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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 805 advanced micro theory 1

Econ 805Advanced Micro Theory 1

Dan Quint

Fall 2007

Lecture 4 – Sept 18 2007


Today necessary and sufficient conditions for equilibrium
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


Today s general results
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) = òtt’ FN-1(s) ds



With symmetric ipv b strictly increasing implies the envelope theorem
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


When must bid functions be increasing
When must bid functions be 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
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
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
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 g x t g x t w x w x t p x p x
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 will almost always be strictly increasing
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
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
Sufficiency except in very weird situations,


What are generally sufficient conditions for optimality in this type of problem
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)


Econ 805 advanced micro theory 1
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
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
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


Proof1
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
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
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 stochastic dominance
A Few Slides on Second-Order this type of problem?Stochastic Dominance


When is one probability distribution less risky than another
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
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
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 x f s ds x g s ds everywhere
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 x f s ds x g s ds everywhere1
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 x f s ds x g s ds everywhere2
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 first order stochastic dominance
(I don’t expect to get to) of revenue in first-price First-Order Stochastic Dominance


When is one probability distribution better than another
When is one probability distribution “better” than another?

  • 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 f x g x everywhere
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 f x g x everywhere1
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