Econ 805 Advanced Micro Theory 1 Dan Quint Fall 2009 Lecture 4

1 Today: Necessary and Sufficient Conditions For Equilibrium  Problem set 1 online shortly  Last lecture: integral form of the Envelope Theorem holds in equilibrium of any Independent Private Value auction where  The highest type wins the object  The lowest possible type gets expected payoff 0  Today: necessary and sufficient conditions for a particular bidding function to be a symmetric equilibrium in such an auction

2 Today’s General Results  Consider a symmetric independent private values model of some auction, and a bid function b : T  R +  Define g(x,t) as one bidder’s expected payoff, given type t and bid x, if all the other bidders bid according to b  Under fairly broad (but not all) conditions: “everyone bidding according to b” is an equilibrium b strictly increasing and g(b(t’),t’) – g(b(t),t) =  t t’ F N-1 (s) ds

3 Necessary Conditions

4  If everyone bids according to the same bid function b,  And b is strictly increasing,  Then the highest type wins,  And so the envelope theorem holds  So what we’re really asking here is when a symmetric bid function must be strictly increasing With symmetric IPV, b strictly increasing implies the envelope theorem

5 When must bid functions be increasing?  Equilibrium strategies are solutions to the maximization problem max x g(x,t)  What conditions on g makes every selection x(t) from x*(t) nondecreasing?

6 When must bid functions be increasing?  Recall supermodularity and Topkis  Strong Set Order: two sets A, B. A  SSO B if for every x’ > x, (x’  B and x  A)  (x  B and x’  A).  (What this means visually.)  A function g : X x T  R has increasing differences if for every x’ > x, the difference g(x’,t) – g(x,t) is nondecreasing in t  Topkis: if g(x,t) has increasing differences and t’ > t, then x*(t’)  SSO x*(t)  This means there exists some selection x(t) from x*(t) which is monotonic  But it does not guarantee that every selection is monotonic, so it doesn’t answer our question  We need something stronger than increasing differences in some ways (although what we use is weaker in others)

7 Single crossing and single crossing differences properties (Milgrom/Shannon)  A function h : T  R satisfies the strict single crossing property if for every t’ > t, h(t)  0  h(t’) > 0 (Also known as, “h crosses 0 only once, from below”)  A function g : X x T  R satisfies the strict single crossing differences property if for every x’ > x, the function h(t) = g(x’,t) – g(x,t) satisfies strict single crossing  That is, g satisfies strict single crossing differences if g(x’,t) – g(x,t)  0  g(x’,t’) – g(x,t’) > 0 for every x’ > x, t’ > t  (When g t exists everywhere, a sufficient condition is for g t to be strictly increasing in x)

8 What single-crossing differences gives us  Theorem. * Suppose g(x,t) satisfies strict single crossing differences. Let S  X be any subset. Let x*(t) = arg max x  S g(x,t), and let x(t) be any (pointwise) selection from x*(t). Then x(t) is nondecreasing in t.  Proof. Let t’ > t, x’ = x(t’) and x = x(t).  By optimality, g(x,t)  g(x’,t) and g(x’,t’)  g(x,t’)  So g(x,t) – g(x’,t)  0 and  g(x,t’) – g(x’,t’)  0  If x > x’, this violates strict single crossing differences * Milgrom (PATW) theorem 4.1, or a special case of theorem 4’ in Milgrom/Shannon 1994

9 Strict single-crossing differences will hold in “most” symmetric IPV auctions  Suppose b : T  R + is a symmetric equilibrium of some auction game in our general setup  Assume that the other N-1 bidders bid according to b; g(x,t) = t Pr(win | bid x) – E(pay | bid x) = t W(x) – P(x)  For x’ > x, g(x’,t) – g(x,t) = [ W(x’) – W(x) ] t – [ P(x’) – P(x) ]  When does this satisfy strict single-crossing?

10 When is strict single crossing satisfied by g(x’,t) – g(x,t) = [ W(x’) – W(x) ] t – [ P(x’) – P(x) ] ?  Assume W(x’)  W(x) (probability of winning nondecreasing in bid)  g(x’,t) – g(x,t) is weakly increasing in t, so if it’s strictly positive at t, it’s strictly positive at t’ > t  Need to check that if g(x’,t) – g(x,t) = 0, then g(x’,t’) – g(x,t’) > 0  This can only fail if W(x’) = W(x)  If b has convex range, W(x’) > W(x), so strict single crossing differences holds and b must be nondecreasing (e.g.: T convex, b continuous)  If W(x’) = W(x) and P(x’)  P(x) (e.g., first-price auction, since P(x) = x), then g(x’,t) – g(x,t)  0, so there’s nothing to check  But, if W(x’) = W(x) and P(x’) = P(x), then bidding x’ and x give the same expected payoff, so b(t) = x’ and b(t’) = x could happen in equilibrium  Example. A second-price auction, with values uniformly distributed over [0,1]  [2,3]. The bid function b(2) = 1, b(1) = 2, b(v i ) = v i otherwise is a symmetric equilibrium.  But other than in a few weird situations, b will be nondecreasing

