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    40 -

APPENDIX

From the rest of the world (U.S.).

(59)

β +β US1 US1 l o g 5 1 * US1 1 4 1 1 1 0 + + U S U S U S Y β β β β US1 β adist β US1 US1

l o g * 1 1 2 1 + U S U S H K R β β

log HK

* US

  • +

    β US1

β US1

log Pat

* US1

3 US1 h β

log S

µ p1 = 0

* h1

3 1 + m U S US1 β β

log S

* m1

(60) 1 1 1 U S U S β β

1 U S b d i s t

1 1 2 1 + U S β β

ρHKUS

1 3 1 + U S h U S β β

1 S h µ

1 3 1 + U S m U S β β

1 4 1 1 + U S U S m β β µ

1 y µ

1 5 1 + U S U S β β

ρHKR

0 1 1 = p U S µ β

Calculations for the solution of system (49)-(60) are lengthy and tedious. Here, we report the results with the specification that the above system is composed of two blocks of equations, one of which is independent of the initial levels. From this block steady state rates of growth are derived and this solution is used to solve for initial levels.

Solutions for steady state rates of growth:

(61)

( ) ( ) ( ) [ { ( ) ( ) ( + + + + + + + + + 1sh 1 + + + + + 1 1sh + = 1 1sm 1 I C T 1 s m i S m 1 I 1 C T s 1 h i S 1sm 1 h H K R i 1 2sh H K i i i 1 2sm s m i s h s m s h I C T s m S m s h S h L K s m s h i s m i s h i i s m i s h s m s h y b d i s 1 t ρ γ β ρ 1 γ β ρ β ρ β β γ β γ 1 2sh β β α α α 1 2sm ρ γ α γ α ρ α ρ α α α γ β γ ρ ICT β β γ β γ β γ α γ α α µ 1 4 1 3 1 4 1 3 1 5 1 2 1 1 1 1 1 3 1 3 1 1 2 1 2 1 1 × 1 4 2 1 4 2 1 4 1 1 3 ) 2 2 1 3 1 3 1 4 1 1 3 1 3 2 2 1 1 1 ) ( ( 1 1

] } β 1 1 1 )

(62) µ Sh1

=γ

1 1Sh

µy1 +γ

1 4Sh

ρ ICT1

1 1 2 p S h µ γ +

(62’) µSm1

=γ

1 1Sm

µy1 + γ

1 4Sm

ρ ICT1

+γ

1 2Sm

µ p1

(63)

( [ 1 1 5 1 2 1 1 1 1 3 2 1 3 1 1 H K R i H K i i i s m s h i s h 1 p b d i s t ρ β ρ β β γ β γ β µ + + = 1 2sm )1

1 3 i γ β +

i1 4Sh

ρ ICT 1

1 3 i γ β +

i1 4Sm

ρ ICT 1

( 1 3 1 4 1 i s h i y γ β β µ + +

1 sh1

1 3 i s m γ β +

1 sm1

)]

where

n

n

dist

i1

i H K ρ

dist1 =

i =1

, ρ HK

=

i =1

n

n

Solutions for initial levels:

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