Econ 805 Advanced Micro Theory 1

1 / 26

Econ 805 Advanced Micro Theory 1 - PowerPoint PPT Presentation

Econ 805 Advanced Micro Theory 1 Dan Quint Fall 2008 Lecture 4 – Sept 11 2008 Today: Necessary and Sufficient Conditions For Equilibrium Today: necessary and sufficient conditions for a particular bidding function to be a symmetric equilibrium Tuesday’s big result was the Envelope Theorem

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

Econ 805 Advanced Micro Theory 1

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

Dan Quint

Fall 2008

Lecture 4 – Sept 11 2008

Today: Necessary and Sufficient Conditions For Equilibrium
• Today: necessary and sufficient conditions for a particular bidding function to be a symmetric equilibrium
Tuesday’s big result was the Envelope Theorem
• Theorem. Suppose that
• For all t, x*(t) is nonempty
• For all (x,t), gt(x,t) exists
• For all x, g(x,-) is absolutely continuous
• gt has an integrable bound: supx Î X| gt(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) + ò0t gt(x(s),s) ds

• Then we applied this to auctions with symmetric independent private values
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 maxxg(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 ¶ 2g / ¶ x¶ t ³0
• 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
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) ³0  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)³0  g(x’,t’) – g(x,t’) > 0

for every x’ > x, t’ > t

• (When gt exists everywhere, a sufficient (but not necessary) condition is for gt to be strictly increasing in x)
What single-crossing differences gives us
• Theorem.* Suppose g(x,t) satisfies strict single crossing differences. Let S Í X be any subset. Let x*(t) = arg maxx Î 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)³g(x’,t) and g(x’,t’)³g(x,t’)
• So g(x,t) – g(x’,t)³ 0 andg(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

• 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) + ò0t gt(b(s),s) ds

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?
When is strict single crossing satisfied byg(x’,t) – g(x,t) = [ W(x’) – W(x) ] t – [ P(x’) – P(x) ] ?
• Assume W(x’)³W(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’) ¹ P(x)(e.g., first-price auction, since P(x) = x), then g(x’,t) – g(x,t) ¹ 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] È [2,3]. The bid function b(2) = 1, b(1) = 2, b(vi) = vi 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
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’) ] FN – 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
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
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?
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) ³ 0  g(x’,t’) – g(x,t’) ³ 0
• gx(x,t) = 0  gx(x,t+d) ³ 0 ³ gx(x,t – d) for all d > 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) = ò0t gt(x(s),s) ds

then x(t) Î arg maxx Î X’ g(x,t)

• (Note that x only guaranteed optimal over X’, not over all X)
• 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

gt(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
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) = gx(x(t),t) x’(t) + gt(x(t),t)

• But we know the envelope condition holds, so

V’(t) = d/dt g(x(t),t) = gt(x(t),t)

• So almost everywhere, gx(x(t),t) x’(t) = 0
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) = òtt’gx(x(s),t) x’(s) ds

• Now, we know from before that at almost every s, gx(x(s),s) x’(s) = 0, so either gx(x(s),s) = 0 orx’(s) = 0
• If x’(s) = 0, then gx(x(s),t) x’(s)= 0 as well
• If gx(x(s),s) = 0, then since x(s) ³x(t), smooth single crossing differences says gx(x(s),t) £ 0 (recall t < s)
• And we know x’(s) ³ 0
• So for almost every s > t, gx(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
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 gx = 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
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
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
• gt 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)Î arg max x Î X* g(x,t) for every t if and only if
• x is nondecreasing
• the envelope condition holds
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
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
• Two probability distributions, F and G
• Ffirst-order stochastically dominatesG if

ò-¥¥ u(s) dF(s) ³ò-¥¥ 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.
Proving FOSD º “F(x)£G(x) everywhere”
• Proof for differentiable u. Rewrite it using a basis consisting of step functionsdq(s) = 0 if s < q, 1 if s ³q
• Up to an additive constant,u(s) = ò-¥¥ u’(q) dq(s) dq
• To see this, calculateu(s’) – u(s) = ò-¥¥ u’(q) (dq(s’) – dq(s)) dq = òss’ u’(q) dq
• So F FOSD G if and only if ò-¥¥dq(s) dF(s) ³ò-¥¥dq(s) dG(s) for every q
Proving FOSD º “F(x)£G(x) everywhere”
• Butò-¥¥dq(s) dF(s) = Pr(s ³q) = 1 – F(q)and similarly ò-¥¥dq(s) dG(s) = 1 – G(q)
• So if F(x) £G(x) for all x, Es~F u(s)³Es~G u(s)

for any increasing u

• “Only if” is because dq(x) is a valid increasing function of x