Hansen s basic RBC model
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1 ansen s basic RBC model George McCandless UCEMA Spring ansen s RBC model ansen s RBC model First RBC model was Kydland and Prescott (98) "Time to build and aggregate uctuations," Econometrica Complicated lagged cumulative investment strange utility function Lots added to look for presistence ansen s model much simpler (98) "Indivisible labor and the business cycle," Journal of Monetary Economics Simple Added indivisible labor to gain persistence and covariance with output Set rules for RBC game Match second moments Newer rule: match impulse response functions ansen s basic model Robinson Crusoe maximizes the discounted utility function max X t u(c t ; l t ) t=
2 The speci c utility functions u(c t ; h t ) = ln c t + A ln( h t ) with A >. The production function is f( t ; k t ; h t ) = t k t h t t is a random technology variable that follows the process t+ = t + " t+ for < <. " t iid, positive, bounded above, E" t =. =) E t is and t+ >. ansen s basic model (continued) Capital accumulation follows the process k t+ = ( )k t + i t The feasibility constraint is Bellmans equation The basic Bellmans equation subject to f( t ; k t ; h t ) c t + i t V (k t ; t ) = max c t;h t [ln c t + A ln( h t ) + E t [V (k t+ ; t+ ) j t ]] Simpler to write as t kt h t c t + i t ; t+ = t + " t+ ; and k t+ = ( )k t + i t : V (k t ; t ) = max ln t kt h t + ( )k t k t+ k t+;h t k t+ and h t are control variables First order conditions +A ln( h t ) + E t [V (k t+ ; t+ ) j t ]]
3 First order conditions are (k t ; t ) = t+ t kt h t + ( )k t k t+ +E t [V k (k t+ ; t+ ) j t (k t ; t t = = ( ) A h t t kt h t t kt h t + ( )k t k t+ The Benveniste-Scheinkman envelope theorem condition (k t ; t ) t t kt h t t k t h t + ( ) + ( )k t k t+ Simplifying the rst order conditions First order conditions can be written as and t kt h t + ( )k t k t+ " # t+ kt+ = E h t+ + ( ) t t+ kt+ h t+ + ( )k j t t+ k t+ ( ) ( h t ) t k t h t In equilibrium, = A t kt h t + ( )k t k t+ c t = t kt h t + ( )k t k t+ Simplifying the rst order conditions (continued) Factor markets give and r t = t kt h t w t = ( First order conditions are simply and c t = E t ) t k t h t rt+ + ( ) j t c t+ ( h t ) w t = Ac t Stationary states
4 Stationary state value of h = h t = h t+ is h = h + A ( ) i; ( ) Stationary state value of k = k t = k t+ = k t+ is ow to study dynamics k = h " # : ( ). Find the approximate Value function and Plan (a) These will describe the dynamics within the precision of the approximation (b) Can be complicated to nd i. Especially if the domain of stochastic variable is large (c) Can be impossible i. If the model is not single agent ii. If the model can not be approximated by social planner. Alternative approachs (a) Log linear approximation of the model i. After the optimization has been done ii. After equilibrium conditions have been imposed (b) Quadratic linear appoximation of the problem Log-linearization techniques Consider a function of the form Taking logs of both side gives F (x t ) = G(x t) (x t ) ln(f (x t )) = ln(g(x t )) ln((x t )) The rst order Taylor series expansion around the stationary state values x
