Transcript Necessary and Sufficient Conditions for Equilibrium

Econ 805
Advanced Micro Theory 1
Dan Quint
Fall 2009
Lecture 4
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
1
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) = tt’ FN-1(s) ds
2
Necessary Conditions
3
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
4
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?
5
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)6
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
7
strictly increasing in x)
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 and g(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
8
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?
9
When is strict single crossing satisfied by
g(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 10
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…) 11
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?
12
Sufficiency
13
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+d)  0  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) = 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)
14
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
15
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.
16
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’) + tt’ 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
17
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
18
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
19
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
20