In a FPSBA, each bidder is characterized by their monetary valuation of the item for sale. Suppose Alice is a bidder and her valuation is a. Then, if Alice is rational: • She will never bid more than a, because bidding more than a can only make her lose net value. • If she bids exactly a, then she will not lose but also not gain any positive value. • If she bids less than a, then she
may have some positive gain, but the exact gain depends on the bids of the others. Alice would like to bid the smallest amount that can make her win the item, as long as this amount is less than a. For example, if there is another bidder Bob and he bids y and y, then Alice would like to bid y+\varepsilon (where \varepsilon is the smallest amount that can be added, e.g. one cent). Unfortunately, Alice does not know what the other bidders are going to bid. Moreover, she does not even know the valuations of the other bidders. Hence, strategically, we have a
Bayesian game - a game in which agents do not know the payoffs of the other agents. The interesting challenge in such a game is to find a
Bayesian Nash equilibrium. However, this is not easy even when there are only two bidders. The situation is simpler when the valuations of the bidders are
independent and identically distributed random variables, so that the valuations are all drawn from a known prior distribution.
Example Suppose there are two bidders, Alice and Bob, whose valuations a and b are drawn from a
continuous uniform distribution over the interval [0,1]. Then, it is a Bayesian-Nash equilibrium when each bidder bids exactly half his/her value: Alice bids a/2 and Bob bids b/2. PROOF: The proof takes the point-of-view of Alice. We assume that she knows that Bob bids f(b) = b/2, but she does not know b. We find the best response of Alice to Bob's strategy. Suppose Alice bids x. There are two cases: • x\geq f(b). Then Alice wins and enjoys a net gain of a-x. This happens with probability f^{-1}(x)=2x. • x. Then Alice loses and her net gain is 0. This happens with probability 1-f^{-1}(x). All in all, Alice's expected gain is: G(x) = f^{-1}(x)\cdot(a-x). The maximum gain is attained when G'(x)=0. The derivative is (see
Inverse functions and differentiation): :G'(x) = - f^{-1}(x) + (a-x)\cdot {1 \over f'(f^{-1}(x))} and it is zero when Alice's bid x satisfies: :f^{-1}(x) = (a-x)\cdot {1 \over f'(f^{-1}(x))} Now, since we are looking for a symmetric equilibrium, we also want Alice's bid x to equal f(a). So we have: :f^{-1}(f(a)) = (a-f(a))\cdot {1 \over f'(f^{-1}(f(a)))} :a = (a-f(a))\cdot {1 \over f'(a)} :a f'(a) = (a-f(a)) The solution of this differential equation is: f(a) = a/2.
Generalization Denote by: • v_i – the valuation of bidder i; • y_i – the maximum valuation of all bidders except i, i.e., y_i = \max_{j\neq i}{v_j}. Then, a FPSBA has a unique symmetric BNE in which the bid of player i is given by: ::E[y_i \mid y_i == Incentive-compatible variant ==