Econ 805 Advanced Micro Theory 1

1
Econ 805Advanced Micro Theory 1

• Dan Quint
• Fall 2009
• Lecture 3

2
First, to finish the thought from last week

• We wanted to show equivalence of two statements
  about
• second-order stochastic dominance
• ò- u(s) dF(s) ³ ò- u(s) dG(s) for every
  incr, concave u
• if and only if
• ò-x F(s) ds ò-x G(s) ds for every x

3
Plan for the proof

• Rewrite u as positive linear combination of basis
  functions h
• u(s) ò- w(q) h(s,q) dq
• (Basis functions are h(x,q) min(x,q) weights
  w(q) areu(q))
• Write E u(x) as a double-integral, flip order of
  integration, 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
• Then integrate by parts to show this is
  equivalent to the integral condition

4
Today Envelope Theorem and Revenue Equivalence

• Last week, we compared the symmetric equilibria
  of the symmetric IPV first- and second-price
  auctions, and found
• The seller gets the same expected revenue in both
• And each type vi of each player i gets the same
  expected payoff in both
• The goal for today is to prove this result is
  much more general. To do this, we will need

5
The Envelope Theorem
6
The Envelope Theorem

• Describes the value function of a parameterized
  optimization problem in terms of the objective
  function
• Aside from allowing us to prove revenue
  equivalence, it will give us
• One-line proof of Shepards Lemma (Consumer
  Theory)
• One-line proof of Hotellings Lemma (Producer
  Theory)
• Easier way to deal with incentive-compatibility
  in mechanism design
• With strong assumptions on derived quantities,
  its trivial to prove well show it from
  primitives today

7
General Setup

• Back away from thinking about multi-player
  Bayesian games, consider a single-agent
  optimization problem
• Choice variable x Î X, parameter t Î T, problem
  is
• maxx Î X f(x,t)
• Define the optimizer
• x(t) arg maxx Î X f(x,t)
• and the value function
• V(t) maxx Î X f(x,t) f(x,t) any x in
  x(t)
• (For auctions, t is your valuation, x is your
  bid, and f is your expected payoff given other
  bidders strategies)
• Well give two versions of the envelope theorem
  one pins down the value of dV/dt when it exists,
  the other expresses V(t) as the integral of that
  derivative

8
An example with X 1,2,3
V(t)maxf(1,t), f(2,t), f(3,t)
f(2,t)
f(1,t)
f(3,t)
t

• For example, f is how good you feel, t is the
  temperature, x 1 is a winter coat, 2 is a
  jacket, 3 is a t-shirt
• V is the upper envelope of all the different
  f(x,-) curves

9
Derivative Version of the Envelope Theorem

• Suppose T 0,1. Recall x(t) arg maxx Î X
  f(x,t).
• Theorem. Pick any t Î 0,1, any x Î x(t), and
  suppose that ft f/ t exists at (x,t).
• If t lt 1 and V(t) exists, then V(t) ³
  ft(x,t)
• If t gt 0 and V(t-) exists, then V(t-)
  ft(x,t)
• If 0 lt t lt 1 and V(t) exists, then V(t)
  ft(x,t)
• The derivative of the value function is the
  derivative of the objective function, evaluated
  at the optimum

10
Derivative Version of the Envelope Theorem
f(x,-)
V(-)
t
11
Proof of the Derivative Version

• Proof. If V(t) exists, then
• V(t) lime ? 0 1/e V(te) V(t)
• lime ? 0 1/e f(x(te),te) f(x,t)
• for any selection x(te) Î x(te)
• By optimality, f(x(te),te) ³ f(x,te), so
• V(t) ³ lime ? 0 1/e f(x,te) f(x,t)
• ft(x, t)
• The symmetric argument shows V(t-) ft(x,t)
  when it exists
• If V(t) exists, V(t) V(t) V(t-), so
• ft(x,t) V(t) ft(x,t)

12
Like I said, this gives us some easy proofs

• Shepherds Lemma (consumer theory)
• hi(u,p) e(u,p) / pi
• e(u,p) is just value function of the minimization
  problem
• minx Î x u(x) ³ u p x
• Envelope theorem e/ pi (px)/ pi xi,
  evaluated at the optimum (hi)
• Hotellings Lemma same result for producer
  theory (firms net supply of an output/input is
  partial derivative of profit function with
  respect to price)

13
The differentiable case (or why you thought you
already knew this)

• Suppose that f is differentiable in both its
  arguments, and x(-) is single-valued and
  differentiable
• Since V(t) f(x(t),t), letting fx and ft denote
  the partial derivatives of f with respect to its
  two arguments,
• V(t) fx(x(t),t) x(t) ft(x(t),t)
• By optimality, fx(x(t),t) 0, so the first term
  vanishes and
• V(t) ft(x(t),t)
• But we dont want to rely on x being
  single-valued and differentiable, or even
  continuous

