EC744 Lecture Notes: Economic Dynamics Prof. Jianjun Miao
1 Deterministic Dynamic System State vector x t 2 R n State transition function x t = g x 0 ; t; ; x 0 = x 0 ; parameter 2 R p A parametrized dynamic system: (X; g) where X R n and g : X R R p! X Di erence equation F (t; x t ; x t+1 ; :::; x t+m ; ) = 0 Or x t+m = f (t; x t ; x t+1 ; :::; x t+m 1 ; ) The solution fx t g is an orbit or trajectory Linear if f is linear. Order m
Focus on rst-order system. Why? x t+1 = f (t; x t ; ) Autonomous system if f does not depend on t: The impact of t typically comes from an exogenous forcing term: K t = (1 ) K t 1 + I t If initial value is known x 0 ; it is easy to obtain a solution. Boundary value problem General solution, particular solution Vector eld d x t = x t+1 x t = f (t; x t ; ) x t = d (t; x t ; )
Steady state of an autonomous system x t+1 = f (x t ) if the xed point x such that x = f (x) (Liapunov Stability) The state x is a stable xed point of the map f if for any > 0; there exists some 2 (0; ) such that for all t s: kx s xk < =) kx t xk < (Asymptotic stability) The state x is asymptotically stable if it is stable and the constant in the last de nition can be chosen so that, if kx s xk < for any s; then kx t xk! 0 as t! 1: Basin of attraction, globally asymptotically stable (Figures) Periodic orbits
2 Scalar Linear Equations First-order di erence equation: x t = ax t 1 + b t : Superposition principle: x g t = xc t + x p t is the general solution to the correspond- 1 and x p t is a particular solution. where the complementary function x c t ing homogeneous equation x t = ax t 2.1 Homogeneous equation x t = ax t 1 Phase diagram Four cases (table): monotonic convergence, damped oscillation, monotone divergence, explosive oscillation
General solution x t = ca t Particular solution x t = x s a t s 2.2 Autonomous equations x t = ax t 1 + b By the superposition principle, x g t = cat + x p t : A natural candidate for the particular solution is the steady state x = b 1 a for a 6= 1: So x g t = cat + x:
Boundary value problem x t = x + (x s x) a t s The solution is asymptotically stable if jaj < 1: 2.3 Non-autonomous equations By the superposition principle, x t = ax t 1 + b t : x g t = cat + x p t : Backward-Looking Solution. Repeated substitution x t = a n x t n + n 1 X i=0 a i b t i Let n = 0 if x 0 is given
Let n! 1 if there is no initial condition Assume jaj < 1 and jb t j < B x g t = cat + 1X i=0 a i b t i Forward-Looking Solution. Repeated substitution x t = 1 n x t+n 1 a a n 1 X j=0 1 a j b t+1+j : Let n! 1 and assume jb t j < B: We must have jaj > 1: x g t = cat 1 a 1X j=0 1 j b t+1+j a Stability
3 Stock Market Bubbles 3.1 Price Dynamics with Adaptive Expectations Under adaptive expectations, Eliminating p e t+1, we have (1 + r) p t = d t + p e t+1 p e t+1 = p t + (1 ) p e t: p t = ap t 1 + b t where R = 1 + r; a = R (1 ) R ; b t = d t (1 ) d t 1 R Backward-looking solution p t = a t p 0 + p t
where p t = R t 1 X i=0 a i d t i : Suppose d t = d > 0 for all t: Then The limit is p = d=r: We have p t = 1 r a t d p t p = a t (p 0 p ) What is wrong with adaptive expectations? Eliminating p t ; we have So p e t+1 = ape t + R d p e t p = a t (p e 0 p ) :
We now have So mistake is systematic! p t p e t = a t (p 0 p e 0 ) 3.2 Price Dynamics with Perfect Foresight Under rational expectations, p e t+1 = p t+1: This is also called perfect foresight in the deterministic case. We have Forward-looking solution is When d t = d for all t; Kill bubble: c = 0! p t = cr t {z} p t+1 = Rp t d t : Bubble + 1 R 1X 1 j d t+j j=0 R {z } Fundamental p t = p + cr t = p + (p 0 p ) R t :
