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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

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Econ 805 advanced micro theory 1 l.jpg

Econ 805Advanced Micro Theory 1

Dan Quint

Fall 2008

Lecture 4 – Sept 11 2008


Today necessary and sufficient conditions for equilibrium l.jpg
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 l.jpg
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 l.jpg
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 l.jpg
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 l.jpg
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


So now given a symmetric equilibrium with the bid function b l.jpg
So now, given a symmetric equilibrium with the bid function (Milgrom/Shannon)b…

  • 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 l.jpg
All these conditions are (Milgrom/Shannon)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 by g x t g x t w x w x t p x p x l.jpg
When is strict single crossing satisfied by (Milgrom/Shannon)g(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 l.jpg
In fact, (Milgrom/Shannon)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 l.jpg
b (Milgrom/Shannon) 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 l.jpg
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?


Sufficient conditions l.jpg
Sufficient Conditions except in very weird situations,


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


Why do we need smooth single crossing differences not just strict s c d l.jpg
Why do we need smooth single crossing differences, not just strict s.c.d.?

  • 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 l.jpg
Let’s prove the sufficiency theorem strict s.c.d.?

  • 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 theorem17 l.jpg
Let’s prove the sufficiency theorem strict s.c.d.?

  • 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 l.jpg
So there you have it… strict s.c.d.?

  • 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 l.jpg
Applying this to auctions, strict s.c.d.?

  • 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 l.jpg
Pulling it all together, strict s.c.d.?

  • 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 l.jpg
And in well-behaved symmetric IPV auctions, strict s.c.d.?

  • 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 l.jpg
Up next… strict s.c.d.?

  • 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


I probably won t get to first order stochastic dominance l.jpg
(I probably won’t get to) strict s.c.d.? First-Order Stochastic Dominance


When is one probability distribution better than another l.jpg
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 l.jpg
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 everywhere26 l.jpg
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


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