The model is solved algebraically, excep for a cubic roo which is solved numerically The mehod of soluion is undeermined coefficiens The noaion in his noe corresponds o he noaion in he program The model is given by: () λ = αγ + η Here, λ is he ex ane excess reurn on he foreign deposi (relaive o he home deposi) * as in he paper γ + is he home less foreign ineres rae differenial (expressed as i i in he paper The + subscrip is useful because we can hink of his as a predeermined iable if we were o use he Blanchard-Kahn soluion echnique) (2) λ = Eq + q γ+ + Eπ+ This is he definiion of he excess reurn π + is home less foreign inflaion π = δ q q + βeπ + (3) ( ) This is he (home less foreign) Phillips curve (4) γ = π + + ργ This is he (home less foreign) Taylor rule The following equaions are he sochasic processes for he exogenous iables: (5) q = ξq + ε (6) η = µη + u The program akes as inpu he values of he parameers α, δ, β,, ρ, ξ, µ, and he iances of q and η, which are enered in lines 8 hrough 26 in he program We posi an undeermined coefficiens soluion for he ineres differenial, he real exchange rae and he relaive inflaion raes: (7) γ = aq + bη + + cγ
(8) q = dq + eη + f γ (9) π = gq + hη + kγ Using equaions (5)-(9), we have: (0) q = dξq + dε + eµη + eu + faq + fbη + fcγ + + + Taking expecaions: Eq = dξ + fa q + eµ + + fb η + fcγ () ( ) ( ) Then again using equaions (5)-(9), we have: (2) π + = gξq + gε + + hµη + hu+ + kaq + kbη + kcγ Taking expecaions: E π = gξ + ka q + hµ + + kb η + kcγ (3) ( ) ( ) Now use (7) and (9) o make subsiuions ino equaion (4): ( ) aq + bη + cγ = gq + hη + kγ + ργ Maching coefficiens, we have: (4) a= g (5) b= h (6) c= k+ ρ Nex use (7), (9) and (3) o make subsiuions ino equaion (3): ( ) (( ) ( ) ) gq + hη + kγ = δ dq + eη + f γ δq + β gξ + ka q + hµ + kb η + kcγ Maching coefficiens, we have: (7) g = δ ( d ) + β( gξ + ka) (8) h = δe + β ( hµ + kb) (9) k = δ f + βkc 2
Eliminae λ by equaing () and (2), hen use equaions (7), (8), () and (3): (20) ( ) (2) ( ) (22) ( α ) ( aq + b + c ) = ( d + fa) q + ( e + fb) + fc ( dq + e + f ) ( aq + bη + cγ ) + ( gξ + ka) q + ( hµ + kb) η + kcγ α η γ η ξ µ η γ η γ Maching coefficiens, we have: + α a = ξd + fa d + gξ + ka + α b = µ e + fb e + µ h + kb + c = fc f + kc Equaions (4)-(22) are nine equaions ha allow for us o solve for he nine undeermined coefficiens: a, b, c, d, e, f, g, h and k Equaions (6), (9) and (22) allow us o solve for c, f, and k Firs, solve for k from (6): c ρ k = Then subsiue his ino (9): c ρ = δ f ( βc), and solve for f: f c ρ = ( βc) δ Then subsiue his ino equaion (22) o ge: c ρ c ρ + c= c c + c δ ( α) ( β ) ( ) This gives us he equaion: ( βc)( c ρ )( c) δ c( α ) δ c( c ρ ) 0= + + 3