3.3 Dividend Taxes General solution p t+1 = Rp t (1 ) d p t () = p () + cr t = p () + (p 0 p ()) R t where p () = (1 ) d : r Unanticipated tax increase from 0 to 1 : to p ( 1 ) and remains this value forever. p ( 1 ) = (1 r 1) d < (1 0) d r The price immediately drops = p ( 0 ) Anticipated tax increase at date T: For t T; the stock price is p t = p ( 1 ) : For t < T; the tax rate is 0 and the solution is p t ( 0 ) = p ( 0 ) + cr t = p ( 0 ) + (p 0 p ( 0 )) R t :
At time T; we must have p T ( 0 ) = p ( 1 ) : So p ( 1 ) = p ( 0 ) + (p 0 p ( 0 )) R T : Solving yields p 0 = p ( 0 ) 1 R t (p ( 0 ) p ( 1 )) : 4 Linear Systems z t+1 = Az t + f t+1 ; where z t 2 R n and A is an n n matrix. We focus on the case of n = 2: " xt+1 y t+1 # = " a11 a 12 a 21 a 22 # " xt y t # + " bt+1 d t+1 # : By the superposition principle z g t = zc t + z p t :
4.1 Homogeneous systems z t+1 = Az t For n = 2; " xt+1 # " a11 a 12 # " xt # y t+1 = a 21 a 22 y t Diagonalizable systems with distinct eigenvalues Find an invertible matrix P such that P 1 AP = D where D is a diagonal matrix. Let 1 and 2 be the eigenvalues of the coe cient matrix A and assume 1 6= 2 : The the corresponding eigenvectors e 1 = (e 11 ; e 12 ) T and e 2 = (e 21 ; e 22 ) T are linearly independent. Let E = (e 1 ; e 2 ) : We have E 1 AE = () AE = E: Now, the system becomes De ne E 1 z t+1 = E 1 z t : ^z t = E 1 z t :
We obtain ^z t+1 = ^z t : The general solution is ^x t = c 1 t 1 ; ^y t = c 2 t 2 : Transform back z t = E^z t Thus, " xt # " e11 e 21 # " c1 t 1 # or y t = e 12 e 22 c 2 t 2 x t = e 11 c 1 t 1 + e 21c 2 t 2 ; y t = e 12 c 1 t 1 + e 22c 2 t 2 We need two boundary conditions to pin down c 1 and c 2 : For complex eigenvalues, we solve for real solutions. sine-cosine terms that generate oscillatory trajectories. The solutions involve Jordan decompo- Repeated eigenvalues and non-diagonalizable systems sition
5 Autonomous systems z t+1 = Az t + b Pick the steady state as the particular solution. z = b (I A) 1 : For n = 2; x t = x + e 11 c 1 t 1 + e 21c 2 t 2 ; y t = y + e 12 c 1 t 1 + e 22c 2 t 2 Stability 1. j 1 j < 1 and j 2 j < 1: Asymptotically stable 2. j 1 j > 1 and j 2 j > 1: Unstable 3. j 1 j > 1 and j 2 j < 1: There is only one saddle path converging to the steady state
To nd the saddle path, we set c 1 = 0: We obtain The saddle path is x t = x + e 21 c 2 t 2 ; y t = y + e 22 c 2 t 2 : x t x = e 21 e 22 (y t y) : The slope of the saddle path is the same as the slope of the eigenvector associated with the stable eigenvalue 2 : Saddle path solutions often make good economic sense. predetermined and x is a jump variable. For example, y is Systems with more than two equations Similar solution method. Suppose we have n variables, s of which are predetermined and k stable eigenvalues. If k = s; then there is a unique saddle path. If k > s; then the equilibrium path is not unique. If k < s; then there is no stable solution.
5.1 Phase Diagrams for Linear Systems Rewrite x t+1 = a + b 1 x t + cy t y t+1 = d + ex t f 1 y t x t = a + bx t + cy t y t = d + ex t fy t where we assume all coe cients are positive. Phaselines: x t = 0 and y t = 0: 5.2 Non-autonomous Systems Let n = 2: Suppose the eigenvalues 1 and 2 are real and distinct. transformed system becomes: The ^x t+1 = 1^x t + ^bt+1 ; ^y t+1 = 2^y t + ^dt+1 :
The general solution is ^x t = c 1 t 1 + ^xp t ^y t = c 2 t 2 + ^yp t where ^x p t and ^yp t are forward or backward-looking solutions. Transforming back, " xt y t # = " e11 e 21 e 12 e 22 # " c1 t 1 + ^xp t c 2 t 2 + ^yp t # : 6 Exchange-Rate Overshooting Dornbusch exchange-rate model. yt d = (e t + p p t ) (r t p t+1 + p t ) (AD) p t+1 p t = yt d y (Phillips curve) m p t = y r t (LM) r t = r + e t+1 e t (UIP) where e t = home/foreign.
