2012-2013 Master 2 Macro I Lecture notes #12 : Solving Dynamic Rational Expectations Models Franck Portier (based on Gilles Saint-Paul lecture notes) franck.portier@tse-fr.eu Toulouse School of Economics Version 1.1 25/11/2012 Changes from version 1.0 are in red 1 / 29
Disclaimer These are the slides I am using in class. They are not self-contained, do not always constitute original material and do contain some cut and paste pieces from various sources that I am not always explicitly referring to (not on purpose but because it takes time). Therefore, they are not intended to be used outside of the course or to be distributed. Thank you for signalling me typos or mistakes at franck.portier@tse-fr.eu. 2 / 29
Introduction Economic agents have to make a decision based on their forecast of some future variable. In the most general case, the entire probability distribution of that variable will affect their optimal choice. The rational expectations hypothesis states that people take into account the actual distribution of the relevant variable they want to forecast. If expectations are rational, we have to solve simultaneously for the equilibrium distribution and for the rest of the model We restrict to linear models in which the distribution of endogenous variables of interest only matters through its first moment, i.e. its mathematical expectation. 3 / 29
1. A simple example Model Agricultural market where producers have to decide how much to produce prior to knowing the equilibrium price yt d = a bp t + ε dt yt s = c + dpt a + ε st, y s t = y d t. Here, p =price,ε d =demand shock, ε s =supply shock,y s =supply, y d =demand,p a = anticipated price. 4 / 29
1. A simple example Rational expectations The Rational Expectations Hypothesis assumes that p a t = E(p t I t ), where E = mathematical expectation operator and I t = information set of the price-setters when they set y s t. Typically, I t = {past values of all the variables, the model, how to solve it} If farmers set y s t at date t 1, : p a t = E t 1 p t. 5 / 29
1. A simple example Solving the model General principle : take expectations in all the equations and then solve for the expected variables, then substitute this in all the equations where expectations enter, this allows to solve for the actual equilibrium variables. Let us try this here. 6 / 29
1. A simple example Solving the model (continued) yt d = a bp t + ε dt yt s = c + dpt a + ε st, y s t = y d t. The model can be collapsed to a bp t + ε dt = c + de t 1 p t + ε st. (1) Take expectations : a be t 1 p t + E t 1 ε dt = c + de t 1 p t + E t 1 ε st. Thus E t 1 p t = a c + E t 1(ε dt ε st ). b + d 7 / 29
1. A simple example Solving the model (continued) substitute this into (1) to solve for p t : p t = 1 b (ε d ε s ) This is the solution. d b(b + d) E t 1(ε d ε s ) + a c b + d. (2) 8 / 29
1. A simple example The method of undetermined coefficients If we know the distribution of those shocks, we can use another method called the method of undetermined coefficients This is akin to the notion of Markov perfect equilibrium in game theory. Method : Identify the relevant state space Look for a solution as a (linear) function of this state vector Given such a function, I can compute all the expectations I need Subsitute this function and the appropriate formula for expectations in all places where it is needed Identify the coefficients on each state variable This gives us equations in the coefficients that allow us to compute them 9 / 29
1. A simple example The method of undetermined coefficients (continued) Assume E t 1 ε dt = E t 1 ε st = 0. The relevant state space is (ε d, ε s ). Look for a solution of the form p t = αε dt + βε st + γ. 10 / 29
