Econ 805 Advanced Micro Theory 1 Dan Quint Fall 2007 Lecture 3 – Sept

1 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 v i 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…

2 The Envelope Theorem

3  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

4 General Setup  Consider an optimization problem with choice variable x  X, parameterized by some parameter t  T: max x  X f(x,t)  Define the optimizer x*(t) = arg max x  X f(x,t) and the value function V(t) = max x  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

5 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

6 Derivative Version of the Envelope Theorem  Suppose T = [0,1]. Recall x*(t) = arg max x  X f(x,t).  Theorem. Pick any t  [0,1], any x*  x*(t), and suppose that f t =  f/  t exists at (x*,t).  If t < 1 and V’(t+) exists, then V’(t+)  f t (x*,t)  If t > 0 and V’(t-) exists, then V’(t-)  f t (x*,t)  If 0 < t < 1 and V’(t) exists, then V’(t) = f t (x*,t)  “The derivative of the value function is the derivative of the objective function, evaluated at the optimum”

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

8 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) ] = f t (x*, t)  The symmetric argument shows V’(t-)  f t (x*,t) when it exists  If V’(t) exists, V’(t+) = V’(t) = V’(t-), so f t (x*,t)  V’(t)  f t (x*,t)

9 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 f x and f t denote the partial derivatives of f with respect to its two arguments, V’(t) = f x (x*(t),t) x*’(t) + f t (x*(t),t)  By optimality, f x (x*(t),t) = 0, so the first term vanishes and V’(t) = f t (x*(t),t)  But we don’t want to rely on x* being single-valued and differentiable, or even continuous…

10 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

11 Several special cases that do guarantee V differentiable…  Suppose X is compact and f and f t are continuous in both their arguments. Then V is differentiable at t, and V’(t) = f t (x*(t),t), if…  x*(t) is a singleton, or  V is concave, or  t  arg max s 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…

12 Integral Version of the Envelope Theorem  Theorem. Suppose that  For all t, x*(t) is nonempty  For all (x,t), f t (x,t) exists  V(t) is absolutely continuous Then for any selection x(s) from x*(s), V(t) = V(0) +  0 t f t (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 f t (x(t),t) whenever it exists

13 When is V absolutely continuous?  Absolute continuity:  > 0,  > 0 s.t. for every finite collection of disjoint intervals {[a i, b i ]} i  1,2,…,K,  i | b i – a i | <    i | V(b i ) – V(a i ) | <   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 |f t (x,t)|  B(t) for all x  X for almost all t  [0,1] Then V is absolutely continuous.

14 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 | b i – a i | <    i | V(b i ) – V(a i ) | <   Assume V increasing (weakly), then we don’t have to carry around a bunch of extra terms

15 Proving V absolutely continuous   i | V(b i ) – V(a i ) | =  i | f(x*(b i ),b i ) – f(x*(a i ),a i ) |  Since f(x*(a i ), a i )  f(x,a i ), this is   i | f(x*(b i ),b i ) – f(x*(b i ),a i ) |  If f(x*(b i ),-) is absolutely continuous in t (assumption 1), this is   i  ai bi | f t (x*(b i ),s) | ds  If f t has an integrable bound (assumption 2), this is   i  ai bi B(s) ds  Now, let L =  i [a i, b i ], 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 = 

16 So to recap…  Corollary. Suppose that  For all t, x*(t) is nonempty  For all (x,t), f t (x,t) exists  For all x, f(x,-) is absolutely continuous  f t has an integrable bound: sup x  X | f t (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) +  0 t f t (x(s),s) ds

17 Revenue Equivalence

18 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

19 To show this, we will…  Show that sufficient conditions for the integral version of the Envelope Theorem hold  x*(t) nonempty for every t  f t =  f/  t exists for every (x,t)  f(x,-) absolutely continuous as a function of t (for a given x)  |f t (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

20 Sufficient conditions for the Envelope Theorem  Let b i : [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 b j (-)  If b i is an equilibrium strategy, b i (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  f t exists at all (x,t)  Fixing x, f is linear in t, and therefore absolutely continuous  f t is everywhere bounded above by B(t) = 1  So the integral version of the Envelope Theorem holds

21 Applying the Envelope Theorem  We know f t (x,t) = Pr(win | bid x) = Pr(all other bids < x)  For the envelope theorem, we care about f t at x = x*(t) = b i (t)  f t (b i (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) = F N-1 (t)  The envelope theorem then gives V(t) = V(0) +  0 t f t (b i (s),s) ds = V(0) +  0 t F N-1 (s) ds  By assumption, V(0) = 0, so V(t) =  0 t F N-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

22 As for the seller…  Since the bidder with the highest value wins the object, the sum of all the bidders’ payoffs is max(v 1,v 2,…,v N ) – Total Payments To Seller  The expected value of this is E(v 1 ) – R, where R is the seller’s expected revenue  By the envelope theorem, the sum of all bidders’ (ex-ante) expected payoffs is N E t V(t) = N E t  0 t F N-1 (s) ds  So R = E(v 1 ) - N E t  0 t F N-1 (s) ds which again depends only on F, not the rules of the auction

23 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 v 2 and therefore expected revenue E(v 2 ); so any auction satisfying these conditions has expected revenue E(v 2 )

24 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

25 Using Revenue Equivalence to Calculate Equilibrium Strategies

26 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) = F N-1 (t) t – b(t)  Revenue equivalence gave us V(t) =  0 t F N-1 (s) ds  Equating these gives b(t) = F N-1 (t) t –  0 t F N-1 (s) ds  Suppose types are uniformly distributed on [0,1], so F(t) = t: b(t) = t N -  0 t F N-1 (s) ds = t N – 1/N t N = (N-1)/N t N

27 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) = F N-1 (t) t – (F N-1 (t) + (N-1)F N-2 (t)(1-F(t)) b(t)  Revenue equivalence gave us V(t) =  0 t F N-1 (s) ds  Equating these gives b(t) = [ F N-1 (t) t –  0 t F N-1 (s) ds ] / (F N-1 (t) + (N-1)F N-2 (t)(1-F(t))