derivation 3

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Derivation and Log-linearization of Otsu (2007)s Small Open Economy Model Chen-chiao Kuo Otsu, Keisuke (2007), "A Neoclassical Analysis of the Asian Crisis: Business Cycle Accounting of a Small Open Economy," IMES Discussion Paper Series No. 07-E-16, Bank of Japan.

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  • Derivation and Log-linearization of Otsu(2007)s Small Open Economy Model

    Chen-chiao Kuo

    Otsu, Keisuke (2007), "A Neoclassical Analysis of the Asian Crisis: Business Cycle Accounting ofa Small Open Economy," IMES Discussion Paper Series No. 07-E-16, Bank of Japan.

  • 1 The benchmark prototype economy

    The representative consumer depends on an expected intertemporal discountedutility function that includes utility from consumption and disutility from labor:

    maxU = E0

    1Xt=0

    tu(ct; lt)

    The utility is:

    u(ct; lt) = log(ct lt )

  • where parameters (> 0) and (> 1) represent the level and curvature of theutility cost of labor respectively.

    subject to the budget constraint:

    wt

    ltlt+rtkt+tt+

    dt+1Rdt

    = ct+xtxt+dt+(dt+1)

    The law of motion for capital:

    kt+1 = xt + (1 )kt

  • where dene = (1 + )(1 + n), is the growth rate of labor augmentingtechnical progress and n is the population growth rate.

    The debt adjustment cost function, (dt+1), as (dt+1d)2

    2 , where d is thesteady state foreign debt.

    Solution of consumersmaximization problem:

    maxfct;lt;kt+1;dt+1g

    U = E0

    1Xt=0

    t log (ct lt )

  • wt

    ltlt + rtkt + tt +

    dt+1Rdt

    = ct + xt [kt+1 (1 )kt] + dt +(dt+1)

    The Lagrangian

    L =1Xt=0

    t log(ct lt ) +1Xt=0

    tt

    8

  • @L@ct

    = tuct tt = 0

    @L@lt

    = tult + ttwt

    lt= 0

    @L@kt+1

    = ttxt + t+1t+1[t+1 + xt+1(1 )] = 0

    @L@dt+1

    = tt

    "

    Rdt 0(dt+1)

    # t+1t+1 = 0

    1

    ct lt= t (A1)

  • l1t

    ct lt= twt

    lt

    Use Equation (A1) substitute t, and we obtain:

    l1t =wt

    lt(A2)

    txt = t+1[t+1 + xt+1(1 )]1

    ct ltxt = Etuct+1[t+1 + xt+1(1 )] (A3)

  • t

    "

    Rdt 0(dt+1)

    #= t+1

    1

    ct lt

    "

    Rdt 0(dt+1)

    #= Etuct+1 (A4)

    The rm produces a nal good (output y with a constant returns to scale) fromcapital and labor using a Cobb-Douglas production function.

    yt = ztkt l1t

    The rms prot maximization problem is:

  • = yt wtlt rtkt

    Solution of rmsmaximization problem:

    maxfkt;ltg

    = ztkt l1t wtlt rtkt

    @

    @kt= ztk

    1t l

    1t rt = 0

    @

    @lt= (1 )ztkt lt wt = 0

  • ztk1t l

    1t = rt

    yt

    kt= rt (A5)

    (1 )ztkt lt = wt

    (1 )ytlt= wt (A6)

    Substitute (A6) into (A2), and we get the labor wedge.

    l1t =wt

    lt= (1 )yt

    lt

    1

    lt

  • l1t = (1 )yt

    lt

    1

    lt

    Substitute (A5) into (A3), and we get the investment wedge.

    xt1

    ct lt= Et

    "uct+1

    yt+1kt+1

    + xt+1(1 )!#

    Substitute 0(dt+1) = (dt+1 d) into (A4), and we get the foreign debtwedge.