1. A simple example The method of undetermined coefficients (continued) Under the assumed solution, we have Substituting into (1) we get E t 1 p t = γ. a b (αε dt + βε st + γ) + ε dt = c + dγ + ε st. This must hold for all realizations of ε d and ε s and therefore we must have a bγ = c + dγ αb + 1 = 0 βb = 1. 11 / 29
1. A simple example The method of undetermined coefficients (continued) Therefore Hence the solution γ = a c b + d α = 1/b β = 1/b. p t = 1 b (ε d ε s ) + a c b + d, which is equivalent to (2) if E t 1 ε dt = E t 1 ε st = 0. 12 / 29
2. A dynamic model Model Suppose now the supply curve is y s t = c + de t p t+1 + ε st Eliminating y between supply and demand : p t = a c b + d b E tp t+1 + ε dt ε st b and assuming for simplicity that the constant terms are zero, we end up with p t = ρe t p t+1 + η t, (3) where η t is a shock. 13 / 29
2. A dynamic model Solution by forward integration (3) implies p t+i = ρe t p t+i+1 + η t Taking expectations i E t p t+i = ρe t p t+i+1 + E t η t+i Substitute forward (forward integration) : k 1 p t = ρ k E t p t+k + ρ j E t η t+j (4) Assume stationary shocks and impose E t p t+k is bounded k. (stochastic equivalent of selecting the non-explosive saddle-path as the unique solution.) Then if ρ < 1, then lim k ρ k E t p t+k = 0, hence p t = ρ j E t η t+j. j=0 j=0 14 / 29
2. A dynamic model Solution by forward integration (continued) p t = ρ j E t η t+j. j=0 This is the unique stationary solution to the rational expectations equation (3). Suppose ρ > 1, then the solution cannot be unique. For any solution p t, p t + x t is also a solution under some conditions on x x t is a random variable (sunspot) satisfying x t+1 = 1 ρ x t + u t+1, (5) with u any shock such that E t u t+1 = 0. Furthermore, we can construct at least one non explosive solution by picking p t+1 = 1 ρ (p t η t ). Hence in this case we have a continuum of equilibria. 15 / 29
2. A dynamic model Undetermined coefficients We can apply the method of undetermined coefficients to this problem if we know the distribution of η. Assume η follows an AR1 : η t = αη t 1 + ω t, with ω t iid and with zero mean. The only relevant state variables are η t 1 and ω t so we look for a solution of the form Then p t = λη t 1 + βω t. E t p t+1 = E t (λη t + βω t+1 ) = λη t = λ(αη t 1 + ω t ). 16 / 29
2. A dynamic model Undetermined coefficients (continued) We can then rewrite (3) as λη t 1 + βω t = ρλ(αη t 1 + ω t ) + αη t 1 + ω t. Identifying, we get λ = ρλα + α β = ρλ + 1 Hence λ = β = α 1 ρα ; 1 1 ρα. 17 / 29
2. A dynamic model Undetermined coefficients (continued) We can check that if ρ < 1, this is indeed correct : ρ j E t η t+j. = j=0 ρ j α j η t j=0 1 = 1 ρα (αη t 1 + ω t ) = λη t 1 + βω t If ρ > 1, this is still a solution, and it is the unique linear Markov solution, i.e. the only one which is a linear combination of the relevant state space. But many other solutions can be constructed by adding a sunspot x satisfying (5). 18 / 29
3. A general model Model General linear model (Blanchard-Kahn) ( Kt X t ) ( Kt 1 = A E t X t+1 ) + U t, (6) K t is an m 1 vector of state (=predetermined) variables, meaning that K t 1 is fixed at the beginning of period t, X t is an n 1 vector of non-predetermined variables, i.e. expected future values of those variables, rather than lagged ones, affect current outcomes. A is an (m + n) (m + n) matrix U an (m + n) 1 vector of shocks. 19 / 29