The program solves his equaion in line 3 There are hree roos o his cubic equaion We use he soluion ha is less han one in absolue value, which is given in line 34 c ρ Then from above, he program solves for k in line 39 using k = And in line 40, he program c ρ k solves for f using f = ( βc) = ( βc) δ δ We have solved for c, f, and k Then equaions (4), (7) and (20) le us solve for a, d and a g From (4) we have g = From (20), we have: ξ d = f + + k+ a, so ( ξ ) ( α) ( f k) ( ξ ) + α ξ d = a Subsiuing hese expressions for g and d ino (7), we ge: a = ( ξ ) ( ) ( ) δ ξ β ξ + + δ + α ξ ( k ) ( ( f k) ) This is he soluion for a in line 4 of he program Then line 42 gives us d as ( + α f k) ξ a d = a, and line 43 gives us g as g = ξ ( ) Nex, equaions (5), (8), and (2) will allow us o solve for b, h and e From equaion b (5), we have h = From (2) we ge ( f + k α ) b+ + µ h= ( µ ) e, so e = ( ( ) ) ( µ ) f + k α + µ b+ Subsiuing hese ino (8) we ge: ( ( ) ) ( ) b f + k α + µ b+ b = δ + β µ + kb µ, or b = δ k f k ( µ )( β ( µ + )) δ ( ( + α ) + µ ) which is line 44 of he program 4
b Line 45 gives us h as h =, and line 46 gives us e as from (8) as ( βµ ) h βkb ( βµ βk ) b e = = δ δ Nex we solve for he real ineres rae differenial We posulae a soluion of he form r = mq + nη + pγ (23) We can use he fac ha r = γ+ Eπ+ and equaions (7) and (3) o wrie: (( ) ( ) ) mq + nη + pγ = aq + bη + cγ gξ + ka q + hµ + kb η + kcγ Maching coefficiens, we have: ( ξ ) m = a g + ka, which is line 47 of he program ( µ ) n = b h + kb, which is line 48 of he program ( ) p = c k, which is line 49 of he program Then we posi a soluion for he ex ane excess reurn: (24) λ = qq + rη + sγ From equaion (), we have λ = αγ η We can hen subsiue from (7) o ge: + qq + rη + sγ = αaq + αη b + αγ c η Maching coefficiens, we have q = αa, which is line 50 of he program r = αb, which is line 5 of he program s = αc, which is line 52 of he program So we have λ = αaq + rη + αγ c Eλ = αaξ q + α b µη + αcγ Then, ( ) + + 5
E λ = αaξ q + rµη + αce γ, and so on 2 2 + 2 + 2 We have: αa r Λ = q + η + se( γ + γ+ + γ+ 2 + ) ξ µ Now, γ = aq + bη + + cγ ec Then ( ) E γ = E aq + bη + cγ = aξ q + bµη + cγ + 2 + + + + ( ) ( ) ( ) E γ = E aq + bη + cγ = aξ + caξ q + bµ + cbµη+ c γ 2 2 2 + 3 + 2 + 2 + 2 + ( ) ( ) ( ) E γ = E aq + bη + cγ = aξ + caξ + c aξ q + bµ + cbµ + c bµη+ c γ 3 2 2 3 2 2 3 + 4 + 3 + 3 + 3 + E We see ha ( γ γ ) + + = + 2 3 2 4 3 2 ( + ξ + ξ + ξ + ξ + ( ξ + ξ) + ξ + ( ξ + ( ξ + ξ) ) + ) 2 3 2 4 3 2 b c c( c ) c c( c ) a c c c c c c q ( ( ) ) + + µ + µ + µ + µ + µ + µ + µ + µ + µ + µ + η c + ( + c) γ + γ c + 2 2 ( ) ( ( )) ( ) E ( γ + γ + ) = a + ξ ( c+ c + ) q + b + µ c+ c + η + + c γ + c γ ξ µ c + + which can be wrien as: 6
