Fall 2009 Problem Set 1 – Solutions

(b) We do several iterations of eliminating strictly dominated strategies. First, note that when v < 75, bidding 75 is strictly dominated: regardless of the other bidder’s strategy, you will win the auction at least half the time, giving strictly negative payoff, while bidding 0 ensures a payoff of 0. Second, assume your opponent never bids 75 unless his value is at least 75. Then a bid of 50 will win with at least probability three-eighths. So when v < 50, a bid of 50 is strictly dominated by a bid of 0. Third, assume your opponent bids below 50 when his value is below 50. Then a bid of 25 will win with at least probability one-fourth. So when v < 25, a bid of 25 is strictly dominated by a bid of 0. (c) A bid of 75 gives an expected payoff of at most v − 75. Equilibrium strategies must survive IESDS. Knowing that your opponent bids below 50 half the time, and below 75 three-fourths of the time, a bid of 50 must win with probability at least 58 , and 5 8(v − 50) > v − 75 for all v ≤ 100. (d) A bid of 0 gets a payoff of 0; a bid of 25 gets a payoff of at least 14(v − 25), which is strictly higher for any v > 25. (e) Argue by strict single-crossing differences. f(x, t) = Pr(win|bid = x)(t− x). Let x′ > x. If Pr(win|bid = x′) > Pr(win|bid = x), then f(x′, t) − f(x, t) is strictly increasing in t. If Pr(win|bid = x′) = Pr(win|bid = x) > 0, then x′ gives strictly lower expected payoff than x and need not be considered. If Pr(win|bid = x′) = Pr(win|bid = x) = 0, then x′ = x = 0 and the assumption that x′ > x is violated.