Problem Set 2: Sketch of Answers HEC Lausanne, Département d économie politique Business Cycles 2003 Prof. Aude Pommeret Ivan Jaccard April 30, 2004 Part I: Open questions 1. Explain why the consensus that prevailed in macroeconomics until the early 190 s broke down and discuss the in uence of the main theoretical contributions that are often associated with this breakdown? (10 lines max.) Example of Answer: Throughout the 1960 s the negative relation between the level of unemployment and the level of in ation provided a reliable guide to the joint movements in unemployment and in ation and this combination of an apparently reliable empirical relation, together with a plausible story to explain it, led to the adoption of the now called "initial Phillips" curve by macroeconomists and policy makers. From 190 on, however, the relation broke down. There are two main reasons. Firstly, in 193, the World economy was hit by a series of negative supply shocks. As a result of the oil embargo stemming from the Arab-Israeli war of 193, the OPEC was able to engineer a quadrupling of oil prices by restricting oil production. Secondly, in the U.S. another factor was the termination of wage and price controls in 193 and 194, which led to a push by workers to obtain wage increases that have been prevented by the controls. The thrust of these events caused the aggregate supply curve to shift sharply leftward, which in turn led to the brokedownof the traditional in ation-unemployment trade-o by generating a high level of unemployment together with a high rate of in ation (stag ation). In the late 1960 s precisely as the original Phillips curve relation was working like a charm, Milton Friedman (1968) and Edmund Phelps (1968) argued that the appearance of a trade-o between in ation and unemployment was an illusion because the equilibrium level of unemployment should depend on labor supply, labor demand and other microeconomic considerations, not on the average rate of money growth. A few years later, the original Phillips curve started to disappear, in exactly the way Friedman and Phelps had
predicted. Moreover, in his famous paper Econometric Policy Evaluation: A Critique, Robert Lucas (196) presented an argument that had devastating implications for the usefulness of conventional econometric models in evaluating policy. Because the way in which expectations are formed changes when the behavior of forecasted variables changes, nowadays most macroeconomists try to build models on rm microeconomic foundations and assume rational expectations. 2. What is the impact of introducing rational expectations on the predictions of the expectations-augmented Phillips curve of Friedman and Phelps? (10 lines max.) Example of Answer: Once rational expectations are introduced in the expectation-augmented Phillips curve, the trade-o between unemployment and in ation in levels disappear. To see this, consider the following speci cation 1 : ¼ t =¼ e t +(¹+z) u t where¼ e t is the expected rate of in ation,u t the level of unemployment,¹a markup due to imperfect competition andz a variable a ecting the labor supply. As long as expectations are formed according to: ¼ e t =µ¼ t 1 where µ measures the sensibility of anticipated in ation to past in ation and when µ is set to 1, we obtain the expectations-augmented Phillips curve which implies that the unemployment rate a ects the change in the in ation rate. ¼ t ¼ t 1 = (¹+z) u t However, once rational expectations are introduced in the model: the model implies that: ¼ e t =¼ t u t =u n = ¹+z whereu n denotes the natural rate of unemployment. As a result, once rational expectations are assumed, there is no way to maintain a low unemployment rate by generating higher rate of in ation and the level of unemployment is solely determined by the structural parameters ¹; z and : Hence, as argued by Lucas(196), once rational expectations are introduced in the expectation-augmented Phillips curve, only unanticipated changes in money should a ect output and predictable movements in money should have no e ect 1 A formal derivation of this condition can be found in Blanchard (2000) p.149. 2
