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ECON 3410/4410. H 2008, ordinary exam. Notes to Q1 1. (a) i. (1) holds in period = 1 and + 1 : 1 2 = 0 + 1 1 + 2 0 + 0 + 1 , = 0 + 1 2 + 2 1 + 1 + 2 , with 0 and 0 as initial values.. A temporary shock to the variable that occurs in period = 1. 1 1 2 1 = 1 , first = 2 + 1 = 2 + 1 , second 1 ii. A permanent change in occurring in period = 1. Set for example 1 = 2 = 1 2 = 1 , first = 1 + 2 + 1 = 1 + 2 + 1 , second Method: sum dynamic multipliers, or define and therefore , as functions of a continuous variable. (b) 1 2 3 = = = = = = 0 + + 0 0 + + 1 0 (1 + ) + (1 + ) + 2 0 ( 0 + )(1 + ) + 2 0 0 + + 2 ( 0 + )(1 + + 2 ) + 3 0 . with = 1 + 2 . (c) “Globally asymptotically stable” means that in the long-run, the solution is independent of the initial condition, and that it converges to a steady-state. If we start in period = 1, and simplify by setting 1 = 0, 2 = 0, , the stable solution converges asymptotically to the steady state ∗ which is independent of 0 , and where ∆∗ = , independent of . The condition for this kind of stability is −1 1. = 1 and = −1 give unstable solutions. || 1 gives explosive solutions. 1 (d) In the stable case, and using = −1 = , = 0 + 1 + 2 + (e) ( + 2 ) = 1 . 1− ¯ ¯ ¯ ¯ (f) Assuming 0 1 , then ¯ ¯ | 1 | requires that 1 and 2 have opposite signs. One example of sign change may be the eﬀect of an increase in the rate of unemployment ¯ on¯ wages (“insider wage ¯ ¯ setting”). If −1 0, we can have ¯ ¯ | 1 | also when 1 and 2 have the same sign. (g) The two graphs in the figure strongly suggest that the two solution paths after a while converges to one single path. This is consistent with stability, since the impact of the two diﬀerent initial situations disappear. That the solutions are trending, must then be due to growth in the exogenous variable. The identical growth rates, asymptotically, is also consistent with stable solutions. 2