r1 = r∗ + ε, plug into(7)∼(10)
β0
1
Q02
0
∗
=
Q1 −
(1 + r0 ) B0 +
> T B10
∗
1 + β0
1 + β0
1+r +ε
"
T B14
#
T B24 =
i
1
β0 (1 + r∗ + ε) h
Q02 −
(1 + r0 ) B0∗ + Q01 < T B20
1 + β0
1 + β0
CA41 =
β0
1 − β0 r0 ∗
1
1
Q01 −
B0 −
Q02 > CA01
∗
1 + β0
1 + β0
1 + r + ε 1 + β0
CA42 =
β0
1
β0
β0
Q02 −
Q01 −
(1 + r0 ) B0∗ < CA02
∗
1 + β0 1 + r + ε
1 + β0
1 + β0
6
5
preference shock: β ↓
β = β1 < β0 ⇒ RßÛÊí¾‘
By
M RS =
1/C1
C2
U1
=
=
U2
β/C2
βC1
β ↓⇒ M RS ↑⇒ Ìæ(‰¢ ⇒ ~õ²‡ (C1 j²) 
Other things being equal, let β = β1 < β0 , plug into(7)∼(10)
∂T B1
1
Q2
1
∗
=
>0
2 Q1 +
2 (1 + r0 ) B0 +
∂β
1 + r∗
(1 + β)
(1 + β)
β ↑⇒ T B1 ↑⇔ β ↓⇒ T B1 ↓
−1
∂T B2
1
=
(1 + r∗ ) [(1 + r0 ) B0∗ + Q1 ] < 0
2 Q2 −
∂β
(1 + β)
(1 + β)2
β ↑⇒ T B2 ↓⇔ β ↓⇒ T B2 ↑
1
1
1
∂CA1
1
∗
=
Q2 > 0
2 Q1 +
2 (1 + r0 ) B0 +
∂β
1 + r∗ (1 + β)2
(1 + β)
(1 + β)
β ↑⇒ CA1 ↑⇔ β ↓⇒ CA1 ↓
∂CA2
−1
1
1
1
∗
=
Q2 −
Q1 < 0
2
2 (1 + r0 ) B0 −
∗
∂β
(1 + β) 1 + r
(1 + β)
(1 + β)2
β ↑⇒ CA2 ↓⇔ β ↓⇒ CA2 ↑
7