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## Econ 805 Advanced Micro Theory 1

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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

- 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

- 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)

- 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

- 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…

- 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

- 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 byg(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

- 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

- 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,

- 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?

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.?

- 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

- 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 theorem

- 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…

- 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,

- 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,

- 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,

- 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…

- 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

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 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 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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