5 gives ln(f (x)) + F (x) F (x) (x t x) ln(g(x)) + G (x) G(x) (x t x) Log-linearization techniques (direct method) In the stationary state ln((x)) (x) (x) (x t x) ln(f (x)) = ln(g(x)) ln((x))) So the rst order Taylor expansion can be written as F (x) F (x) (x t x) G (x) G(x) (x t x) (x) (x) (x t x) Remember that this holds only near x An example using a Cobb-Douglas production function Y t = t K t t Take logs ln Y t = ln t + ln K t + ( ) ln t rst order Taylor expansion gives ln Y + Y Y t Y ln + t + ln K + K K t K + ( ) ln + ( ) t Since in a stationary state ln Y = ln + ln K + ( ) ln get Y Y t Y t + K K t K ( ) + That reduces to t Y t Y + t + K t K + ( ) t
6 Log-linearization techniques (Uhlig s method) Write the original variable as X t = Xe f X t or ex t = ln X t ln X bring together all the exponential terms that you can becomes A t B t C t AB = Ae e A t B e e B t C e e C t C e ea t+ eb t ec t Reference:Uhlig, arald, (999) "A toolkit for analysing nonlinear dynamic stochastic models easily", in Ramon Marimon and Andrew Scott, Eds., Computational Methods for the Study of Dynamic Economies, Oxford University Press, Oxford, p.-. Log-linearization techniques (Uhlig s method) The Taylor series expansion (linear) gives e e A t+ e B t e C t e e A+ e B ec + e e A+ e B e C e At e A +e e A+ e B e C e Bt e B = + e A t + e B t e C t ; e e A+ e B e C e Ct e C So e e A t+ e B t e C t + e A t + e B t e C t The approximation is A t B t C t AB C + e A t + e B t e C t Log-linearization techniques (Uhlig s method) Some rules from Uhlig X e e t+ay e t + X e t + ay e t ; ex tyt e ; h i h i X E t ae e t+ a + ae ext+ t h i E t [X t+ ] = X + E ext+ t
7 Log linear version of ansen s model The ve equations of the ansen model are (adjusted) Ct = E t (r t+ + ( )) C t+ AC t = ( ) ( t ) Y t t C t = Y t + ( )K t K t+ Y t = t K t t r t = Y t K t We will do the log-linearization equation by equation Log linear version of ansen s model First equation = E t Ct C t+ (r t+ + ( )) " # C Ce e t = E t CeC e re ert+ + ( ) Ce ec t t+ CeC e t+ h i C = E t re e t Ct++er e t+ C + ( ) e e t Ct+ e re t h + C e i t Ct+ e + er t+ + ( ) h + C e i t Ct+ e h = E t + C e i t Ct+ e + rer t+ ; or (after cancelling the s and cleaning up the expections) Log linear version of ansen s model e C t E t e Ct+ + re t er t+ Second equation AC t = ( ) ( t ) Y t t AC + C e t ( " ACe e C t = ( ) Y e e Y t e t ( ) Y e e Y t AC e C t ) Y + e Y t e t ( ) Y # ( ) Y + Y e t ey t ( ) Y e t
8 given that in the stationary state Log linear version of ansen s model This becomes AC = ( ) Y ec t e Y t ( ) Y ( ) ( )Y e t = e Y t e t so Log linear version of ansen s model = e C t e Yt + e t The next three equations (in their Log-linear form) are Y Y e t CC e t + K h( ) K e i t Kt+ e e t + K e t + ( ) e t Yt e Y e t Kt e er t where r = Y =K Log linear version of ansen s model The stochastic process is t+ = t + " t+ putting in the log di erence of the s the linerar approximation is + e t+ So the simple version is e e t+ = e e t + " t+ = + e t + " t+ e t+ = e t + t+ The log-linear version of the model 8
9 The equations of the full log-linear model are = C e t E tct+ e + re t er t+ = C e t Yt e + e t = Y Y e t CC e t + K h( ) K e i t Kt+ e = e t + K e t + ( ) e t Yt e = Y e t Kt e er t and e t+ = e t + t+ Solving the log-linear version of the model n o The variables of the model are ekt+ Yt e Ct e t e er t plus the stochastic variables t De ne the state variables as h i x t = ekt De ne the "jump" variables as De ne the stochastic variable as y t = Y t C t t r t z t = [ t ] Solving the log-linear version of the model The model can be written as = Ax t + Bx t + Cy t + Dz t ; = E t [F x t+ + Gx t + x t + Jy t+ + Ky t + Lz t+ + Mz t ] ; z t+ = Nz t + " t+, E t (" t+ ) = : Where A = K B = K ( + ) 9