  • 1ct lt

    "

    R

    1

    dt (dt+1 d)

    #= Et[uct+1]

    The government collects tax revenues at date t, and the government budgetconstraint:

    tt = (1 1 lt)wtlt + (xt 1)xt

    Substitute the government budget constraint into the consumer budget con-straint and we can get the economys resource constraint at last:

  • wt

    ltlt+ rtkt+ (1 1

    lt)wtlt+ (xt 1)xt+ dt+1

    Rdt= ct+ xtxt+ dt+(dt+1)

    rtkt + wtlt xt + dt+1Rdt

    = ct + dt +(dt+1)

    rtkt + wtlt = xt + ct + dt dt+1Rdt

    +(dt+1)

    yt = ct + xt + dt dt+1Rdt

    +(dt+1 d)2

    2

    where the trade balance is:

  • tbt = dt dt+1Rdt

    +(dt+1 d)2

    2

    The equilibrium conditions in the detrended per-capita form:

    yt = ct + xt + dtdt+1Rdt

    +(dt+1 d)2

    2(1)

    yt = ztkt l1t (2)

    l1t = (1 )yt

    lt

    1

    lt(3)

  • xt1

    ct lt= Et

    "uct+1

    yt+1kt+1

    + xt+1(1 )!#

    (4)

    1

    ct lt

    "

    R

    1

    dt (dt+1 d)

    #= Et[uct+1] (5)

    Endogenous variables are : ct ,xt ,yt ,kt ,dt ,lt.

    Exogenous variables are: zt, dt, xt, and lt.

    We need the capital accumulation law to close the model.

  • kt+1 = xt + (1 )kt (6)

    2 Steady state of the prototype model

    From equation(5), the calculation process is below:

    1

    ct lt

    "

    R

    1

    dt (dt+1 d)

    #= Et[uct+1]

  • 1c l [

    R

    1

    d (d d)] = uc

    1

    d= R (ss1)

    From equation(4), the calculation process is below:

    xt1

    ct lt= Et

    "uct+1

    yt+1kt+1

    + xt+1(1 )!#

    x =

    "z

    k

    l

    1+ x(1 )

    #

  • x

    = z

    k

    l

    1+ x(1 )

    k

    l

    1=1

    z

    "x

    x(1 )

    #

    k

    l=

    (1

    z

    "x

    x(1 )

    #) 11

    (ss2)

    From equation(6), the calculation process is below:

    kt+1 = xt + (1 )kt

  • k = x+ (1 )k

    x = [ (1 )]k (ss3)

    From equation(2), the calculation process is below:

    yt = ztkt l1t

    y = zkl1 = zk

    l

    1k = z

    k

    l

    l

    From equation(1), the calculation process is below:

  • yt = ct + xt + dtdt+1Rdt

    +(dt+1 d)2

    2

    y = c+ x+ d dRd

    +(d d)2

    2

    c = y x d+ dRd

    (d d)2

    2= z

    k

    l

    1k [ (1 )]k d+ d

    Rd

    c = zk

    l

    l [ (1 )]

    k

    l

    l d+ d

    Rd

    c = zk

    l

    l [ (1 )]

    k

    l

    l d+ d

  • c = zk

    l

    l [ (1 )]

    k

    l

    l + ( 1)d (ss4)

    From equation(3), the calculation process is below:

    l1t = (1 )yt

    lt

    1

    lt

    l1 = (1 )yl

    1

    l

    l = (1 )zk

    l

    1 l

  • l = (1 ) z

    k

    l

    1 l

    l =

    "(1 ) z

    k

    l

    1 l

    #1

    (ss5)

    We obtain kl rst, and in order obtain l, then k, then c, then y, then x.

    k =k

    ll

    c = zk

    l

    l [ (1 )]

    k

    l

    l + d( 1)

