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# Econ 805 Advanced Micro Theory 1 - PowerPoint PPT Presentation

Econ 805 Advanced Micro Theory 1. Dan Quint Fall 2007 Lecture 4 – Sept 18 2007. Today: Necessary and Sufficient Conditions For Equilibrium. Problem set 1 online (due 9 a.m. Wed Oct 3); email list

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### Econ 805Advanced Micro Theory 1

Dan Quint

Fall 2007

Lecture 4 – Sept 18 2007

• 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

• 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) = òtt’ FN-1(s) ds

With symmetric IPV, b strictly increasing implies the envelope theorem

• 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

• 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 condition is for gt to be strictly increasing in x)

What single-crossing differences gives us (Milgrom/Shannon)

• 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

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?

When is strict single crossing satisfied by symmetric IPV auctionsg(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, b will be nondecreasing

b symmetric IPV auctions 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…)

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?

Sufficiency except in very weird situations,

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)

But… this type of problem?

• 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

Claim this type of problem?

• 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) = ò0t FN-1(s) ds

then g(b(t),t) ³ 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.

Proof. this type of problem?

• 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’) + òtt’ FN-1(s) ds

• = òtt’ [ FN-1(s) – FN-1(t’) ] ds

• But F is increasing (weakly), so FN-1(t’)³FN-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

Proof. this type of problem?

• 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

So basically, in well-behaved symmetric IPV auctions, this type of problem?

• 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… this type of problem?

• 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…

A Few Slides on Second-Order this type of problem?Stochastic Dominance

When is one probability distribution this type of problem?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) ³ E u(Y) for every nondecreasing, concave u, or ò-¥¥ u(s) dF(s) ³ò-¥¥ 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,ò-¥x F(s) ds £ò-¥x G(s) ds

• Rothschild-Stiglitz (1970): 1« 2 «3

What does this tell us? this type of problem?

• 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

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) = ò0ts 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.

A cool proof SOSD of revenue in first-priceº“ò-¥x F(s) ds £ò-¥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) = ò-¥¥ w(q) h(s,q) dq

with w(q) ³ 0, where these basis functions are themselves increasing and concave

• Then we’ll show that X SOSD Y if and only if

ò-¥¥ h(x,q) dF(x) ³ò-¥¥ h(y,q) dG(y)

for all the basis functions

• (“Only if” is trivial, since h(s,q) is increasing and concave; “if” just involves multiplying this inequality by w(q) and integrating over q)

A cool proof SOSD of revenue in first-priceº“ò-¥x F(s) ds £ò-¥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 functionsh(x,q) = min(x,q)

• Claim is thatu(x) = a + bx + ò0¥ (-u’’(q)) h(x,q) dq

• Note that -u’’(q) is nonnegative, since u is concave

• To see the equality, integrate by parts, with db = -u’’ dq, a = h:ò a db = a b – ò b da = –h(x,q)u’(q)|q=-¥¥ – ò-¥¥ –u’(q) 1q<x dq= –xu’(¥) + constant + ò-¥x u’(q) dq

• Since X and Y have the same mean,

ò-¥¥ (a+bx) dF(x) =ò-¥¥ (a+by) dG(y)

A cool proof SOSD of revenue in first-priceº“ò-¥x F(s) ds £ò-¥x G(s) ds everywhere”

• So all that’s left is to determine when

ò-¥¥ h(s,q) dF(s) ³ò-¥¥ h(s,q) dG(s)

• Integrate by parts: u = h(s,q),dv = dF(s), LHS becomesh(¥,q) F(¥) – h(-¥,q) F(-¥) – ò-¥¥ F(s) hs(s,q) ds= q – 0 – ò-¥¥ F(s) 1s<q ds = q – ò-¥ q F(s) ds

• Similarly, the right-hand side becomes q – ò-¥ q G(s) ds

• So Es~F h(s,q) ³ Es~G h(s,q)«ò-¥ q F(s) ds £ò-¥ q G(s) ds

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

(I don’t expect to get to) of revenue in first-price First-Order Stochastic Dominance

• 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 another?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 another?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