# Econ 805 Advanced Micro Theory 1

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```							      Econ 805

Dan Quint
Fall 2008
Lecture 3 – Sept 9 2008
First, to finish the thought from last week:

We wanted to show equivalence of two statements about
second-order stochastic dominance:

- u(s) dF(s)  - u(s) dG(s) for every incr, concave u

if and only if

-x F(s) ds  -x G(s) ds for every x

1
Plan for the proof

 Rewrite u as positive linear combination of basis functions h:
u(s) = - w(q) h(s,q) dq

 Show that X SOSD Y if and only if
- h(x,q) dF(x)  - h(y,q) dG(y)
for all the basis functions

 (Basically an exercise in integration by parts, but we ran out of time)

 Once we have that…
 “Only if” is one step, since each h(s,q) is itself increasing and concave
 “If” is three steps: multiply by w(q), and integrating over q, change order
of integrals

2
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…

3
The Envelope Theorem

4
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 Shepard’s Lemma (Consumer Theory)
   One-line proof of Hotelling’s Lemma (Producer Theory)
   Easier way to deal with incentive-compatibility in mechanism
design

 With strong assumptions on derived quantities, it’s trivial
to prove; we’ll show it from primitives today
5
General Setup

 Consider an optimization problem with choice variable
x  X, parameterized by some parameter t  T:
maxx  X f(x,t)

 Define the optimizer
x*(t) = arg maxx  X f(x,t)
and the value function
V(t) = maxx  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)

 We’ll 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                            6
An example with X = {1,2,3}

V(t)=max{f(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 7
Derivative Version of the Envelope Theorem

 Suppose T = [0,1]. Recall x*(t) = arg maxx  X f(x,t).

 Theorem. Pick any t  [0,1], any x*  x*(t), and
suppose that ft =  f/ t exists at (x*,t).
 If t < 1 and V’(t+) exists, then V’(t+)  ft(x*,t)
 If t > 0 and V’(t-) exists, then V’(t-)  ft(x*,t)
 If 0 < t < 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”

8
Derivative Version of the Envelope Theorem

f(x*,-)
V(-)
t

9
Proof of the Derivative Version

 Proof. If V’(t+) exists, then
V’(t+)      =        lime  0 1/e [ V(t+e) – V(t) ]
=        lime  0 1/e [ f(x(t+e),t+e) – f(x*,t) ]
for any selection x(t+e)  x*(t+e)

 By optimality, f(x(t+e),t+e)  f(x*,t+e), so
V’(t+)              lime  0 1/e [ f(x*,t+e) – 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)            10
The differentiable case (or why you thought
 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 don’t want to rely on x* being single-valued and
differentiable, or even continuous…                             11
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                                   12
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  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 don’t need differentiability everywhere – all we
actually need is differentiability “most of the time”      13
Absolute Continuity

 Definition: V is absolutely continuous if " e > 0,
\$ d > 0 such that for every finite collection of disjoint
intervals {[ai, bi]}i  1,2,…,K ,
Si | bi – ai | < d  Si | V(bi) – V(ai) | < e
 Lemma. Suppose that
 f(x,-) is absolutely continuous (as a function of t) for all
x  X, and
 There exists an integrable function B(t) such that for almost
all t  [0,1],
|ft(x,t)|  B(t) for all x  X
Then V is absolutely continuous.
(We’ll prove this in a moment.)               14
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) + 0t ft(x(s),s) ds

 Even if V(t) isn’t differentiable everywhere, absolute
continuity means it’s 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
15
Proving f(x,-) abs cont and |ft| has an
integrable bound  V abs cont
 First: since B is integrable, limx    { t : B(t) > 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  B(s) ds =  B(s) ds
   But the difference between  Bk(s) ds and  B(s) ds is exactly the
integral above, which must therefore converge to 0 as x  

 Given e, find M such that  { t : B(t) > M } B(s) ds < e /2,
and let d = e /2M                                                      16
Proof, cont’d

 Need to show that for nonoverlapping intervals,
Si | bi – ai | < d  Si | V(bi) – V(ai) | < e

 Assume V increasing (weakly), then we don’t 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)  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 aibi ft(x*(bi),s) ds

 If ft has an integrable bound (assumption 2), this is
 Si aibi B(s) ds
17
Proof, cont’d

 Trying to show Si aibi B(s) ds < e

 Let L = i [ai, bi], J = { t : B(t) > M }, and K be the set with
|K|  d that maximizes K B(s) ds

 Recall that J B(s) ds < e/2

 Now, |K – J|  |K|  d ; and B(t)  M for all t in K – J ; so

L B(s) ds  K B(s) ds  J B(s) ds + K-J B(s) ds < e /2 + d M = e

 QED
18
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  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) + 0t ft(x(s),s) ds

19
Revenue Equivalence

20
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 seller’s expected revenue, is the same
across all such auctions

21
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

22
Sufficient conditions for the Envelope
Theorem
 Let bi : [0,T]  R+ be bidder i’s equilibrium strategy
 Let f(x,t) be i’s 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)  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

23
Applying the Envelope Theorem

   We know ft(x,t) = Pr(win | bid x) = Pr(all other bids < 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) + 0t ft(bi(s),s) ds
=       V(0) + 0t FN-1(s) ds
   By assumption, V(0) = 0, so V(t) = 0t 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                                                      24
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 seller’s
expected revenue

 By the envelope theorem, the sum of all bidders’ (ex-ante)
expected payoffs is
N Et V(t) = N Et 0t FN-1(s) ds

 So
R = E(v1) – N Et 0t FN-1(s) ds
which again depends only on F, not the rules of the auction
25
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 seller’s 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)
26
Next lecture…

 Next lecture, we’ll formalize necessary and sufficient
conditions for equilibrium strategies

 In the meantime, we’ll show how today’s results
make it easy to calculate equilibrium strategies

27
Using Revenue Equivalence
to Calculate Equilibrium Strategies

28
Equilibrium Bids in the All-Pay Auction

 All-pay auction: every bidder pays his bid, high bid wins

 Bidder i’s 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) = 0t FN-1(s) ds

 Equating these gives
b(t) = FN-1(t) t – 0t FN-1(s) ds

 Suppose types are uniformly distributed on [0,1], so F(t) = t:
b(t) = tN - 0t FN-1(s) ds = tN – 1/N tN = (N-1)/N tN         29
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 i’s
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) = 0t FN-1(s) ds

 Equating these gives
b(t) = [ FN-1(t) t – 0t FN-1(s) ds ] / (FN-1(t) + (N-1)FN-2(t)(1-F(t))

30

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