Desktop version

Home arrow Business & Finance arrow Financial Liberalisation: Past, Present and Future

Appendix

Deriving the Linearized Budget Constraint

First, to derive Eq. (8.5), start with:

then note that Rt, 1 = 1, factor out CR and NOt on the LHS and RHS, respectively:

Then divide the whole expression by NOt

Now recall that 1/ (+1 ( + r^j and take logs and exponential in the infinite summations:

Applying the log to the interest rate factorial, it is easy to see that:

We can then add and subtract infinitely many lnCf s and lnNOt’s:

which is Eq. (8.6) in the main text:

To linearize this expression, we use the standard formulaf(x) = f(a) +f (x)

(x—a)

“1Г к!Г 1 "

Now note that 1 + УК = 1 + кУК = 1 + = and

_ м J L J L 1 -К L1-К-

that in the steady state, when all the hat variables take the value of 0,

да да

c 1 + ?K = 1 + ?K , then it is easy to obtain the final linearized form:

_ i=1 _ _ i=1 _

Derivation of the Current Account Reaction Function with External Habits and a Constant World Real Interest Rate

We start by linearizing the definition of aggregate consumption

C = (1 -y)CR + yCNr to obtain

Following Obstfeld and Rogoff (1996), the current account can be expressed as:

which, noting that Bt = (1 -y)Bp + BGt and r = eln(1+r) — 1, can be linearized as the following the current account to net output ratio:

CA

where cat = . Now substitute Eq. (8.13) into Eq. (8.14), then into

' NOt 1

Eq. (8.15). Simplifying and using — = 1 and (e“ — 1) = r as in Kano (2008), yields: e

The constant world real interest rate assumption implies ln(1 + rt) = 0 V i. We also add and subtract (1 — y)hcat_ 1 to obtain

Furthermore, log-linearizing the Euler equation gives

which can be expressed as

Substituting Eq. (8.17) into Eq. (8.16) gives

which corresponds to Eq. (8.8) in the main text. To further solve this expression, it is necessary to assume a stochastic process for Д1пЛ^О(. Under the assumption of Д1пЛ^О,=рД1п^О,_ 1 + et Equation (8.8) can be expressed as follows:

да

where we replaced -vc) ~y){1 ~)( - Et-1))lnNOt+i with

i=0

hK(1~y)(( ~r) 2nd (1 _^)1 _ hKyKiEt Д ln NOt+i with

  • (l-к) i=0
  • (l -y )(l - hie) ^no . After further simplification, one can then obtain 1 - рк 1

Eq. (8.9) in the main text:

To derive Eq. (8.10) note that under the assumption that ln NO t=p ln ln NO t _ 1 + et, Д ln NO t = (p _ 1)

ln NO't _ 1 + e t and follow the steps above. In this case

 
Source
< Prev   CONTENTS   Source   Next >

Related topics