econ 805 advanced micro theory 1 dan quint fall 2007 lecture 3 – sept 11 2007

Econ 805Advanced Micro Theory 1

Dan Quint

Fall 2007

Lecture 3 – Sept 11 2007

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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…

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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}

f(1,t)

f(2,t)

f(3,t)

V(t)=max{f(1,t), f(2,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+) = lim 0 1/ [ V(t+) – V(t) ]

= lim 0 1/ [ f(x(t+),t+) – f(x*,t) ]

for any selection x(t+) x*(t+)

By optimality, f(x(t+),t+) f(x*,t+), so

V’(t+) lim 0 1/ [ f(x*,t+) – 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

f(1,t)

f(2,t)

f(3,t)

V(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

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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 will actually need is…

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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: > 0, > 0 s.t. for every finite collection of disjoint intervals {[ai, bi]}i 1,2,…,K ,

i | bi – ai | < i | V(bi) – V(ai) | <

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 < /2; find this M, and let = /2M

Need to show that for nonoverlapping intervals, i | bi – ai | < i | V(bi) – V(ai) | <

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

i | V(bi) – V(ai) | = i | f(x*(bi),bi) – f(x*(ai),ai) |

Since f(x*(ai), ai) f(x,ai), this is

i | f(x*(bi),bi) – f(x*(bi),ai) |

If f(x*(bi),-) is absolutely continuous in t (assumption 1), this is

i aibi | ft(x*(bi),s) | ds

If ft has an integrable bound (assumption 2), this is

i aibi B(s) ds

Now, let L = i [ai, bi], J = { t : B(t) > M }, and K be the set with |K| that maximizes K B(s) ds

If |K| |J|, K J, so L B(s) ds K B(s) ds J B(s) ds < /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 < /2 + M =

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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 Equivalenceto 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))