14
Of course, V need not be differentiable everywhere
V(t)
f(2,t)
f(1,t)
f(3,t)
t

• Even in this simple case, V is only
  differentiable most of the time
• This will turn out to be true more generally, and
  good enough for our purposes

15
Several special cases that do guarantee V
differentiable

• Suppose X is compact and f and ft are continuous
  in both their arguments. Then V is
  differentiable at t, and V(t) ft(x(t),t), if
• x(t) is a singleton, or
• V is concave at t, or
• t Î arg maxs V(s)
• (In most auctions we look at, all interior
  types will have a unique best-response, so V will
  pretty much always be differentiable)
• But we dont need differentiability everywhere
  all we actually need is differentiability most
  of the time

16
Absolute Continuity

• Definition V is absolutely continuous if " e gt
  0, d gt 0 such that for every finite collection
  of disjoint intervals ai, bii Î 1,2,,K ,
• Si bi ai lt d ? Si V(bi) V(ai) lt
  e
• Lemma. Suppose that
• f(x,-) is absolutely continuous (as a function of
  t) for all x Î X, and
• There exists an integrable function B(t) such
  that for almost all t Î 0,1,
• ft(x,t) B(t) for all x Î X
• Then V is absolutely continuous.
• (Well prove this in a moment.)

17
Integral Version of the Envelope Theorem

• Theorem. Suppose that
• For all t, x(t) is nonempty
• For all (x,t), ft(x,t) exists
• V(t) is absolutely continuous
• Then for any selection x(s) from x(s),
• V(t) V(0) ò0t ft(x(s),s) ds
• Even if V(t) isnt differentiable everywhere,
  absolute continuity means its differentiable
  almost everywhere, and continuous so it must be
  the integral of its derivative
• And we know that derivative is ft(x(t),t)
  whenever it exists

18
Proving f(x,-) abs cont and ft has an
integrable bound ? V abs cont

• First since B is integrable, limx ? ò t
  B(t) gt x B(s) ds 0
• If B is integrable, it is finite almost
  everywhere
• Let B(s) B(s) when B(s) finite, 0 otherwise
• B and B differ on a set of measure zero, so have
  same integral
• Let Bk(s) B(s) when B(s) k, 0 otherwise
• So B1, B2, increasing sequence of functions
  that converge to B
• So their integrals converge to ò B(s) ds ò
  B(s) ds
• But the difference between ò Bk(s) ds and ò B(s)
  ds is exactly the integral above, which must
  therefore converge to 0 as x ?
• Given e, find M such that ò t B(t) gt M
  B(s) ds lt e /2, and let d e /2M

19
Proof, contd

• Need to show that for nonoverlapping intervals,
  Si bi ai lt d ? Si V(bi) V(ai) lt
  e
• Assume V increasing (weakly), then we dont have
  to deal with multiple cases
• Si ( V(bi) V(ai) ) Si ( f(x(bi),bi)
  f(x(ai),ai) )
• Since f(x(ai), ai) ³ f(x,ai), this is Si (
  f(x(bi),bi) f(x(bi),ai) )
• If f(x(bi),-) is absolutely continuous in t
  (assumption 1), this is
• Si òaibi ft(x(bi),s) ds
• If ft has an integrable bound (assumption 2),
  this is
• Si òaibi B(s) ds

20
Proof, contd

• Trying to show Si òaibi B(s) ds lt e
• Let L Èi ai, bi, J t B(t) gt M , and K
  be the set with K d that maximizes òK B(s)
  ds
• Recall that òJ B(s) ds lt e/2
• Now, K J K d and B(t) M for all t
  in K J so
• òL B(s) ds òK B(s) ds òJ B(s) ds òK-J B(s)
  ds lt e /2 d M e
• QED

21
So to recap

• Corollary. Suppose that
• For all t, x(t) is nonempty
• For all (x,t), ft(x,t) exists
• For all x, f(x,-) is absolutely continuous
• ft has an integrable bound supx Î X ft(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 ft(x(s),s) ds

22
Revenue Equivalence
23
Back to our auction setting from last week

• Independent Private Values
• Symmetric bidders (private values are i.i.d.
  draws from a probability distribution F)
• Assume F is atomless and has support 0,T
• Consider any auction where, in equilibrium,
• The bidder with the highest value wins
• The expected payment from a bidder with the
  lowest possible type is 0
• The claim is that the expected payoff to each
  type of each bidder, and the sellers expected
  revenue, is the same across all such auctions