  • y = zk

    l

    l

    x = [ (1 )]k

    d = d

    3 Log-linearization

    c^t = log ct log c

    x^t = log xt log x

  • y^t = log yt log y

    k^t = log kt log k

    l^t = log lt log l

    z^t = log zt log z

    d^t =dtd 1

    ^dt = log dt log d

  • ^xt = log xt log x

    ^ lt = log lt log l

    Equilibrium conditions in the detrended per-capita form:

    yt = ct + xt + dtdt+1Rdt

    +(dt+1 d)2

    2(1)

    yt = ztkt l1t (2)

  • l1t = (1 )yt

    lt

    1

    lt(3)

    xt1

    ct lt= Et

    "uct+1

    yt+1kt+1

    + xt+1(1 )!#

    (4)

    1

    ct lt

    "

    R

    1

    dt (dt+1 d)

    #= Et[uct+1] (5)

    kt+1 = xt + (1 )kt (6)

    From equation (1)

  • yt = ct + xt + dtdt+1Rdt

    +(dt+1 d)2

    2

    yey^t = cec^t + xex^t +d^t + 1

    dhd^t+1 + 1

    di

    Rde^dt

    +hd^t+1 + 1

    d d

    i22

    y(1+y^t) = c (1 + c^t)+x(1+x^t)+d^t + 1

    dd^t+1d+ d

    Rd(1 + ^dt)

    +d^t+1d+ d

    d

    22

    y + yy^t = c+ cc^t + x+ xx^t + dd^t + d d^t+1d+ dRd(1 + ^dt)

    +d^t+1d

    22

  • where y = c+ x+ d dRd andd^t+1d

    22 0

    y + yy^t = c+ cc^t + x+ xx^t + dd^t + d d^t+1d+ dRd(1 + ^dt)

    dRd

    +d

    Rd

    Rd(1 + ^dt)yy^t = Rd(1 + ^dt)cc^t +Rd(1 + ^dt)xx^t +Rd(1 + ^dt)dd^td^t+1dd+Rd(1 + ^dt)

    d

    Rd

    Rdyy^t = Rdcc^t +Rdxx^t +Rddd^tdd^t+1 d+ d+ d^dt

    y^t =c

    yc^t +

    x

    yx^t +

    d

    yd^t d

    Rdyd^t+1 +

    d

    Rdy^dt (1.c)

  • equation (2)

    yt = ztkt l1t

    yey^t = zez^t(kek^t)(lel^t)1 = (zkl1)ez^t+k^t+(1)l^t

    ey^t = ez^t+k^t+(1)l^t

    (1 + y^t) = 1 + z^t + k^t + (1 )l^ty^t = z^t + k^t + (1 )l^t (2.c)

    Equation (3)

  • l1t = (1 )zt kt

    lt

    !1

    lt

    lel^t

    1= (1 )zez^t

    kek^t

    lel^t

    1 le

    ^ lt

    l1e(1)l^t = (1 )zez^tkek^tlel^t 1 le

    ^ lt

    ll1e(1)l^t+^ lt = (1 )zez^tkek^tlel^t

    ll1 h1 + ( 1) l^t + ^ lti = (1 )zkl 1 + z^t + k^t l^t

    llh1 + ( 1) l^t + ^ lt

    i= (1 )zkl1

    1 + z^t + k^t l^t

  • where l l = (1 )y and y = zkl1

    (1 )yh1 + ( 1) l^t + ^ lt

    i= (1 )y

    1 + z^t + k^t l^t

    1 + ( 1) l^t + ^ lt =

    1 + z^t + k^t l^t

    ^ lt = z^t + k^t l^t ( 1) l^t

    ^ lt = z^t + k^t + (1 ) l^t (3.c)

    equation (4)

  • xt1

    ct lt= Et

    "uct+1

    yt+1kt+1

    + xt+1(1 )!#

    xe^xt

    1

    cec^t lel^t= Et

    24 1cec^t+1 lel^t+1

    yey^t+1k1ek^t+1+xe^xt+1(1 )