10 C = Y C Solving the linear version of the model D = F = [] ; G = [] ; = [] ; J = r ; K = ; L = [] M = [] N = [] Solving the linear version of the model We look for a solution of the form x t = P x t + Qz t y t = Rx t + Sz t Note that here C is of full rank and has a well de ned inverse C The solutions can be found from = (F JC A)P (JC B G + KC A)P KC B + R = C (AP + B); N (F JC A) + I k (JR + F P + G KC A) vec(q) = vec JC D L)N + KC D M and S = C (AQ + D) Explaining the solution We look for the laws of motion of the model x t = P x t + Qz t ; y t = Rx t + Sz t :
11 We begin by substituting the laws of motion into the two equations of the model Reduce each equation to one in which there are only two variables: x t and z t : Use the stochastic process in the expectational equation to replace z t+ = Nz t + " t+ Taking expectations, the " t+ = disappear Explaining the solution Begin with the model = Ax t + Bx t + Cy t + Dz t = E t [F x t+ + Gx t + x t + Jy t+ + Ky t + Lz t+ + Mz t ] Substitute in In the rst equation this gives x t = P x t + Qz t ; y t = Rx t + Sz t : = A [P x t + Qz t ] + Bx t + C [Rx t + Sz t ] + Dz t Explaining the solution In the second equation = E t [F [P x t + Qz t+ ] + G [P x t + Qz t ] + x t +J [Rx t + Sz t+ ] + K [Rx t + Sz t ] + Lz t+ + Mz t ] Substitute one more time in the second equation = E t [F [P [P x t + Qz t ] + Q [Nz t + " t+ ]] + G [P x t + Qz t ] +x t + J [R [P x t + Qz t ] + S [Nz t + " t+ ]] +K [Rx t + Sz t ] + L [Nz t + " t+ ] + Mz t ] This simpli es to (because E t " t+ = ) and we remove the expectations operator = F [P [P x t + Qz t ] + QNz t ] + G [P x t + Qz t ] +x t + J [R [P x t + Qz t ] + SNz t ] +K [Rx t + Sz t ] + LNz t + Mz t
12 Explaining the solution The two equations can be rearranged to give and = [AP + B + CR] x t + [AQ + CS + D] z t ; = [F P P + GP + + JRP + KR] x t + [F P Q + F QN + GQ + JRQ + JSN + KS + LN + M] z t : Since these equations need to hold for all x t and z t, it must be that = AP + B + CR = AQ + CS + D = F P P + GP + + JRP + KR = F P Q + F QN + GQ + JRQ + JSN + KS + LN + M Explaining the solution The third equation is = F P + GP + JRP + + KR and the rst is (if the inverse of C exists) Combining these one gets R = C AP C B = F P + GP J C AP + C B P + K C AP + C B = F P JC AP + GP JC AP JC BP + KC AP KC B = F JC A P JC B + KC A G P Explaining the solution KC B + ere F is a matrix (a scalar) Finding the solution to the quadratic equation = F JC A P JC B + KC A G P KC B + can be done using = ap + bp + c
13 The solution to this equation is found from P = b p b ac a There are usually two di erent solutions to this problem. We use jp j < in order to choose the stable root. Once P is known, nding R is simple using Explaining the solution R = C AP C B Finding Q (with P and R already known, from above) Use the equations = F P Q + F QN + GQ + JRQ + JSN + KS + LN + M and S can be written as = AQ + CS + D S = C AQ C D Substitute this into the rst equation = F P Q + F QN + GQ + JRQ JC AQN JC DN KC AQ KC D + LN + M Rearrange to get F P + G + JR KC A Q + F JC A QN Explaining the solution = JC DN + KC D LN + M This equation F P + G + JR KC A Q + F JC A QN = JC DN + KC D LN + M has Q in two di erent places on the left hand side Q in the nal position in F P + G + JR KC A Q Q in the second to the last position in F JC A QN