24
To show this, we will

• Show that sufficient conditions for the integral
  version of the Envelope Theorem hold
• x(t) nonempty for every t
• ft f/ t exists for every (x,t)
• f(x,-) absolutely continuous as a function of t
  (for a given x)
• ft(x,t) B(t) for all x, almost all t, for
  some integrable function B
• Use the Envelope Theorem to calculate V(t) for
  each type of each bidder, which turns out to be
  the same across all auctions meeting our
  conditions
• Revenue Equivalence follows as a corollary

25
Sufficient conditions for the Envelope Theorem

• Let bi 0,T ? R be bidder is equilibrium
  strategy
• Let f(x,t) be is expected payoff in the auction,
  given a type t and a bid x, assuming everyone
  else bids their equilibrium strategies bj(-)
• If bi is an equilibrium strategy, bi(t) Î x(t),
  so x(t) nonempty
• f(x,t) t Pr(win bid x) E(p bid x)
• so f/ t (x,t) Pr(win bid x), which gives
  the other sufficient conditions
• ft exists at all (x,t)
• Fixing x, f is linear in t, and therefore
  absolutely continuous
• ft is everywhere bounded above by B(t) 1
• So the integral version of the Envelope Theorem
  holds

26
Applying the Envelope Theorem

• We know ft(x,t) Pr(win bid x) Pr(all other
  bids lt x)
• For the envelope theorem, we care about ft at x
  x(t) bi(t)
• ft(bi(t),t) Pr(win in equilibrium given type t)
• But we assumed the bidder with the highest type
  always wins Pr(win given type t) Pr(my type is
  highest) FN-1(t)
• The envelope theorem then gives
• V(t) V(0) ò0t ft(bi(s),s) ds
• V(0) ò0t FN-1(s) ds
• By assumption, V(0) 0, so V(t) ò0t FN-1(s) ds
• The point this does not depend on the details of
  the auction, only the distribution of types
• And so V(t) is the same in any auction satisfying
  our two conditions

27
As for the seller

• Since the bidder with the highest value wins the
  object, the sum of all the bidders payoffs is
• max(v1,v2,,vN) Total Payments To Seller
• The expected value of this is E(v1) R, where R
  is the sellers expected revenue
• By the envelope theorem, the sum of all bidders
  (ex-ante) expected payoffs is
• N Et V(t) N Et ò0t FN-1(s) ds
• So
• R E(v1) N Et ò0t FN-1(s) ds
• which again depends only on F, not the rules of
  the auction

28
To state the results formally

• Theorem. Consider the Independent Private
  Values framework, and any two auction rules in
  which the following hold in equilibrium
• The bidder with the highest valuation wins the
  auction (efficiency)
• Any bidder with the lowest possible valuation
  pays 0 in expectation
• Then the expected payoffs to each type of each
  bidder, and the sellers expected revenue, are
  the same in both auctions.
• Recall the second-price auction satisfies these
  criteria, and has revenue of v2 and therefore
  expected revenue E(v2) so any auction satisfying
  these conditions has expected revenue E(v2)

29
Next lecture

• Next lecture, well formalize necessary and
  sufficient conditions for equilibrium strategies
• In the meantime, well show how todays results
  make it easy to calculate equilibrium strategies

30
Using Revenue Equivalenceto Calculate
Equilibrium Strategies
31
Equilibrium Bids in the All-Pay Auction

• All-pay auction every bidder pays his bid, high
  bid wins
• Bidder is expected payoff, given type t and
  equilibrium bid function b(t), is
• V(t) FN-1(t) t b(t)
• Revenue equivalence gave us
• V(t) ò0t FN-1(s) ds
• Equating these gives
• b(t) FN-1(t) t ò0t FN-1(s) ds
• Suppose types are uniformly distributed on 0,1,
  so F(t) t
• b(t) tN - ò0t FN-1(s) ds tN 1/N tN
  (N-1)/N tN

32
Equilibrium Bids in the Top-Two-Pay Auction

• Highest bidder wins, top two bidders pay their
  bids
• If there is an increasing, symmetric equilibrium
  b, then is expected payoff, given type t and bid
  b(t), is
• V(t) FN-1(t) t (FN-1(t)
  (N-1)FN-2(t)(1-F(t)) b(t)
• Revenue equivalence gave us
• V(t) ò0t FN-1(s) ds
• Equating these gives
• b(t) FN-1(t) t ò0t FN-1(s) ds / (FN-1(t)
  (N-1)FN-2(t)(1-F(t))