3. A general model Model transformation Let then ( A00 A A = 01 A 10 A 11 ), U t = ( U0t U 1t ), K t = A 00 K t 1 + A 01 E t X t+1 + U 0t X t = A 10 K t 1 + A 11 E t X t+1 + U 1t. We can solve for E t X t+1 and write the whole system with backward-looking dynamics : ) ( A00 A = 01 A 1 11 A 10 A 01 A 1 ) ( ) 11 Kt 1 E t X t+1 A 1 11 A 10 A 1 11 X t ( U0t A + 01 A 1 11 U ) 1t ( Kt = M A 1 11 U 1t ( Kt 1 X t ) + V t. (7) 20 / 29
3. A general model The diagonal case ( Kt E t X t+1 ) = M ( Kt 1 X t ) + V t. (7) Let us start with the simple case where M is diagonal, say ( ) d 1 0... 0 D0 0 M =, where D 0 D 0 = 0 d 2...... 1......... 0 and 0... 0 d m D 1 = d 1 0... 0 0 d 2............... 0 0... 0 d n and V t = ( V0t V 1t ) 21 / 29
3. A general model The diagonal case (continued) Then, iterating (7), K t+i = D i+1 0 K t 1 + i j=0 For K t+i to remain bounded, we need Similarly D i j 0 V 0t+j. d i < 1, i = 1,..., m. (8) i 1 E t X t+i = D1X i t + D i 1 j 1 E t V 1t+j j=0 i 1 = D1(X i t + D 1 j 1 E t V 1t+j ). j=0 22 / 29
3. A general model The diagonal case (continued) Consider first a k such that d k < 1. Let x kt (resp. v kt ) be the k-th term in X t (resp. V 1t ). Then i 1 E t x t+i = (d k )i x t + (d k )i 1 j E t v k,t+j (d 1 k j=0 i 1 ) i E t x t+i j=0 (d 1 k ) 1+j E t v k,t+j = x t. This is similar to (4) and since d 1 k > 1 there is no unique initial value of x t yielding a non-explosive path for subsequent expectations. 23 / 29
3. A general model The diagonal case (continued) Second, assume conversely that d k > 1 for k = 1,..., n. (9) Then all the terms in D1 i go to infinity as i. Thus it must be that lim i + X t + i 1 j=0 D 1 j 1 E t V 1t+j = 0. Therefore, ( ) X t = D 1 1+j 1 Et V 1t+j. j=0 24 / 29
3. A general model The diagonal case (continued) d i < 1, i = 1,..., m. (8) d k > 1 for k = 1,..., n. (9) To summarize To have a unique non-explosive solution, (8) and (9) must hold If (8) is violated, all solutions are explosive If (9) is violated, there is a continuum of non-explosive solutions. 25 / 29
3. A general model The non diagonal case Assume now that the matrix M is not diagonal. The idea is to transform the model into a diagonal one. Generically, M is diagonalizable in the complex plane C. Thus there exists P invertible such that M = P 1 DP, with D diagonal : d 0 0............ 0 0........................ d m 0......... ( D =...... 0 d m+1 0...... D0 0 =......... 0......... 0 D 1.................. 0 0............ 0 d m+n ) 26 / 29
3. A general model The non diagonal case (continued) The diagonal coefficients are the eigenvalues of M. Eigenvalues are ranked by ascending module : d 0 d 1... d m d 1 d 2... d n. (7) rewrite P ( Kt E t X t+1 Let Z t = P ( Kt 1 X t ) ; ) = D ( ( Kt P X t )) + PV t. We are back to the diagonal case with this change of variable. 27 / 29
4. How general is the general model? More lags for K Suppose that we have terms in K t k, k > 1. The solution is to add the variables (K t 1,..., K t k+1 ) to the state vector. Suppose we have this equation K t = ak t 1 + bk t k Let us assume to save on notation that K is a scalar. K t Then the vector K t 1... satisfies K t k+1 K t K t 1... K t k+1 = a 0... b 1 0...... 0......... 0 0 1 0 which only involves the first lag of this vector. K t 1 K t 2... K t k, 28 / 29
4. How general is the general model? More lags for X Suppose we have terms in E t X t+k, k > 1. Let s add E t X t+1,..., E t X t+k 1 to the jump variables Suppose that we have X t = ae t X t+1 + be t X t+k. Then the vector Ψ t = (X t, E t X t+1,..., E t X t+k 1 ) satisfies X t a 0... b E t X t+1 Ψ t = E t X t+1... = 1 0...... E t X t+2 0............ E t X t+k 1 0 0 1 0 E t X t+k = a 0... b 1 0...... 0......... 0 0 1 0 E tψ t+1. 29 / 29