Transcript Lecture 3 – The Envelope Theorem and Revenue Equivalence

Econ 805
Advanced Micro Theory 1
Dan Quint
Fall 2007
Lecture 3 – Sept 11 2007
Today: Revenue Equivalence
 Reminder: no class on Thursday
 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…
1
The Envelope Theorem
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The Envelope Theorem
 Describes the value function of a parameterized
optimization problem in terms of the objective function
 Leads to one-line proofs of Shepard’s Lemma (Consumer
Theory) and Hotelling’s Lemma (Producer Theory)
 Leads to straightforward proof of Revenue Equivalence
and other results in auction theory and mechanism design
 With strong assumptions on derived quantities, it’s trivial
to prove; we’ll show it from primitives today
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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),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
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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
 Think of the function V as the “upper envelope” of all the
different f(x,-) curves
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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”
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Derivative Version of the Envelope Theorem
f(x*,-)
V(-)
t
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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)
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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 don’t want to rely on x* being single-valued and
differentiable, or even continuous…
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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
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enough for our purposes
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, 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 – what we
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will actually need is…
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
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When is V absolutely continuous?
 Absolute continuity: "e> 0, $d > 0 s.t. 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
|ft(x,t)|  B(t) for all x  X for almost all t  [0,1]
Then V is absolutely continuous.
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Proving V absolutely continuous
 Since B is integrable, there is some M s.t.
 { t : B(t) > M } B(s) ds < e/2; find this M, and let d = e/2M
 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
carry around a bunch of extra terms
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Proving V absolutely continuous
 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
 Now, let L = i [ai, bi], J = { t : B(t) > M }, and K be the set with
|K|  d that maximizes K B(s) ds
 If |K|  |J|, K J, so L B(s) ds  K B(s) ds  J B(s) ds < e/2
 If |K| > |J|, then 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
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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
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Revenue Equivalence
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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,V]
 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
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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
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Sufficient conditions for the Envelope
Theorem
 Let bi : [0,V]  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
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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
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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
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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)
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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
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Using Revenue Equivalence
to Calculate Equilibrium Strategies
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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
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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))
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