GRAD MUSICIANSHIP I
GRAD MUSICIANSHIP I MUS 204A
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Date Created: 10/22/15
Note on Linearized Solutions to the Optimal Growth Model Econ 204A Prof Bohn This note reviews the linearized dynamics of the optimal growth model and derives loglinearized solutions General Problem Linearization Linearization is a common approach in macroeconomics to obtain approximate solutions The basic principle is the Taylor series approximation Consider rst the univariate case Suppose y hx is a twice differentiable function that you want to linearize around point 7c Taylor s Theorem says that hx h70WOTquot96 9hquot 3996 932 where 5 is between X and 7c The last term is known as the error term and it is often written as Ox f2to highlight that the error grows with the square of the distance between X and 7c A linear approximation is obtained by omitting the error term hx 2 M h39Gc x f Note that this relationship not an equation but an approximation Taylor s theorem applies similarly to multivariate functions Suppose y hxlxn is a twice differentiable function in ER that you want to linearize at a point 7c ENHWTCH Taylor s theorem says 8h 3x hxlxn ho 0 x 7c OHx 2H where the last term an unspeci ed error that is a bounded multiple of the Euclidian distance between x1 x and 7c that is has the same order of magnitude O As approximation this is written as Bh ax x39x x hxlxn zh7c Application to Optimal Growth In the optimal growth model the nonlinear differential equations for k and c are l fk c ng6k c f k5p9gc The steady state conditions are f k 6 p 9g and cfk ng5k In logarithms one can write I k fkk ck ng6 ew e W eWHW n g 6 hl 1nk 1nc c39 c d2 grew 6 p eg hz Once The Taylor series approximation at c k c k simpli es because hl 11109 1118 0 and hz 1nk 0 The relevant partial derivatives are 9quot B 111k 10 1 1k 175 IHUV 10 1 1k Bln1k31nkfe 9 n enc n l fi39en e fe quot e quot f f39 lnClnr elncelnr f and 91mg 2 9135 6 an n u n n u BlnZr 913k if 81 km f 910359103 f kk Evaluating them at CiCl one obtains gzf39 1n 1nzr 3 lingo 31mg and fz dmzr Collecting c and kterms the loglinearized equations can be written in matrix form as I k i 1 k l kquot N 1 kkquot i 3 16 n n A n forAB I5 c39c K 0 Inc Inc 1ncc K 0 where K o f quotkk gt0 This is a system of homogenous linear differential equations with constant coefficients General Problem Solving Systems of Linear Differential Equations In general consider a system of n linear differential equations written in vector notation as y A y Here y denotes an n dimensional vectors of variables y1 yn ydot the corresponding vector of time derivatives and A is an nbyn matrix of xed coef cients One can show that the general solution for each of the variables is a linear combination of exponential functions that involve the eigenvalues and eigenvectors of the matrix A The proof would require too much linear algebra to be worth doing See the appendix of BarroSalaiMartin De nitions An eigenvector v of a matrix A is a nonzero vector such that Av is proportional to v with some proportionality factor 1 which is a scalar number The value 11 is called the eigenvalue associated with the eigenvector v In algebraic terms Avuv implies A lLb Iv0 which implies that the nbyn matrix A ILL I must have a zero determinant This determinant is an n th order polynomial in it which has n roots real or complex perhaps some identical ones For this reason an ndimensional matrix u generally has n 39 and 39 The 39 39 are commonly obtained by solving the characteristic equation I A ILL I 0 Math Result If the eigenvalues are real and distinct denoted u1 uquot the general solution for each variable yj is a sum of exponentials yjt2L1vljeuquott where the coef cients vij depend on boundary conditions Sketch of the Proof The key to the proof is to transform the vector system into a set of separate univariate linear differential equations each of which is solved by an exponential function That is assemble the eigenvectors of A in a matrix V v1vn and assemble the associated eigenvalues into a diagonal matrix M Because A Vl Ji Vl Vl ul Read lbyn vector A vi lbyn vector vi ui for all i one obtains the matrix equationA V V M Read nbyn matrix A V n byn matrix V M Hint Write out what the elements are in the example Let me assert without proof that V is invertible ifthe eigenvalues are distinct Then VT1 A V M also VilV V V71 I De ne the vector z V71 y which is a set of 11 linear combinations of yvariables and consider the original linear system dydt A y Premultiply the system by V71 and insert V V71 as needed one nds with brackets inserted for clarity V 1V 1AVV 1y V 1AVV 1yV 1V MzV 1VMz M z a tit equations 21 ul 21 The solutions are zit zi0 eul39t Transforming back z V71 y implies Recall that M is a diagonal matrix Hence M z is simply a list of 11 separate differential y V z which means that each yvariable is a linear combination of zvariables ie a linear combination of eXponential functions Application to Optimal Growth In the growth model we have n2 y1lnk lnk y2 lnc lnc and A Aalj where a11 gt0 an c k lt0 a21 K lt0 and an 0 The condition IA u I 0 yields a quadratic equation for the eigenvalues p4 il2 an Xazz I a21a12 H2 an a21a12 0 The solutions are a i 2 I 132 12 i an 61216112 7iquot 7 78 Because Kgt0 the root is greater than 32 in absolute value Hence u1lt0 and u2gtB The general solutions can be written as t W 2 d t W Hzf