Econ 805 Advanced Micro Theory 1

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## Econ 805 Advanced Micro Theory 1

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**Econ 805Advanced Micro Theory 1**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: ò-¥¥ u(s) dF(s) ³ò-¥¥ 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 • Show that X SOSD Y if and only if ò-¥¥ h(x,q) dF(x) ³ò-¥¥ 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**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**• 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**• Consider an optimization problem with choice variable x Î X, parameterized by some parameter t Î T: maxx Î X f(x,t) • Define the optimizer x*(t) = arg maxx Î Xf(x,t) and the value function V(t) = maxxÎXf(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 maxxÎXf(x,t). • Theorem. Pick any tÎ[0,1], any x* Îx*(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)Îx*(t+e) • By optimality, f(x(t+e),t+e) ³ f(x*,t+e), so V’(t+) ³lime 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)**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 • tÎ arg maxsV(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 xÎX, and • There exists an integrable function B(t) such that for almost all tÎ[0,1], |ft(x,t)|£B(t) for all xÎX 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 | < dSi| 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 =Siòaibift(x*(bi),s) ds • If ft has an integrable bound (assumption 2), this is £SiòaibiB(s) ds**Proof, cont’d**• Trying to show SiòaibiB(s) ds <e • Let L = Èi [ai, bi], J = { t : B(t) > M }, and K be the set with |K| £dthat maximizes òKB(s) ds • Recall thatòJB(s) ds < e/2 • Now, |K – J| £ |K| £d ;and B(t) £ M for all t in K – J ; so òLB(s) ds £òKB(s) ds £òJB(s) ds + òK-JB(s) ds < e /2 + d M = e • QED**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**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)Îx*(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) = NEtò0t FN-1(s) ds • So R = E(v1) – N Etò0t 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 Equivalenceto 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))