Econ 805Advanced Micro Theory 1

Dan Quint

Fall 2007

Lecture 4 – Sept 18 2007

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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

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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?

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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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

Meanwhile, since there’s time…

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A Few Slides on Second-OrderStochastic Dominance

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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

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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

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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 v2, and revenue in a first-price auction is E(v2 | v1)

Let X, Y, and Z be random variables derived from bidders’ valuations, as follows:

X = g(v1) Z = v2 – g(v1) Y = v2

where g(t) = 0t s dFN-1(s) / FN-1(t) = E(v2 | v1 = t)

Note that Y = X + Z, andE(Z | X=g(t)) = E(v2 | v1 = t) – E(v2 | v1 = 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.

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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 )

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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 thatu(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)

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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 becomesh(,) F() – h(-) F(-) – -

F(s) hs(s,) ds= – 0 – -

F(s) 1s< ds = – -

F(s) ds

Similarly, the right-hand side becomes – -G(s) ds

So Es~F h(s,) Es~G h(s,) -F(s) ds -

G(s) ds

So X SOSD Y if and only if this holds for every

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

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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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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, calculateu(s’) – u(s) = -

u’() ((s’) – (s)) d = ss’ u’() d

So F FOSD G if and only if - (s) dF(s) -

(s) dG(s) for every

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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, Es~F u(s) Es~G u(s)

for any increasing u

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