1. Econ 805 Advanced Micro Theory 1 Dan Quint Fall 2007 Lecture 4 – Sept 18 2007

2. 1 Today: Necessary and Sufficient Conditions For Equilibrium  Problem set 1 online (due 9 a.m. Wed Oct 3); email list  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  Time permitting, stochastic dominance

3. 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

4. 3 Necessary Conditions

5. 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

6. 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?  Recall supermodularity and Topkis  If g(x,t) has increasing differences in (x,t), then the set x*(t) is increasing in t (in the strong set order)  For g differentiable, this is when   g /  x  t  0  But let t’ > t; if x* is not single-valued, this still allows some points in x*(t) to be above some points in x*(t’), so it wouldn’t rule out equilibrium strategies which are decreasing at some points

7. 6 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. 7 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. 8 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. 9 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. 10 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. 11 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. 12 Sufficiency

14. 13 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. 14 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. 15 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. 16 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. 17 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. 18 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. 19 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  Meanwhile, since there’s time…

21. 20 A Few Slides on Second-Order Stochastic Dominance

22. 21 When is one probability distribution less risky than another?  Two random variables X and Y with the same mean, with distributions F and G  Three conditions to consider: 1. “Every risk-averse utility maximizer prefers X to Y”, i.e., E u(X)  E u(Y) for every nondecreasing, concave u, or  -   u(s) dF(s)   -   u(s) dG(s) (also called SOSD) 2. “Y is a mean-preserving spread of X”, or “Y = X + noise”:  r.v. Z s.t. Y = d X + Z, with E(Z|X) = 0 for every value of X 3. For every x,  -  x F(s) ds   -  x G(s) ds  Rothschild-Stiglitz (1970): 1  2  3

23. 22 What does this tell us?  Risk-averse buyers greatly impact auction design – changes equilibrium strategies – we’ll get to that in a few lectures (Maskin and Riley)  Risk-averse sellers have less impact – equilibrium strategies are the same, all that changes is seller’s valuation of different distributions of revenue  Claim. With symmetric IPV, a risk-averse seller prefers a first-price to a second-price auction

24. 23 Proof: we’ll show revenue in second-price auction is MPS of revenue in first-price  Recall that revenue in a second-price auction is v 2, and revenue in a first-price auction is E(v 2 | v 1 )  Let X, Y, and Z be random variables derived from bidders’ valuations, as follows:  X = g(v 1 )  Z = v 2 – g(v 1 )  Y = v 2  where g(t) =  0 t s dF N-1 (s) / F N-1 (t) = E(v 2 | v 1 = t)  Note that Y = X + Z, and E(Z | X=g(t)) = E(v 2 | v 1 = t) – E(v 2 | v 1 = t) = 0  So Y is a mean-preserving spread of X, so any risk-averse utility maximizer prefers X to Y  But X is the revenue in the first-price auction, and Y is the revenue in the second-price auction – Q.E.D.

25. 24 A cool proof SOSD  “  -  x F(s) ds   -  x G(s) ds everywhere”  We’ll use the “extremal method” or “basis function method”  We’ll rewrite our generic (increasing concave) function u(s) as a positive sum of basis functions u(s) =  -   w(  ) h(s,  ) d  with w(  )  0, where these basis functions are themselves increasing and concave  Then we’ll show that X SOSD Y if and only if  -   h(x,  ) dF(x)   -   h(y,  ) dG(y) for all the basis functions  (“Only if” is trivial, since h(s,  ) is increasing and concave; “if” just involves multiplying this inequality by w(  ) and integrating over  )

26. 25 A cool proof SOSD  “  -  x F(s) ds   -  x G(s) ds everywhere”  We’ll do the special case of u twice differentiable. Our basis functions will be a constant, a linear term, and the functions h(x,  ) = min(x,  )  Claim is that u(x) = a + bx +  0  (-u’’(  )) h(x,  ) d   Note that -u’’(  ) is nonnegative, since u is concave  To see the equality, integrate by parts, with db = -u’’ d , a = h:  a db = a b –  b da = –h(x,  )u’(  )|  =-   –  -   –u’(  ) 1  <x d  = –xu’(  ) + constant +  -  x u’(  ) d   Since X and Y have the same mean,  -   (a+bx) dF(x)   -   (a+by) dG(y)

27. 26 A cool proof SOSD  “  -  x F(s) ds   -  x G(s) ds everywhere”  So all that’s left is to determine when  -   h(s,  ) dF(s)   -   h(s,  ) dG(s)  Integrate by parts: u = h(s,  ), dv = dF(s), LHS becomes h( ,  ) F(  ) – h(-  ) F(-  ) –  -   F(s) h s (s,  ) ds =  – 0 –  -   F(s) 1 s<  ds =  –  -   F(s) ds  Similarly, the right-hand side becomes  –  -   G(s) ds  So E s~F h(s,  )  E s~G h(s,  )   -   F(s) ds   -   G(s) ds  So X SOSD Y if and only if this holds for every 

28. 27 (I don’t expect to get to) First-Order Stochastic Dominance

29. 28 When is one probability distribution “better” than another?  Two probability distributions, F and G  F first-order stochastically dominates G if  -   u(s) dF(s)   -   u(s) dG(s) for every nondecreasing function u  So anyone who’s maximizing any increasing function prefers the distribution of outcomes F to G  (Very strong condition.)  Theorem. F first-order stochastically dominates G if and only if F(x)  G(x) for every x.

30. 29 Proving FOSD  “F(x)  G(x) everywhere”  Proof for differentiable u. Rewrite it using a basis consisting of step functions   (s) = 0 if s < , 1 if s    Up to an additive constant, u(s) =  -   u’(  )   (s) d   To see this, calculate u(s’) – u(s) =  -   u’(  ) (   (s’) –   (s)) d  =  s s’ u’(  ) d   So F FOSD G if and only if  -     (s) dF(s)   -     (s) dG(s) for every 

31. 30 Proving FOSD  “F(x)  G(x) everywhere”  But  -     (s) dF(s) = Pr(s   ) = 1 – F(  ) and similarly  -     (s) dG(s) = 1 – G(  )  So if F(x)  G(x) for all x, E s~F u(s)  E s~G u(s) for any increasing u  “Only if” is because   (x) is a valid increasing function of x