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Market Design and Analysis Lecture 4

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Market Design and Analysis Lecture 4

Lecturer: Ning Chen (陈宁)

Email: ningc@ntu.edu.sg

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Question: For first price and second price auction, which generates more revenue?

Second price auction may look bad - the winner’s price may be much lower than his bid.

(An extreme once happened in a New-Zealand government-auction: One firm bid NZ$100,000 for a license, and paid the second-highest price of only NZ$6)

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In Economics, the (revenue) analysis of auctions is usually based on the assumption that the valuations of the buyers are drawn from a known distribution (e.g. [0,1]).

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Let’s consider a simple case – there are 2 buyers with values v1,v2 drawn independently and uniformly from [0,1].

For second price auction, they bid bi = vi, hence the expected revenue is E[min(v1,v2)].

For first price auction, what do they bid? How can we compute the expected revenue without knowing bi?

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A Bayesian game has, in addition to a normal game, for each player i, a type space Ti. There is a probability distribution P generating the type for each player. Player i

knows his (fixed) type ti∈Ti

only knows the probability distribution of types of other players

has a payoff functionui(s1,…,sn; ti): S1 x …x Sn x Ti R

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In a Bayesian game,

a strategy for player i is a function si(ti), i.e. si: Ti Si. For a given type ti, si(ti) gives the strategy selected by player i.

a strategy profile s* = (s*1,…,s*n) is called a Bayesian Nash equilibrium if for each player i and his type ti, s*i(ti) is the solution ofwhere

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For each buyer i,

there is a privately know true value (i.e. type) vi (assume drawn uniformly at random from [0,1])

decide bi, a function of vi, as his submitted bid

For first price auction, his utility is

ui(b1,…,bn;vi) = vi – biif i has largest bid and is a winner,

ui(b1,…,bn;vi) = 0 otherwise.

For second price auction, his utility is

ui(b1,…,bn;vi) = vi – maxj≠iif i has largest bid and is a winner,

ui(b1,…,bn;vi) = 0 otherwise.

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Claim. Bidding bi = vi (n-1)/n for each buyer i is a Bayesian Nash equilibrium for first price auction.

Proof. Consider any buyer i, assume all other buyers use this strategy.

For a bid b ≤ (n-1) / n, the probability of winning is (b∙n/(n-1))n-1. So the expected utility is (vi-b) (b∙n/(n-1))n-1

Derivative w.r.t b gives – (b∙n/(n-1))n-1 + (vi-b)(n-1)bn-2(n/(n-1))n-1 which should equal zero

Hence, –b + (vi-b)(n-1) = 0, which gives b = vi(n-1)/n.

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Let’s consider a simple case – there are 2 buyers with values v1,v2 drawn independently and uniformly from [0,1].

For second price auction, they bid bi = vi, hence the expected revenue is E[min(v1,v2)].

For first price auction, assume they bid according to Bayesian Nash equilibrium, i.e. bi = vi / 2. Then the expected revenue is E[max(v1,v2)] / 2.

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Fact. A non-negative random variable X satisfies

Proof. Let fX(z) be the density function of X. The we have Using the fact that

we have

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By uniform distribution, Pr[vi < x] = x Pr[vi ≥ x] = 1-x.

Since v1,v2 are independent, Pr[min(v1,v2) ≥ x] = Pr[v1 ≥ x & v2 ≥ x] = (1-x)2

Hence,

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E[max(v1,v2)] = 2/3.

Hence, E[max(v1,v2)] / 2 = 1/3.

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The Vickrey auction with reservation price r, VAr, sells the item if any buyer bids above r. The price the winning buyer pays is the maximum of the second highest bid and r.

Consider the revenue obtained when

VA0.5, i.e. r = 0.5

there are 2 buyers, and their values v1,v2 are drawn independently at random from the uniform distribution in [0,1].

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Analysis:

Case 1. both v1,v2 < 0.5. No one wins and no revenue

Case 2. one bid < 0.5, one bid ≥ 0.5. Get revenue 0.5

Case 3. both v1,v2 ≥ 0.5. This is the same as the above analysis except that the two variables are drawn uniformly from [0.5,1]. Thus, the expected revenue is 0.5+0.5/3 = 2/3

Therefore, the expected revenue is 0 * 1/4 + 0.5 * 1/2 + 2/3 * 1/4 = 5/12

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There are n agents, each desires some item or service.

