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Econ 805 Advanced Micro Theory 1 Dan Quint Fall 2008 Lecture 4 – Sept 11 2008

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1 Today: Necessary and Sufficient Conditions For Equilibrium Today: necessary and sufficient conditions for a particular bidding function to be a symmetric equilibrium

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2 Tuesday’s big result was the Envelope Theorem Theorem. Suppose that For all t, x*(t) is nonempty For all (x,t), g t (x,t) exists For all x, g(x,-) is absolutely continuous g t has an integrable bound: sup x X | g t (x,t) | B(t) for almost all t, with B(t) some integrable function Then for any selection x(s) from x*(s), V(t) = V(0) + 0 t g t (x(s),s) ds Then we applied this to auctions with symmetric independent private values

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3 Another fairly general necessary condition: monotonicity In symmetric IPV auctions, equilibrium bid strategies will generally be increasing in values; how to prove? 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

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4 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 (but not necessary) condition is for g t to be strictly increasing in x)

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5 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

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6 So now, given a symmetric equilibrium with the bid function b… Define g(x,t) as the expected payoff, given bid x and value t, when everyone else uses the equilibrium bid function If g satisfies strict single crossing differences, then b must be (weakly) increasing And (from Tuesday), if g is absolutely continuous and differentiable in t with an integrable bound, then V(t) = V(0) + 0 t g t (b(s),s) ds

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7 All these conditions are almost always satisfied by 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) So g(x,t) is absolutely continuous and differentiable in t, with derivative bounded by 1 What about strict single crossing differences? 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?

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8 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, g will satisfy strict single crossing differences, so we know b will be nondecreasing

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9 In fact, 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

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10 b will almost always be strictly increasing In first-price auctions, equilibrium bid distributions don’t have point masses even when type distributions do When there is a positive probability of each bidder having a particular value, they play a mixed strategy at that value Otherwise, there’d be a positive probability of ties, and the same logic would hold – either prefer to increase by epsilon to win those ties, or decrease by epsilon if you’re indifferent about winning them In second-price auctions, if the type distribution has a point mass, bidders still bid their valuations Still a dominant strategy So in that case, there are positive-probability ties

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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 hold Next: when are these sufficient conditions for a bid function b to be a symmetric equilibrium?

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12 Sufficient Conditions

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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)

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14 Why do we need smooth single crossing differences, not just strict s.c.d.? Consider g(x,t) = (x – t) 3, x(t) = x, X = [0,1] Clearly, x(t) is nondecreasing And it satisfies the envelope theorem: since g t (x(t),t) = – (x(t) – t) 2 = 0 and g(x(t),t) = 0 But g(x,t) is maximized at x = 1, so x(t) = t is not a solution

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15 Let’s prove the sufficiency theorem Lots of extra terms due to the possibility of discontinuities – we’ll just do the case where x(t) is continuous (and therefore absolutely contin) x(t) absolutely continuous means it’s differentiable almost everywhere So almost everywhere, by the chain rule, d/dt g(x(t),t) = g x (x(t),t) x’(t) + g t (x(t),t) But we know the envelope condition holds, so V’(t) = d/dt g(x(t),t) = g t (x(t),t) So almost everywhere, g x (x(t),t) x’(t) = 0

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16 Let’s prove the sufficiency theorem Suppose at type t, instead of x(t), I played x(t’), with t’ > t My gain from the change is g(x(t’),t) – g(x(t),t) = t t’ g x (x(s),t) x’(s) ds Now, we know from before that at almost every s, g x (x(s),s) x’(s) = 0, so either g x (x(s),s) = 0 or x’(s) = 0 If x’(s) = 0, then g x (x(s),t) x’(s) = 0 as well If g x (x(s),s) = 0, then since x(s) x(t), smooth single crossing differences says g x (x(s),t) 0 (recall t < s) And we know x’(s) 0 So for almost every s > t, g x (x(s),t) x’(s) ds 0 So integrating up, the switch from x(t) to x(t’) can’t be good The symmetric argument rules out t’ < t

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17 So there you have it… If g is differentiable and satisfies smooth single crossing differences, then x weakly increasing and satisfying the Envelope condition implies x(t) is optimal AMONG THE RANGE OF x We haven’t said anything about other possible x outside the range of x Now, when will smooth single crossing differences be satisfied in auctions? Well, g(x,t) = t W(x) – P(x), so g x = t W’(x) – P’(x) is weakly increasing in t as long as W is nondecreasing in x So as long as higher bids win more often, the only condition we have to worry about is g being differentiable wrt x

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18 Applying this to auctions, Let b be a bid function, g the implied expected payoff function If g is differentiable with respect to x b is weakly increasing b and g satisfy the envelope condition then b(t) is a best-response among the range of b Still need to check separately for deviations outside the range of b If the range of b is convex, that is, T is convex and b is continuous, only really have to worry about highest type deviating to higher bids, lowest type deviating to lower bids

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19 Pulling it all together, Theorem (Constraint Simplification). Let g(x,t) be a parameterized optimization problem. Suppose that g is differentiable in both its arguments g t has an integrable bound g satisfies strict and smooth single crossing differences Let x : [0,1] X be the sum of an absolutely continuous function and a jump function, and let X* be the range of x Then x(t) arg max x X* g(x,t) for every t if and only if x is nondecreasing the envelope condition holds

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20 And 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

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21 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

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22 (I probably won’t get to) First-Order Stochastic Dominance

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23 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.

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24 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

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25 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

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