on activity. The implication of rational expectations is thus that Keynesian models are not able to explain business uctuations. 2. What are the main empirical weaknesses of the RBC benchmark model? Example of Answer: A detailed answer can be found in the article of Jean-Pierre Danthine (199) In a search of a successor to IS-LM, section III, anomalies and puzzles References: ² O. Blanchard (2000), Macroeconomics, Prentice Hall ed. ² M. Friedman (1968), The Role of Monetary Policy, March 1968, American Economic Review 58-11, 1-1 ² R. Lucas (196), Econometric Policy Evaluation: A Critique, Carnegie-Rochester Conference on Public Policy pp. 19-46 ² G. Mankiw (1990), A Quick Refresher Course inmacroeconomics NBER Working Paper 3256 ² S. Mishkin (2000), The Economics of Money, Banking and Financial Markets, Addison Wesley Longman ed. ² E. Phelps (1968), Money-Wage Dynamics and Labor-Market Equilibrium, Journal of Political Economy, August 1968 68-11 3
Part II: Samuelson s accelerator-multiplier model a) Characterize the combinations of and that make the steady-state: (i) stable and unstable, (ii) those for which the model displays oscillations The model can be written as: y t =( + )y t 1 y t 2 This second order di erence equationcan be written as a linear rst-order homogenous system of equations ony t = (y t ;y t 1 ) 0 : with: A = Y t =AY t 1 + 1 0 With two initial conditions on Y the solution to this system can be computed by recursive substitution to yield: Y t =A t Y 0 The stability of this solution depends on the eigenvalues of A. These can be obtained by solving the characteristic equation: 0 = 0 = ja Ij + 1 0 = 2 ( + ) + 1; 2 = ( + ) p ( + ) 2 4 2 Case 1: Distinct real roots ( + ) 2 4 >0 Considering only the set of parameters for which ; >0(remember that this is the accelerator model), we can see that 1; 2>0: If we call dominant root the root with the higher absolute value, the time path will be convergent-whatever the initial conditions may be-if and only if the dominant root is less than 1 in absolute value. 4
If 1; 2<1 the system is stable from all initial conditions (this is sometimes called a sink). If 2< 1 1 there is part of the system dynamics that is stable and part that is unstable from arbitrary initial conditions (this is sometimes called a saddle path). Case 2: Repeated real roots Then: Therefore if: ( + ) 4 = 0 = p 1 both roots are unstable (the system is sometimes called a source ). But both roots are stable otherwise. Case 3: Complex roots ( + ) 4 <0 It can be shown that if < 1 then the system has an oscillatory convergence towards the steady state. But if 1 the system is a source with unstable oscillations. b) Consider the special case where = 0:9 and = 0:05:What are the eigenvalues? The eigenvalues of the system are 0.8941 and 0.0559. c) For this set of parameter, nd the set of initial conditions ony 1 andy 0 that yield convergence towards a steady-state. Is this set of solution unique? The general solution is: y t =A 1 t1 +A 2 t2 Since both eigenvalues are less than unity, the system is a sink and any two initial conditions ony 1 andy 0 lead to the same steady state. Therefore, the set of conditions ona 1 anda 2 is in nite. d) The investment function becomes: i t = (y t 1 y t 2 )+u t The dynamic equation that characterizes the system is: 5
y t = ( + )y t 1 y t 2 +u t To obtain an autoregressive representation for output, we substitute foru t to get: y t = ( + )y t 1 y t 2 +½u t 1 +e t e) Write the 2 above 3 process 2for output 3 in matrix 2 3form as: S2 t =MS t 1 +be t y t y t 1 1 LetS t = 6y t 1 4y t 2 5 ;S t 1 = 6y t 2 4y t 3 5 ;b = 6 0 4 0 5 andm = 6 4 u t u t 1 1 ( + ) 0 ½ 1 0 0 0 0 1 0 0 0 0 0 ½ f) Use your de nition of the vector S above to write the Samuelson model in matrix form. De ne: = 2 4 0 0 0 0 1 1 0 0 0 g) Describe how to compute the conditional responses toe t : Using the notation above, we can write: 3 5 3 5 Therefore: S t+k =M k+1 S t 1 + kx M i be t+k i i=0 and: E t (S t+k ) =M k+1 S t 1 +M k be t So that: E t 1 (S t+k ) =M k+1 S t 1 and: E t (S t+k ) E t 1 (S t+k ) =M k be t E t (Z t+k ) E t 1 (Z t+k ) = M k be t See The mathlab codes in annex and the response of consumption, investment and output to a unit investment shock. 6