14 Need to use a theorem from advanced matrix algebra Theorem Let A, B, and C be matrices whose dimensions are such that the product ABC exists. Then vec(abc) = (C A) vec(b) where the symbol denotes the Kronecker product. Explaining the solution Think of as F P + G + JR KC A Q + F JC A QN = JC DN + KC D LN + M (notice that we added I) where W QI + XQN = Z W = F P + G + JR KC A X = F JC A Z = JC DN + KC D LN + M Take vec of both sides of the equation, so vec (W QI) + vec (XQN) = vec (Z) This equals (I W ) vec (Q) + (N X) vec (Q) = vec (Z) or (I W + N X) vec (Q) = vec (Z) If (I W + N X) is invertible vec (Q) = (I W + N X) vec (Z) Explaining the solution What are vec and (the Kronecker product)
15 First vec a a vec a = a a a a a a a a a : the columns are made into a vector Explaining the solution The Kronecker product is A B = a a b b b a a b a B = a b b B a b a b a b a b = a b a b a b a b a b a b a b a b a b a b a b a b : a b a b a b a b a b a b a b a b Calibration Solution to model is numerical Need values for parameters Some we borrow from literature (quarterly) = :99 = : = : Need a value for A a B a B Choose A so that = = Use stationary state equation for = h + A ( ) i ( ) A = : for = : K = :9 and using the production function, Y = :
16 r = = = : From data for US use = :9 Matices for Calibrated model A = C = : Matices for Calibrated model Numerical solution for model B = : : : : :98 : D = F = [] G = [] = [] J = :8 K = L = [] M = [] N = [:9] The quadratic equation gives the solutions P = :9 and P = :9 The stable value is The value for Q is P = :9 Q = :
17 The matrices R and S are R = : :9 : :9 Numerical solution for model The laws of motion are Recall that e t follows the process and S = ek t+ = :9 e K t + : e t ; ey t = : e K t + : e t ; ec t = :9 e K t + :9 e t ; : :9 : : e t = : e K t + : e t ; er t = :9 e K t + : e t : e t = :9 e t + t Two ways of nding the variances of the variables of the model Simulations Run lots of simulated economies Calculate the variances from this "data" Calculate variances from laws of motion See book for detains Need to calibrate var( t ) so that var( e Y t ) = :% gets standard error of t = : Tables of second moments Standard errors as fraction of output ey t e Ct e t er t e It Standard error :8 " : " : " :9 " : " As % of output % : % 9:9 % : % :% Standard errors from the data ey t Ct e t e It e As % of output % :% 9: % 88: %
18 Does well for consumption Badly for hours worked and investment % stationary state values are found in another program A=[ -kbar ] ; B=[ (-delta)*kbar theta -] ; C=[ - -/(-hbar) ybar -cbar - -theta -]; D=[ ] ; F=[]; G=F; =F; J=[ - beta*rbar]; K=[ ]; L=F; M=F; N=[.9]; Cinv=inv(C); a=f-j*cinv*a; b=-(j*cinv*b-g+k*cinv*a); c=-k*cinv*b+; P=(-b+sqrt(b^-*a*c))/(*a); P=(-b-sqrt(b^-*a*c))/(*a); if abs(p)< P=P; else P=P; end R=-Cinv*(A*P+B); Q=(J*Cinv*D-L)*N+K*Cinv*D-M; QD=kron(N,(F-J*Cinv*A))+(J*R+F*P+G-K*Cinv*A); Q=Q/QD; S=-Cinv*(A*Q+D); ansen s model with indivisible labor Objective: increase variance of hours worked Make labor indivisible one works X hours per week or not at all Add unemployment since some fraction of the population will not be working 8
