Econ 805 Advanced Micro Theory 1

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Econ 805Advanced Micro Theory 1

• Dan Quint
• Fall 2007
• Lecture 4 Sept 18 2007

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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

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Todays 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 bidders 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

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Necessary Conditions
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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 were really asking here is when a
  symmetric bid function must be strictly increasing

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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?
• 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
  ³ 0
• 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 wouldnt rule out
  equilibrium strategies which are decreasing at
  some points

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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) ³ 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) ³ 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 strictly increasing in
  x)

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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) ³ g(x,t) and g(x,t) ³
  g(x,t)
• So g(x,t) g(x,t) ³ 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
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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 bg(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?

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When is strict single crossing satisfied
byg(x,t) g(x,t) W(x) W(x) t
P(x) P(x) ?

• Assume W(x) ³ W(x) (probability of winning
  nondecreasing in bid)
• g(x,t) g(x,t) is weakly increasing in t, so if
  its strictly positive at t, its 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 theres 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

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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)

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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?

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Sufficiency
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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,td) 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)

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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 were doing on Thursday anyway

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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.

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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) ³
  FN-1(s) for every s in the integral, so this is
  (weakly) negative
• Symmetric argument holds for t
• So the envelope formula is exactly the condition
  that there is never a gain to deviating to a
  different types equilibrium bid

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Proof.

• All thats 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 types
  bid isnt 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 cant gain by bidding
  above b(T), no one can
• By the symmetric argument, we only need to check
  the lowest types 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

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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

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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 dont know mechanism design, dont worry,
  well go over it
• Meanwhile, since theres time

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A Few Slides on Second-OrderStochastic Dominance
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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) ³
  ò- 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(ZX) 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

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What does this tell us?

• Risk-averse buyers greatly impact auction design
  changes equilibrium strategies well 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 sellers valuation of different
  distributions of revenue
• Claim. With symmetric IPV, a risk-averse seller
  prefers a first-price to a second-price auction

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Proof well 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) ò0t s dFN-1(s) / FN-1(t) E(v2 v1
  t)
• Note that Y X Z, andE(Z Xg(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.

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A cool proof SOSD º ò-x F(s) ds ò-x G(s) ds
everywhere

• Well use the extremal method or basis
  function method
• Well 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 well show that X SOSD Y if and only if
• ò- h(x,q) dF(x) ³ ò- 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)

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A cool proof SOSD º ò-x F(s) ds ò-x G(s) ds
everywhere

• Well do the special case of u twice
  differentiable. Our basis functions will be a
  constant, a linear term, and the
  functions h(x,q) min(x,q)
• Claim is that u(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 hò a db a b ò b da
  h(x,q)u(q)q- ò- u(q) 1qxu() constant ò-x u(q) dq
• Since X and Y have the same mean,
• ò- (abx) dF(x) ò- (aby) dG(y)

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A cool proof SOSD º ò-x F(s) ds ò-x G(s) ds
everywhere

• So all thats left is to determine when
• ò- h(s,q) dF(s) ³ ò- 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) 1sò- q F(s) ds
• Similarly, the right-hand side becomes q ò- q
  G(s) ds
• So EsF h(s,q) ³ EsG 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

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(I dont expect to get to) First-Order
Stochastic Dominance
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When is one probability distribution better
than another?

• Two probability distributions, F and G
• F first-order stochastically dominates G if
• ò- u(s) dF(s) ³ ò- u(s) dG(s)
• for every nondecreasing function u
• So anyone whos 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.

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Proving FOSD º F(x) G(x) everywhere

• Proof for differentiable u. Rewrite it using a
  basis consisting of step functions dq(s)
  0 if s
• Up to an additive constant, u(s) ò- u(q)
  dq(s) dq
• To see this, calculate u(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) ³
  ò- dq(s) dG(s) for every q

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Proving FOSD º F(x) G(x) everywhere

• But ò- dq(s) dF(s) Pr(s ³ q) 1 F(q)and
  similarly ò- dq(s) dG(s) 1 G(q)
• So if F(x) G(x) for all x, EsF u(s) ³ EsG
  u(s)
• for any increasing u
• Only if is because dq(x) is a valid increasing
  function of x