11 b will almost always be strictly increasing  Suppose b(-) were constant over some range of types [t’,t’’]  Then there is positive probability (N – 1) [ F(t’’) – F(t’) ] F N – 2 (t’) of tying with one other bidder by bidding b* (plus the additional possibility of tying with multiple bidders)  Suppose you only pay if you win; let B be the expected payment, conditional on bidding b* and winning  Since t’’ > t’, either t’’ > B or B > t’, so either you strictly prefer to win at t’’ or you strictly prefer to lose at t’  Assume that when you tie, you win with probability greater than 0 but less than 1  Then you can strictly gain in expectation either by reducing b(t’) by a sufficiently small amount, or by raising b(t’’) by a sufficiently small amount  (In addition: when T has point mass… second-price… first-price…)

12 So to sum up, in “well-behaved” symmetric IPV auctions, except in very weird situations,  any symmetric equilibrium bid function will be strictly increasing,  and the envelope formula will therefore hold  Next: when are these sufficient conditions for a bid function b to be a symmetric equilibrium?

13 Sufficiency

14 What are generally sufficient conditions for optimality in this type of problem?  A function g(x,t) satisfies the smooth single crossing differences condition if for any x’ > x and t’ > t,  g(x’,t) – g(x,t) > 0  g(x’,t’) – g(x,t’) > 0  g(x’,t) – g(x,t)  0  g(x’,t’) – g(x,t’)  0  g x (x,t) = 0  g x (x,t+  )  0  g x (x,t –  ) for all  > 0  Theorem. (PATW th 4.2) Suppose g(x,t) is continuously differentiable and has the smooth single crossing differences property. Let x : [0,1]  R have range X’, and suppose x is the sum of a jump function and an absolutely continuous function. If  x is nondecreasing, and  the envelope formula holds: for every t, g(x(t),t) – g(x(0),0) =  0 t g t (x(s),s) ds then x(t)  arg max x  X’ g(x,t)  (Note that x only guaranteed optimal over X’, not over all X)

15 But…  Establishing smooth single-crossing differences requires a bunch of conditions on b  We can use the payoff structure of an IPV auction to give a simpler proof  Proof is taken from Myerson (“Optimal Auctions”), which we’re doing on Thursday anyway

16 Claim  Theorem. Consider any auction where the highest bid gets the object. Assume the type space T has no point masses. Let b : T  R + be any function, and define g(x,t) in the usual way. If  b is strictly increasing, and  the envelope formula holds: for every t, g(b(t),t) – g(b(0),0) =  0 t F N-1 (s) ds then g(b(t),t)  g(b(t’),t), that is, no bidder can gain by making a bid that a different type would make. If, in addition, the type space T is convex, b is continuous, and neither the highest nor the lowest type can gain by bidding outside the range of b, then everyone bidding b is an equilibrium.

17 Proof.  Note that when you bid b(s), you win with probability F N-1 (s); let z(s) denote the expected payment you make from bidding s  Suppose a bidder had a true type of t and bid b(t’) instead of b(t)  The gain from doing this is  g(b(t’), t) – g(b(t), t) = t F N-1 (t’) – z(t’) – g(b(t),t)  = (t – t’) F N-1 (t’) + t’ F N-1 (t’) – z(t’) – g(b(t),t)  = (t – t’) F N-1 (t’) + g(x(t’),t’) – g(x(t),t)  Suppose t’ > t. By assumption, the envelope theorem holds, so  = (t – t’) F N-1 (t’) +  t t’ F N-1 (s) ds  =  t t’ [ F N-1 (s) – F N-1 (t’) ] ds  But F is increasing (weakly), so F N-1 (t’)  F N-1 (s) for every s in the integral, so this is (weakly) negative  Symmetric argument holds for t’ < t  So the envelope formula is exactly the condition that there is never a gain to deviating to a different type’s equilibrium bid

18 Proof.  All that’s left is deviations to bids outside the range of b  With T convex and b continuous, the bid distribution has convex support, so we only need to check deviations to bids above and below the range of b  Assume (for notational ease) that T = [0,T]  If some type t deviated to a bid B > b(T), his expected gain would be  g(B,t) – g(b(t),t) = [ g(B,t) – g(b(T),t) ] + [ g(b(T),t) – g(b(t),t) ]  The second term is nonpositive (another type’s bid isn’t a profitable deviation)  We also know g(x,t) = t Pr(win | bid x) – z(x) has increasing differences in x and t, so for B > b(T), if g(B,t) – g(b(T),t) > 0, g(B,T) – g(b(T),t) > 0  So if the highest type T can’t gain by bidding above b(T), no one can  By the symmetric argument, we only need to check the lowest type’s incentive to bid below b(0)  (If b was discontinuous or T had holes, we would need to also check deviations to the “holes” in the range of b)  QED

19 So basically, in well-behaved symmetric IPV auctions,  b : T  R + is a symmetric equilibrium if and only if  b is increasing, and  b (and the g derived from it) satisfy the envelope formula

20 Up next…  Recasting auctions as direct revelation mechanisms  Optimal (revenue-maximizing) auctions  Might want to take a look at the Myerson paper, or the treatment in one of the textbooks  If you don’t know mechanism design, don’t worry, we’ll go over it