Econ 805Advanced Micro Theory 1

Dan Quint

Fall 2009

Lecture 4

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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

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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’ FN-1(s) ds

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Necessary Conditions

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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

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When must bid functions be increasing?

Equilibrium strategies are solutions to the maximization problem maxx g(x,t)

What conditions on g makes every selection x(t) from x*(t) nondecreasing?

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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)

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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 gt exists everywhere, a sufficient condition is for gt to be strictly increasing in x)

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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 maxx 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 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

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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?

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When is strict single crossing satisfied byg(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(vi) = vi otherwise is a symmetric equilibrium.

But other than in a few weird situations, b will be nondecreasing

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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

(In addition: when T has point mass… second-price… first-price…)

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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?

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Sufficiency

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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 gx(x,t) = 0 gx(x,t+) 0 gx(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) = 0t gt(x(s),s) ds

then x(t) arg maxx X’ g(x,t)

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

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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

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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) = 0t FN-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.

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Proof.

Note that when you bid b(s), you win with probability FN-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 FN-1(t’) – z(t’) – g(b(t),t) = (t – t’) FN-1(t’) + t’ FN-1(t’) – z(t’) – g(b(t),t) = (t – t’) FN-1(t’) + g(x(t’),t’) – g(x(t),t) Suppose t’ > t. By assumption, the envelope theorem holds, so = (t – t’) FN-1(t’) + t

t’ FN-1(s) ds

= tt’ [ FN-1(s) – FN-1(t’) ] ds

But F is increasing (weakly), so FN-1(t’) FN-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

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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

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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

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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