    !35

    Ignore Et for a moment

    xe^xt

    cec^t+1 lel^t+1

    =

    yey^t+1k1ek^t+1 + xe^xt+1(1 )

    cec^t lel^t

  • LHS

    xe^xtcec^t+1 xe^xtlel^t+1

    = xce^xt+c^t+1 xle^xt+l^t+1

    = xc (1 + ^xt + c^t+1) xl1 + ^xt + l^t+1

    = xc+ xc^xt + xcc^t+1 xl xl ^xt xll^t+1

    = (c l) x^xt + (c l) x + xcc^t+1 xll^t+1

    RHS

  • yey^t+1k1ek^t+1 + xe^xt+1(1 )

    cec^t lel^t

    =

    2664yey^t+1k1ek^t+1cec^t yey^t+1k1ek^t+1lel^t

    +xe^xt+1(1 )cec^t xe^xt+1(1 )lel^t

    3775

    =

    2664cyke

    y^t+1k^t+1+c^t lykey^t+1k^t+1+l^t

    +(1 )cxe^xt+1+c^t (1 )lxe^xt+1+l^t

    3775

    =

    24 cyk 1 + y^t+1 k^t+1 + c^t lyk 1 + y^t+1 k^t+1 + l^t+(1 )cx (1 + ^xt+1 + c^t) (1 )lx

    1 + ^xt+1 + l^t

    35

  • =

    2664cyk

    1 + y^t+1 k^t+1

    + cyk c^t lyk

    1 + y^t+1 k^t+1

    lykl^t + (1 )cx (1 + ^xt+1) + (1 )cxc^t(1 )lx (1 + ^xt+1) (1 )xll^t

    3775

    =

    24 (c l) yk 1 + y^t+1 k^t+1+ hyk + (1 )xi cc^t(1 ) (c l) x (1 + ^xt+1)

    hyk + (1 )x

    ill^t

    35

    =

    24 (c l) yk 1 + y^t+1 k^t+1+ xcc^t(1 ) (c l) x (1 + ^xt+1) xll^t

    35= (c l)

    hyk

    1 + y^t+1 k^t+1

    + (1 )x (1 + ^xt+1)

    i+xcc^txll^t

    = (c l)"

    yk + yk y^t+1 ykk^t+1

    +(1 )x + (1 )x^xt+1

    #+ xcc^t xll^t

  • = (c l) x + (c l)"yk y^t+1 ykk^t+1+(1 )x^xt+1

    #+ xcc^t xll^t

    = (c l) x + 0@ (c l) yk y^t+1 (c l) ykk^t+1+(c l) (1 )x^xt+1 + x cc^t x ll^t

    1A

    where x = hyk + x(1 )

    iand x =

    yk + x(1 )

    LHS=RHS(c l) x^xt + (c l) x + xcc^t+1 xll^t+1

    = (c l) x+ 0@ (c l) yk y^t+1 (c l) ykk^t+1+(c l) (1 )x^xt+1 + x cc^t x ll^t

    1A

  • (c l) x^xt = 0@ (c l) yk y^t+1 (c l) ykk^t+1+(c l) (1 )x^xt+1 + x cc^t x ll^t

    1Axcc^t+1 + xll^t+1

    x^xt =

    0@ yk y^t+1 ykk^t+1 + (1 )x^xt+1+ x(cl)cc^t x(cl)ll^t

    1A xc(c l)c^t+1 +

    xl

    (c l) l^t+1

  • ^xt =

    x

    y

    ky^t+1

    x

    y

    kk^t+1 +

    (1 )

    ^xt+1 +c

    (c l)c^t l

    (c l) l^t

    c(c l)c^t+1 +

    l

    (c l) l^t+1

    Add back expectation

    ^xt =

    x

    y

    kEty^t+1 k^t+1

    +

    c

    (c l) (c^t Etc^t+1) (4.c)

    l

    (c l)l^t Etl^t+1

    +

    (1 )Et^xt+1

  • equation (5)

    1

    ct lt

    "

    R

    1

    dt (dt+1 d)

    #= Et[uct+1]

    1

    cec^t lel^t

    "

    R

    1

    de^dt

    d^t+1 + 1

    d d

    #= Et

    1

    cec^t+1 lel^t+1

    Ignore Et for the moment.

