401ps8answers-10 - ECON 401 Autumn 2010 ANSWERS FOR PROBLEM...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
ECON 401 Hartman Autumn 2010 ANSWERS FOR PROBLEM SET VIII 1.a. The Hamiltonian is 1/2 (2 ) bt a Hec k cg k  . We choose c to maximize H , which gives 1 /0 bt a Hc e a c   , and the evolution of is given by 1/2 /( ) Hk k     . The first of these implies that 1 bt a ea c  so that 12 (1 ) bt a bt a be ac e a a c c  . We can substitute into 1/2 () k  to get 1 1 / 2 ) ( ) bt a bt a bt a be ac e a a c c e ac k  , which simplifies to 1/2 ) c ab k c  or 1/2 1 cb k c a  . To use the calculus of variations, solve 1/2 2 kk c g k  for c and substitute into the expression under the integral to get 1/2 0 ) bt a ekg k k d t  . If we let 1/2 (, , ) ) bt a Gtkk e k g k k  and remember that 1/2 2 ck gk k , then 11 / 2 ) bt a Gkea c k , 1 / bt a Gk ea c , and ) bt a bt a dG be ac e a a c c dt k    . The Euler equation requires that Gd G kd t k , and we can substitute to see that the Euler equation is / 2 1 2 ( 1 ) bt a bt a bt a e ac k be ac e a a c c   . The Euler equation simplifies to 1/2 ) c k c or 1/2 1 k c a , which is the same thing we found using the Hamiltonian. The transversality condition is 0.1 1 lim( ) 0 ta t c  . From the c equation, we see that if 0 c , then 1/2 kb  or 2 . In the graph on the next page, the 0 c locus is the vertical line above *2 kk b . If we move horizontally to the right of the line (i.e., if we hold c fixed at any level and increase k ), then c decreases from 0 c to a negative value because 3/2 1 / 2 ) / ( 1 ) 0 c k a     . (Remember that 01 a .) It follows that c is decreasing to the right of the 0 c line and increasing to the left of the line. From the k equation, we see that if 0 k , then 1/2 2 . For this equation, 1/2 / dc dk k and 22 3 / 2 1 / 2 ) 0 dc d k k . It follows that the equation 1/2 2 has a maximum when 2 k  . Now, b  and b  ; it follows that the maximum of the 0 k curve occurs to the right of . Moreover, if 0 k in this equation, then we see that 1/2 2 crosses the vertical axis at g . The 0 k locus has been drawn and labeled in the graph on the next page; note that the maximum for the 0 k curve occurs to the right of the 0 c line. If we move vertically up from the 0 k curve, then k decreases from 0 k to a negative number because /1 0 kc   . It follows that k increases below the 0 k curve and decreases above that curve. The optimal trajectory is the saddle point trajectory indicated in the graph. The steady state at the saddle point is stable along that trajectory.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/04/2011 for the course ECON 401 taught by Professor Staff during the Spring '08 term at University of Washington.

Page1 / 5

401ps8answers-10 - ECON 401 Autumn 2010 ANSWERS FOR PROBLEM...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online