145hw4W10answer

145hw4W10answer - Econ 145: Problem Set IV Suggested Answer...

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Econ 145: Problem Set IV Suggested Answer March 13, 2010 1. (A.15-1, SSS) You are one of n bidders who draw i.i.d. valuations from F ( x ). The probability that your draw x is the highest is F ( x ) n 1 . Then what is the probability that your draw is k th highest? What are these probabilities when each draw follow the uniform distribution on [0 , 1]? Answer: Denote { v 1 ,v 2 , ··· n 1 n } are descendingly ordered draws among n bidders. I.e. v 1 denotes the highest draw and v n denotes the lowest draw. Assuming your draw is k -th highest and denote your draw as x (i.e. x = v k ). This means there are ( k 1) draws which are above than yours, and there are ( n k )draws wh icharebe lowthanyours . v n v n 1 v k +1 | {z } ( n k ) draws below your draw x |{z} your draw = k -th highest dras = v k v k 1 v 2 v 1 | {z } ( k 1) drasws above your draw First, we consider following two probabilities without considering combinations. (i) Probability of ( k 1) draws above your draw x Pr [( k 1) draws are above your draw x ]= [( v k 1 >x )and and ( v 2 )and( v 1 )] = [( v k 1 )] ····· [( v 2 )] · [( v 1 )] = { 1 F ( x ) } k 1 Note that the second equality comes from the independence of draw. (ii) Probability of ( n k ) draws are below your draw x [( n k ) draws are below your draw x [( v n x v n 1 x and ( v k 1 x )] = [( v n x )] · [( v n 1 x )] [( v k 1 x )] = { F ( x ) } n k Again, the second equality comes from the independence of draw. Second, we consider the combinations. We know that there are ( k 1) draws above your draw x .out of totally ( n 1) draws (note: excluding your draw). To choose such ( k 1) draws out of ( n 1), there are μ n 1 k 1 combinations to do so 1 . Once you calculate “above” combinations, ”below” combinations are automati- cally determined. Since we have to consider all combinations, the probability of your draw is k th highest 1 Some math textbook use the notation of n 1 C k 1 to express combinations 1
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is Pr [your draw x is k -th highest] = μ n 1 k 1 ·{ 1 F ( x ) } k 1 1 F ( x ) } n k See footnote 2 . 2. Consider a second price auction with n bidders. (a) Compute the seller’s expected revenue (in symmetric Bayesian Nash equilibrium) when each bidder’s value is independently and uniformly distributed on [0 , 2] . Note: CDF of Order Statistic There are n bidders. Assume valuation v is drawn independently from the Cumulative Distribution Function (henceforce CDF) F ( x ). (1) G 1 ,n ( x ): CDF of highest valuation among n draws G 1 ,n ( v )= { F ( x ) } n = F ( x ) | {z } prob that a draw is below x n The interpretation is (very) easy. To make the highest draw be less than or equal to x ,a l l n inde- pendent draws have to be smaller than x . (2) G 2 ,n ( x ): CDF of highest valuation among n draws G 2 ,n ( x { F ( x ) } n + n (1 F ( x )) { F ( x ) } n 1 = { F ( x ) } n | {z } (i) + n (1 F ( x )) { F ( x ) } n 1 | {z } (ii) = { F ( x ) } n + n |{z} n possibilities , · (1 F ( x )) | {z } one draw is above x, F ( x ) } n 1 | {z } ( n 1) draws are below x The interpretation of this CDF is little bit tricky.
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145hw4W10answer - Econ 145: Problem Set IV Suggested Answer...

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