y1 p11 9 p12 9 an y2 p21 8 p22 8 with undetermined coefficients pm that are determined by the problem s boundary conditions Because eW diverges to in nity with u2gt0 convergence to a steady state immediately implies zero coefficients on the second eXponential term ie p12 p22 0 The given initial value k0 C0 determines y10 1nk0 1nk p11 e p11 hence y1t lnko 1nkquot W for all t The remaining coefficient p21 can be determined by combining the solution for capital which implies 1 ln with the original differential equation for capital 3 lnZQ lmze Lt1lne gt lnzef7 u1nze At t0 one obtains p21 1nc0 1ncquot gm tn lnk0 1nkquot gm Lt1p11 Because 3 1 2 this can be written more concisely as p21 p11 Leuz This ratio can also be e interpreted as the slope of the saddle path More Intuition What are the transformed variables in the Growth Model application This section is supplementaliintendedfor students who don tfind matrix equations insightful You may skip this section ifyou don tfind it insightful In the linearized growth model the question answered by the zvariables is the following Can we nd linear combinations of y1lnklnk and y2lnclnc such that the timederivative of each linear combination depends only on its own level and not on the other variable To explore if this is possible let s de ne variables 21 y1 51 y2 for some yetundetermined coef cients il2 We are looking for values so that dzidt is a function of 2 only If we can such values then each 2 is a singlevariable differential equation that we can solve To nd such linear combinations we use the method of undetermined coe icients that is we postulate that linear combinations with unknown coef cients exist and then compute what they are This approach is a commonlyused general method here introduced by example Drop subscripti in the following when a generic z is considered In the application we know that y any1 any2 and y any1 Hence 2 y1 2 a11 a21y1 alzyz a11 a21z y2 alzyz a11 a21z 6112 a11 a21 l39y2 The variable y drops out if and only if a11 a21 a12 0 This condition is a quadratic equation in g which has two solutions 512 i 2 g Tii1T jc2 De ne the variables zl lnk 51 lnc and zz lnk 52 lnc as the linear combinations of lnk and lnc with these speci c values De ne lul all i a21 il2 Then 21 and 22 satisfy separate differential equations 21 1 21 and 22 uz 22 Note that the coef cients 1L1 and Hi can also be expressed as lul an l and that 11 2 5 5 2 l 12 a11 12a21 7i Vail a21a12 5i 3 The two equations 21 1 21 and 22 uz 22 are each homogeneous linear differential equations with xed coef cients Their solutions have the form z1t z10 em and 12t 120 eJzquot Thus we have shown that two different linear combinations of lnk and lnc have exponential solutions Now we can transform back to the original variables Because z1y1 1y2 and Z2 3 1 2 y2 one obtains Y2 and Y1 Z1 139YZ 75351 ffiglzz Hence y1t lnk lnk Eff z10e 1t Efjg1 z20e zt 1 and 3 201I1C 1I1C z10eWEZL 1 z20e 2t 2 This con rms that the solutions for capital and consumption are indeed linear combinations of exponential functions Moreover one can see how the weights one the exponentials are related to initial conditions Finally one may con rm that 1L1 and 112 are the eigenvalues of A To see this write out the equation for eigenvalues 11 lvl 12 2 0IA lvl39 J lvl 11J 12 21 21 This quadratic equation is indeed solved by the values computed above Check for yourself Also note that the product of the eigenvalues equals the determinant of A here A a12a21 uluz Back to matrix analysis When we de ned 21 22 in terms of some unknowns 1 2 we were in effect 7 7 l assuming that V 1 could be wr1tten as a 11st of rowvectors Lg For V 1 1 1 matrix 1nvers1on Z yields V E 1 21 71151 Looking back to the solutions 1 and 2 one can see that the weights on 27 l i the exponential functions in y1 and yz can be interpreted as the elements of V To con rm that V is indeed the matrix of eigenvectors let s compute VT1 A V M and con rm that the result is a diagonal matrix with eigenvalues one nds Vil39Az 1 3951 11 12 11E1 21 12 12 51 12 1 3952 21 0 11E2 21 12 12 52 12 using a11 ia21 6112 an l Moreover 71 39 1 12 51 12 52 quot51 12 l12 51 1 0 12 51 0 1 0 V A V E2 E1 121252 212 71 1 152751 0 175152 0 a gzj 0 HZ using 5112 ul In the direct calculations we in effect ended up computing the eigenvectors through the method of undetermined coef cients Going through these calculations one should suf ce to obtain the intuition The standard approach to solving systems of differential equations is to calculate the eigenvalues of A from the characteristic equation IA uI 0 and then assert immediately that the solutions have the form y jt2in1vijeu quott This avoids the tedious calculations required to compute V The last step is to exploit the boundary conditions In the growth model u1lt0 and u2gt0 Hence convergence to a steady state requires 2200 The initial capital stock implies yl01nk01nkz10 There are two alternative ways to compute yz If one has calculated V and already knows 1 2 it s easy to calculate 3220 lnc0 lnC E L z10 Indeed note that Z l y2t39y1t39y1t y1ltrgt holds for all t This provides the slope of the loglinearized saddlepath If V has not been computed most common in applications one computes y20 by combining the solution to the other variable With one of the original differential equations For example L121 y1t ul y2t also implies 3 20 11393 10 lla139y1t ll39 1I7115ing H1J2 12 21