  • cec^t+1 lel^t+1

    R

    1de

    ^dt

    d^t+1d+ d d

    =

    cec^t lel^t

    cec^t+1 R1

    de^dtcec^t+1dd^t+1+lel^t+1 R 1de^dt+l

    el^t+1dd^t+1 = cec^t lel^t

    LHS

    cec^t+1 R1

    de^dt cec^t+1dd^t+1 + lel^t+1 R 1de^dt + l

    el^t+1dd^t+1

    = cdec^t+1^dt R cec^t+1dd^t+1 + l

    del^t+1^dt R + l

    el^t+1dd^t+1

    =

    24 cd (1 + c^t+1 ^dt) R c (1 + c^t+1)dd^t+1ld

    1 + l^t+1 ^dt

    R + l

    1 + l^t+1

    dd^t+1

    35

  • =24 cd (1 + c^t+1) R cd ^dt R c (1 + c^t+1)d^t+1

    ld

    1 + l^t+1

    R +

    l

    dR^dt + l

    1 + l^t+1

    dd^t+1

    35

    where R1d= or R d =

    1

    = c (1 + c^t+1) c^dt cdd^t+1 l1 + l^t+1

    + l ^dt+ l

    dd^t+1

    = hc+ cc^t+1 c^dt 1cdd^t+1 l ll^t+1 + l ^dt + 1ldd^t+1

    i

    = h(c l) + cc^t+1 (c l) ^dt 1 (c l)dd^t+1 ll^t+1

    i

    RHS

  • cec^t lel^t

    = hc (1 + c^t) l

    1 + l^t

    i

    = c+ cc^t l ll^t

    LHS=RHS

    "(c l) + cc^t+1 (c l) ^dt1 (c l)dd^t+1 ll^t+1

    #=

    c+ cc^t l ll^t

    cc^t+1 (c l) ^dt 1 (c l)dd^t+1 ll^t+1 = cc^t ll^t

  • (c l) ^dt = cc^t+1 cc^t ll^t+1 + ll^t 1 (c l)dd^t+1

    ^dt =c

    (cl) (c^t+1 c^t)l

    (cl)l^t+1 l^t

    1dd^t+1

    Add back expectation.

    ^dt =c

    (c l) (Etc^t+1 c^t)l

    (c l)Etl^t+1 l^t

    1dEtd^t+1 (5.c)

    equation (6)

  • kt+1 = xt + (1 )ktkek^t+1 = xex^t + (1 )kek^t

    k(1 + k^t+1) = x(1 + x^t) + (1 )k(1 + k^t)kk^t+1 = xx^t + (1 )kk^t

    k^t+1 =x

    kx^t + (1 )k^t (6.c)

    We will solve the model by the linearizing the equations that characterize thecompetitive equilibrium around the steady state.

    Here again the log-linearized model :

  • y^t =c

    yc^t +

    x

    yx^t +

    d

    yd^t d

    Rdyd^t+1 +

    d

    Rdy^dt (1.c)

    y^t = z^t + k^t + (1 )l^t (2.c)

    ^ lt = z^t + k^t + (1 ) l^t (3.c)

    ^xt =

    x

    y

    kEty^t+1 k^t+1

    +

    c

    (c l) (c^t Etc^t+1) (4.c)

    l

    (c l)l^t Etl^t+1

    +

    (1 )Et^xt+1

    ^dt =c

    (c l) (Etc^t+1 c^t)l

    (c l)Etl^t+1 l^t

    1dEtd^t+1 (5.c)

  • k^t+1 =x

    kx^t + (1 )k^t (6.c)