# Equilibria in an AuctionSuppose a single item will be sold to one of two bidders. Each bidder’s value is drawn uniformly at random from the interval \([0, 1]\). Each bidder will simultaneously name a price, and the item will be sold to the higher bidder for the price that bidder named. If this price is \(p\) and the winning bidder has a value of \(v\), then the utility of the winning bidder is \(v - p\) and the utility of the losing bidder is 0. This situation is more complex than the normal form games we studied in class for two reasons. First, the set of strategies is infinite. Second, rather than known utilities, there is uncertainty over them, so we need a Bayesian model. In general, the equilibrium strategy will depend on (i.e., be a function of) \(v\). This assignment walks you through the derivation of an equilibrium in this setting.### (a)Suppose your value is \(v\) and the bid of the other bidder is uniformly distributed on \([0, 0.5]\). If you bid \(p\), what is your expected utility?### (b)Suppose your value is \(v\) and the bid of the other bidder is uniformly distributed on \([0, 0.5]\). Show that your optimal bid is \(0.5v\).### (c)Suppose you bid following your strategy from part (b). What is the distribution of your bids? (Recall that your value is uniform on \([0, 1]\)).### (d)Use the analysis from the preceding parts to argue that each player bidding half their value is a (Bayes) Nash equilibrium of this game.

Let the bids made by two professors be : b1 & b2 Also the their valuations for the bid be : v1 & v2 Reservation Valuation (v) => [0 , 1 ] Uniformly Hence , probability of point valuation = 0 , which simply means that as the valuation is uniformly distributed so having same valuation has probability close to zero.a .) There are two bidders 1 & 2 . & b1 & b2 => [ 0 , infinity ] If b1 > b2 , Utility of 1 = V1 - p Utility of 2 = 0 And vice - versa for b2 < b1 Since , v1 , v2 ∈ [0 , 0.5] Probability density function : F(z) = z/0.5 Also , we assume that bids are linear to valuation : b1 = av1 , b2 = av2 Probability of winning = Probability of (p > b2 ) => P( p > av2 ) = > P(p/a > v2 ) = > Probability of (v2 < p/a ) = p/a ( From Probability density function of valuation of person 1 [0 ,1 ] ) Probability of loosing = Probability of (p < b2 ) = p/0.5a (From density function of person 2 ) # Expected Utility = Probability of winning * (Utility of winning ) + Probability of loosing *(Utility of loosing ) => Probability of winning * (V1 -p ) + Probability of loosing *(0 ) => p(V1 - p ) /a b ) Now as depicted above similar would be the expected utility for player 2 - Optimal value of p is given by the point where rate of change of expected utility with respect to (p ) is equal to zero. d(EU)/dp = 0 V/A - 2P/a =0 P = v/2 (Hence proved ) c ,) Distribution of bids would be : P ∈ ( 0 , 0.5 ) As we put the maximum and minimum values of each bidders valuation which is in the range [0 ,1 ] . d ) For Bayesian Nash Equilibrium : b1(v1) = b2(v2) -- ( Bayesian Nash equilibrium is achieved at the point where bidding functions which are a function of their respective valuations are equal ) For Person 1 - Probability of winning = Probability of (p > b2 ) => P( p > av2 ) = > P(p/a > v2 ) = > Probability of (v2 < p/a ) = p/a ( From Probability density function of valuation of person 1 [0 ,1 ] ) Probability of loosing = Probability of (p < b2 ) = p/0.5a (From density function of person 2 ) # Expected Utility = Probability of winning * (Utility of winning ) + Probability of loosing *(Utility of loosing ) => Probability of winning * (V1 -p ) + Probability of loosing *(0 ) => p(V1 - p ) /a Optimal value : d(EU)/dp = 0 p = v1 /2 Similarly , For Person 2 - => p = v2/2