Problem 2: Order Statistics Suppose that we have n = 2 bidders whose values v; are...

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Problem 2: Order Statistics Suppose that we have n = 2 bidders whose values v; are independently distributed uniformly on [0, 100]. (This is typically denoted vid U[0, 100].) Let v() denote the highest value and v(2) denote the second-highest value (the so-called first and second order statistics). Since the values are themselves random, v() and v(2) are also random variables. We saw in class that, in any efficient auction, E [v()] determines the total surplus, E [v()] determines the seller's expected revenue, and E [v()] -E[v(2)] determines the bidder's expected surplus. We also stated special cases of the general fact that and E [v()] = = 2 100 1 E[v(2)] = 100. 3 Your task: Derive these formulae. You may find the following facts helpful: (i) Recall that, for vi ~ U[0, 100], we have F(x) = Pr(v < x) = x/100 and f(x) = Pr(v = x) = 1/100. = Fm (x) (1 F(x))2-m (where a = 1 (ii) Since the vi's are iid, we have Pr (exactly m of the vi's are below x) for any number a). (iii) For each fixed bidder i, we have Pr (vi = = v() = x) = f(x). F(x). (iv) E [v()] = Pr(v = v())E [v() | v = v()] + Pr(v2 = v())E [v() | v = v()]. Bonus Problem: Derive the generalizations of the above formulae when there are n > 2 bidders (and justify your answer). Problem 2: Order Statistics Suppose that we have n = 2 bidders whose values v; are independently distributed uniformly on [0, 100]. (This is typically denoted vid U[0, 100].) Let v() denote the highest value and v(2) denote the second-highest value (the so-called first and second order statistics). Since the values are themselves random, v() and v(2) are also random variables. We saw in class that, in any efficient auction, E [v()] determines the total surplus, E [v()] determines the seller's expected revenue, and E [v()] -E[v(2)] determines the bidder's expected surplus. We also stated special cases of the general fact that and E [v()] = = 2 100 1 E[v(2)] = 100. 3 Your task: Derive these formulae. You may find the following facts helpful: (i) Recall that, for vi ~ U[0, 100], we have F(x) = Pr(v < x) = x/100 and f(x) = Pr(v = x) = 1/100. = Fm (x) (1 F(x))2-m (where a = 1 (ii) Since the vi's are iid, we have Pr (exactly m of the vi's are below x) for any number a). (iii) For each fixed bidder i, we have Pr (vi = = v() = x) = f(x). F(x). (iv) E [v()] = Pr(v = v())E [v() | v = v()] + Pr(v2 = v())E [v() | v = v()]. Bonus Problem: Derive the generalizations of the above formulae when there are n > 2 bidders (and justify your answer).

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To derive the formulae for the expected value of the first and second order statistics we first need ... View the full answer