Download presentation

Presentation is loading. Please wait.

Published byJarrod Rhoden Modified about 1 year ago

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

2
1 Today: Necessary and Sufficient Conditions For Equilibrium Today: necessary and sufficient conditions for a particular bidding function to be a symmetric equilibrium

3
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

4
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

5
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)

6
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

7
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

8
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?

9
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

10
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

11
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

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

13
12 Sufficient Conditions

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 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

16
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

17
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

18
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

19
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

20
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

21
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

22
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

23
22 (I probably won’t get to) First-Order Stochastic Dominance

24
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.

25
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

26
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

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google