19 Problem of non-convexity of consumption set In general, maximization is only valid over convex sets Def of a convex set straight lines between any two points in set are also in set Example of a non-convex set ow non-convexity is xed in ansen s model The problem is the jump in income between working and not working ansen invented an "unemployment insurance" Lump sum transfers that make income equal for all solves non-convexity problem consumption increases smoothly with wage since all receive same income (based on wages) solve problem of too much heterogenity ousehold problem maximize X max t u(c t ; t ) t= subject to c t + i t = w t h t + r t k t t = probability in time t of supplying h units of labor 9
20 Expected utility u(c t ; t ) = ln c t + h t A ln( h ) h + A( u(c t ; t ) = ln c t + h t A ln( h ) h h t h ) ln() ousehold problem Maximization problem becomes X max t [ln c t + Bh t ] t= with subject to constraints B = A ln( h ) h t kt h t = c t + k t+ ( )k t and ln t+ = ln t + " t+ ousehold problem First order conditions = ( ) t kt h t + B; c t = c t + E t c t+ t+ k t h t + ( ) With equilibrium condition, these simplify to = E t Ct C t+ (r t+ + ( C t = ( ) Y t B t : )) ; Full model = E t Ct C t+ (r t+ + ( C t = ( ) Y t B t ))
21 C t + K t+ = Y t + ( )K t r t = t Kt t Y t = t Kt t Stationary state Equations = r + ( ) ( ) Y C = B r = K Y = K C = Y K solve to give = B ( ) ( ) and K = + ( ) Stationary state Comparing to basic ansen model To get same stationary state, need the same in both cases Then other variables will be the same Old stationary state equation = h + A ( ) i ( ) Set the two equal h + A ( ) i = ( ) ( ) A ln( h ) h ( ) ; We replaced B with A ln( h) h Need to determine h that make the two SS the same
22 Stationary state Solve to get h ln( h ) = h A ( ) + A ( ) h i ( ) ( ) i = G G is a constant To nd h Get h = :8, = :, and = : Log-linear model Taking the log-linear approximation of the model gives Solution method Use Uhlig s method e C t E t e Ct+ + re t er t+ e C t + e t e Yt Y e Y t C e C t + ( )K e K t K e K t+ e Y t e t e K t ( ) e t e Y t e Kt er t = Ax t + Bx t + Cy t + Dz t ; = E t [F x t+ + Gx t + x t + Jy t+ + Ky t + Lz t+ + Mz t ] ; z t+ = Nz t + " t+, E t (" t+ ) = ; h i h where, x t = ekt, y t = eyt ; C e t ; e i, h i t ; er t and zt = et
23 Solve for x t = P x t + Qz t y t = Rx t + Sz t Results C = A = K B = Y C ( ) K ( + ) D = F = [] ; G = [] ; = [] J = r K = L = [] ; M = [] ; and N = [] : The linear policy functions are ek t+ = :98 e K t + : t and where Results R = : : : :9 y t = R e K t + S t S = : 9 8 : : : 9 Using this model, we calculate the variances of the variables ey t e Ct e t er t e It Standard errors : " :8 " : " : " : " As % of output % : % : % : 9% :% Increased variance in hours worked Slight increase in investment Lower variance in consumption (compared to data)
24 Impulse response functions ow does the economy respond to a one time shock to technology " t = except " = : Recall that = :9 e t = e t + " t Response of technology (path of e t ) Impulse response functions Then calculate the time path of capital with K e = using ek t+ = P K e t + Q e t get path of e K t. Use this to nd path of other variables using y t = R e K t + S t Impulse response functions: Basic ansen model Impulse response functions: ansen with indivisible labor Comparing impulse responses Both models get same impulse Put each set of responses on di erent axis Get Comparing impulse response Rotating so that we don t see the time axis
25 x Y K C r 8 9 Figure : Responses of ansen s basic model x Y K C r 8 9 Figure : Responses for ansen s model with indivisible labor
26 .. Y K C.. r Figure : Responses for both ansen models x Y K C r x Figure : Comparing the response of the two models
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