Econ 805 Advanced Micro Theory 1 Dan Quint Fall 2009 Lecture 3

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

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) are–u’’(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

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…

The Envelope Theorem

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

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) 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

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

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 < 1 and V’(t+) exists, then V’(t+) ³ 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”

Derivative Version of the Envelope Theorem f(x*,-) V(-) t

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) Î x*(t+e) By optimality, f(x(t+e),t+e) ³ f(x*,t+e), so V’(t+) ³ 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)

Like I said, this gives us some easy proofs… Shepherd’s 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 = ¶ (p*x)/ ¶ pi = xi, evaluated at the optimum (hi) Hotelling’s Lemma – same result for producer theory (firm’s net supply of an output/input is partial derivative of profit function with respect to price)

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…

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

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 don’t need differentiability everywhere – all we actually need is differentiability “most of the time”

Absolute Continuity Definition: V is absolutely continuous if " e > 0, $ d > 0 such that for every finite collection of disjoint intervals {[ai, bi]}i Î 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 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. (We’ll prove this in a moment.)

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) 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

Proving f(x,-) abs cont and |ft| has an integrable bound  V abs cont First: since B is integrable, limx  ¥ ò { 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 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) > M } B(s) ds < e /2, and let d = e /2M

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) 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

Proof, cont’d Trying to show Si òaibi B(s) ds < e Let L = Èi [ai, bi], J = { t : B(t) > M }, and K be the set with |K| £ d that maximizes òK B(s) ds Recall that òJ B(s) ds < 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 < e /2 + d M = e QED

V(t) = V(0) + ò0t ft(x(s),s) ds 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

Revenue Equivalence

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

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

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) 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

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) + 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

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 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

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)

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

Using Revenue Equivalence to Calculate Equilibrium Strategies

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) = 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

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) = 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))