a c b c c E( γ + γ+ + ) = + ξ q + + µ η + ( + c) γ + γ+ ξ c µ c c a c b c c = + ξ q + + µ η + + c γ + aq + bη + cγ ξ c µ c c a b = q + η + γ c c c ( ξ )( ) ( µ )( ) ( ) ( ) Then we ge: αa r a b Λ = q + η + αc q + η + γ ξ µ ( ξ )( c) ( µ )( c) c αa αcb αc = q + r+ η + γ ξ c µ c c ( )( ) αa r+ c = q + η + αc γ c c c ( ξ )( ) ( µ )( ) So we can wrie (25) vq wη xγ Λ = + +, where v = w = αa ( ξ )( c) r+ c ( µ )( c), which is line 53 of he program, which is line 54 of he program αc x =, which is line 55 of he program c Nex, we calculae some iances and coiances ( γ q ) ( aq bη cγ ξq ε ) aξ ( q ) cξ ( γ q ) cov, = cov + +, + = + cov, + + + Hence, cov ( γ, q ) ( q ) aξ =, which is line 56 of he program cξ 7
= + + + = + Then, cov ( γ, η ) cov ( aq bη cγ, µη u ) bµ ( η ) cµ cov ( γ, η ) Hence, cov ( γ, η ) ( η ) + + + bµ =, which is line 57 of he program cµ ( γ+ ) = ( aq + bη + cγ) 2 2 2 = c ( γ ) + a ( q ) + b ( η ) + 2ac cov ( q, γ ) + 2bc cov ( η, γ ) This is line 58 of he program 2 2 a b 2ac 2bc cov, cov, 2 2 2 2 ( q ) ( η ) ( q γ ) ( η γ ) = + + + c c c c ( λ r) = ( qq + rη + sγ mq + nη + pγ) ( ) ( η ) ( γ ) ( ) cov (, γ ) ( ) cov ( η, γ ) cov, cov, = qm q + rn + ps + pq + ms q + pr + ns which is line 59 of he program ( Λ r) = ( vq + wη + xγ mq + nη + pγ) ( ) ( η ) ( γ ) ( ) cov (, γ ) ( ) cov ( η, γ ) cov, cov, = vm q + wn + xp + pv + mx q + pw + nx which is line 60 of he program ( r) = ( mq + nη + pγ) 2 ( ) 2 ( η ) 2 ( γ ) 2 cov (, γ ) 2 cov ( η, γ ) = m q + n + p + mp q + np which is line 6 of he program ( λ) = ( qq + rη + sγ) 2 ( ) 2 ( η ) 2 ( γ ) 2 cov (, γ ) 2 cov ( η, γ ) = q q + r + s + qs q + rs \ which is line 62 of he program Then he regression coefficien for regressing λ on r is given by line 63 of he program The regression coefficien for regressing which is line 64 of he program ( λ r) ( r ) cov, Λ on r is given by, which is ( Λ r) ( r ) cov, 8
Nex, we have: ( π) = ( gq + hη + kγ) 2 ( ) 2 ( η ) 2 ( γ ) 2 cov (, γ ) 2 cov ( η, γ ) = g q + h + k + gk q + hk which is line 65 of he program cov( π+ π) = cov ( gq+ + hη+ + kγ+, gq + hη + kγ) = cov ( gξq + gε + + hµη + hu+ + kaq + kbη + kcγ, gq + hη + kγ ) = cov (( gξ + ka) q + ( hµ + kb) η + kcγ, gq + hη + kγ) 2 = g ( gξ + ka) ( q) + h( hµ + kb) ( η) + k c ( γ) + ( k ( gξ + ka) + gkc) cov ( q, γ ) + ( h( hµ + kb) + hkc) cov ( η, γ ) which is line 66 of he program The serial correlaion of inflaion is given by cov ( π+ π) ( π ), which is line 67 of he program Then, cov ( γ+, γ) = cov ( aq + bη + cγ, γ) = c ( γ) + a cov ( γ, q) + bcov ( γ, η), which is line 68 of he program The serial correlaion of he nominal ineres rae is given by of he program cov ( γ+ γ) ( γ ), which is line 69 ( γ+ r) = ( aq + bη + cγ mq + nη + pγ) ( ) ( η ) ( γ ) ( ) cov (, γ ) ( ) cov ( η, γ ) cov, cov, = am q + bn + cp + pa + mc q + pb + nc which is line 70 of he program 9