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14.123 Sample Exam

14.123 Sample Exam - 14.123 Microeconomic Theory III Final...

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14.123 Microeconomic Theory III Final Make Up Exam (80 Minutes) 1. (30 points) This question assesses your understanding of expected utility theory. (a) Show that there exists a preference relation on preferences that satis fi es the in- dependence axiom but is discontinuous. (For an example, you can take the set of consequences as { x, y, z } and consider lexicographic preferences.) Answer: Denote the lotteries by ( p x , p y , 1 p x p y ) . Consider the lexicographic relation p º q [( p x > q x ) or [ p x = q x and p y > q y ]] . ⇐⇒ This is a discontinuous preference relation because the upper counter set for p is { q | q x > p x } { q | q x = p x , q y p y } , which is clearly not a closed set when p is in the interior. To check that it satis fi es the independence axiom, take any p, q, r P and a (0 , 1] . If p q , then p = q (as the indi ff erence sets are singletons), and hence ap +(1 a ) r = aq +(1 a ) r , showing that ap +(1 a ) r aq +(1 a ) r . If p  q , then either p x > q x , in which case ap x + (1 a ) r x > aq x + (1 a ) r x , showing that ap + (1 a ) r  aq + (1 a ) r , or p x = q x and p y > q y , in which case ap x + (1 a ) r x = aq x + (1 a ) r x and ap y + (1 a ) r y > aq y + (1 a ) r y , showing once again that ap + (1 a ) r  aq + (1 a ) r . This shows that if p º q = ap + (1 a ) r º aq + (1 a ) r . Conversely, if p º 6 q , then q  p (by completeness) and as we just shows this imples that aq +(1 a ) r  ap +(1 a ) r , showing that ap + (1 a ) r º 6 aq + (1 a ) r . (b) Under Postulates P1-5 of Savage, consider the as likely as relation ˙ between events, derived from betting preferences as in the class. Consider events A and B such that A ˙ S \ A and B ˙ S \ B , where S is the state space. Show that A ˙ B . Answer: Recall from the class that under P1-5, º ˙ is a qualitative probability. In particular, if C, D, E are disjoint events, C  ˙ D ⇐⇒ C E  ˙ D E. (1) Now, for simplicity of notation, let W = A B , X = B \ C , Y = A \ B , and Z = S \ ( A B ) . Suppose for a contradiction that A  ˙ B . Then, by (1), Y  ˙ X .
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