- By
**paul2** - Follow User

- 351 Views
- Uploaded on

Download Presentation
## Econ 805 Advanced Micro Theory 1

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

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

- 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

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)

- 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

- 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 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, b will be nondecreasing

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…)

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?

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…

- 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

- 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.

- 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.

- 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,

- 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
- Meanwhile, since there’s time…

When is one probability distribution 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?

- 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 º“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 º“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 º“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

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

Download Presentation

Connecting to Server..