Eliminating y d t and r t yields a system for (e t ; p t ) : (e t+1 e t ) = p t + y m r ; (1 ) (p t+1 p t ) = (e t + p p t ) y + (y m + p t) Solving for the steady state p = r + m y; e = p p + 1 (y + r ) : : Transforming this system: " xt+1 y t+1 where x t = e t e; y t = p t p and A = " # = A " xt y t 1 1= # 1 1 (+=) 1 # : Assume Derive the characteristic equation < 1 < (1 + 1=) : z () = 2 T + D = 0
where T is the trace of A and D is the determinant. We can check z (1) < 0: We need z ( 1) > 0; or assume The general solution: < 4 (2 + 1=) ( + 2) : e t = e + c 1 h 11 t 1 + h 21c 2 t 2 ; p t = p + c 1 h 12 t 1 + h 22c 2 t 2 ; where h i = (h i1 ; h i2 ) T is the eigenvector for i : Assume 1 > 1: Then 1 > 2 > 1: Saddle path solution: e t = e + h 21 c 2 t 2 ; p t = p + h 22 c 2 t 2 ; and the saddle path is e t e = h 21 h 22 (p t p) : Use the initial condition to pin down c 2. What is the e ect of a permanent increase in money supply m? Suppose the economy is initially at the steady state. First, compute the new steady state (^p; ^e). Next, study the transition path.
Suppose the price is sticky or predetermined (p 0 = p) but the exchange rate is free to jump. The transition path is p t = ^p (^p p) t 2 ; ^p p e t = ^e + (1 2 ) t 2 : A monetary expansion results in an immediate depreciation of the domestic currency and sustained in ation. The exchange rate overshoots its long-run equilibrium, so that in ation is actually accompanied by a gradual appreciation of the currency during the transition. 7 Nonlinear Systems Consider x t = f (x t 1 ) We say the steady state x = f (x) is hyperbolic if none of the eigenvalues of the Jacobian matrix of the partial derivatives Df (x) has modulus exactly 1.
Linear approximation x t+1 = x + Df (x) (x t x) If all eigenvalue of Df (x) have moduli strictly less than 1, then x is asymptotically stable (a sink). If all eigenvalues have moduli greater than 1, then x is unstable (a source). Otherwise, a saddle. 8 The Structure of Growth Models 8.1 Descriptive Growth: Solow (1956) Model Model Y t = F (K t ; L t )
L t+1 = (1 + n) L t K t+1 = (1 ) K t + I t I t = sy t De ne x t = X t =L t and f (k) = F (K=L; 1) : We then have: k t+1 = h (k t ) where h (k) = (1 ) k + sf (k) 1 + n Two steady states 0 and (n + ) k = sf (k)
8.2 Optimal Growth subject to max 1X t=0 t u (c t ) ; = 1 1 + c t + k t+1 (1 ) k t = f (k t ) Equilibrium system for (k t ; c t ) k t+1 = f (k t ) + (1 ) k t c t u 0 (c t ) = u 0 (c t+1 ) f 0 (k t+1 ) + 1 Steady state c = f (k) k f 0 (k) = + Draw phase diagram
At steady state, we compute Jacobian J = 2 4 @k t+1 @k t @c t+1 @c t @c t+1 @k t+1 @k t @c t 3 5 Total di erentiation dk t+1 = f 0 + 1 dk t dc t f 0 + 1 u 00 dc t+1 + u 0 f 00 dk t+1 = u 00 dc t So dc t+1 = f 00 A dk t + 1 where A = u 00 =u 0 f 00! A dc t Thus, J = " 1= 1 f 00 A 1 f 00 A #
J has trace and determinant T = 1 + 1= f 00 A > 2 D = 1= > 1 Hence T 2 4D (1 + 1=) 2 4= > 0 and the characteristic equation has two positive real roots 2 > 1 > 1 > 0 Saddle 8.3 Overlapping Generations: Diamond (1965) Individuals live for two periods and supply labor inelastically at 1. Saving of the young in period t generates capital that is used in period t + 1 Population growth N t = N 0 (1 + n) t
Individual solves subject to max u (c 1t ) + u (c 2t+1 ) c 1t + s t = w t c 2t+1 = R t+1 s t Solving yields saving function s t = s (w t ; R t ) Firms so max F (K t ; N t ) w t N t r t K t r t = f 0 (k t ) w t = f (k t ) kf 0 (k t ) where r t is the rental rate of capital. So R (k) = f 0 (k) + 1 w (k) = f (k) kf 0 (k)
Equilibrium K t+1 = N t s (w t ; R t+1 ) or (1 + n) k t+1 = s (w (k t ) ; R (k t+1 )) : Stability depends on dk t+1 j ss dk t With government debt or bubbles, we have system for (k t ; b t ) (1 + n) (k t+1 + b t+1 ) = s (w (k t ) ; R (k t+1 )) (1 + n) b t+1 = R (k t ) b t Two types of equilibrium: inside-money equilibrium b t = 0 for all t: Golden rule steady state f 0 (k ) = n + (1 + n) b = s (w (k ) ; 1 + n) (1 + n) k
Draw phase diagram Three steady states: (0; 0) k; 0 ; G = (k ; b ) The steady state G exists if and only if f 0 k < n + or the economy is dynamically ine cient. If f 0 k n +, the economy is dynamically e cient and there cannot be bubbles! The bubble removes ine ency and the economy is at the golden rule.