Agent i’s value for receiving service is vi, and 0 otherwise.

The mechanism solicits sealed bids (b1,…,bn) from agents and computes an outcome, consisting of

allocation x = (x1,…,xn)

price / payment p = (p1,…,pn) ≥ 0

The utility of agent i is ui(bi) = vi∙xi(bi) – pi(bi).

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Assume there is a cost c(x) of producing the outcome x.

The surplus (or social welfare)of an outcome is defined by ∑i vixi - c(x)

The profit of an outcome is defined by ∑i pi - c(x)

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Single item auction:

Combinatorial auction: There are n various items and each agent i desires a bundle of items Si. Then

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We say a mechanism is truthful if truth-telling is a dominant strategy for every agent. That is, for any agent i, vi, biand b-i, ui(vi,b-i) ≥ ui(bi,b-i)

We assume all mechanisms are voluntary participation, i.e. no agent has negative expected utility for participation pi(bi) ≤ bi∙xi(bi).

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Theorem. A mechanism is truthful if and only if, for any agent i and bids of other agents b-i fixed,

xi(bi) is a monotone non-decreasing

pi(bi) = bi ∙xi(bi) – ∫0bi xi(z) dz

For a fixed monotone allocation rule x(∙), the payment rule p(∙) is also fixed. Hence, when specifying a truthful mechanism, we need only specify a monotone allocation rule.

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Proof. Assume the mechanism is truthful, i.e. for any value vi and possible lies bi, vi ∙xi(vi) – pi(vi) ≥ vi ∙xi(bi) – pi(bi)

Consider two possible values z1 and z2 with z1 < z2, and two scenarios:

vi = z1, bi = z2 z1 ∙xi(z1) – pi(z1) ≥ z1 ∙xi(z2) – pi(z2)

vi = z2, bi = z1 z2 ∙xi(z2) – pi(z2) ≥ z2 ∙xi(z1) – pi(z1)

Adding them gives z1 ∙ xi(z1) + z2 ∙ xi(z2) ≥ z1 ∙ xi(z2) + z2 ∙ xi(z1) Hence, (z2 – z1)∙(xi(z2) – xi(z1)) ≥ 0 xi(z2) ≥ xi(z1)

This is true for any z2 > z1, thus xi(∙) is monotone.

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Next we will show that the payment must bepi(bi) = bi ∙xi(bi) – ∫0bi xi(z) dz

Given ui(bi) = vi ∙xi(bi) – pi(bi), consider its partial derivative w.r.t bi:u’i(bi) = vi ∙x’i(bi) – p’i(bi)

Since truthfulness implies that ui(bi) is maximized at vi = bi, we have 0 = u’i(vi) = vi ∙x’i(vi) – p’i(vi)which holds for any value of vi. In particular, when vi = z, p’i(z) =z ∙ x’i(z)

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Hence,

By voluntary participation assumption, pi(0) = 0, and we are done (for the first direction).

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Proof. The other direction. Assume that xi(bi) is monotone and payment is pi(bi) = bi ∙xi(bi) – ∫0bi xi(z) dz

Consider two possible values z1 and z2 with z1 < z2. Suppose vi = z2, we will show that agent i does not benefit by shading his bid down to z1.

(The other case when vi = z1and agent i does not benefit by shading his bid up to z2 is left as an assignment.)

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The following figure shows xi(∙) is monotone by assumption.

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bidding bi = vi = z2

valuation vi ∙ xi(vi)

payment pi(vi) = vi ∙xi(vi) – ∫0vi xi(z) dz

utility ui(vi) = vi ∙ xi(vi) – pi(vi)

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bidding bi = z1

valuation vi ∙ xi(bi)

payment pi(bi) = bi ∙xi(bi) – ∫0bi xi(z) dz

utility ui(bi) = vi ∙ xi(bi) – pi(bi)

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The difference of utility when bidding bi = vi = z2and bidding bi = z1 < z2. (The monotonicity of the allocation implies this difference is always non-negative.)

Hence, the mechanism is truthful.

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Theorem. A mechanism is truthful if and only if, for any agent i and bids of other agents b-i fixed,

xi(bi) is a monotone non-decreasing

pi(bi) = bi ∙xi(bi) – ∫0bi xi(z) dz

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For any deterministic mechanism xi(∙)∈{0,1}, monotonicity implies that xi(∙) is a step function.If agent i wins, his payment is pi(vi) = vi ∙xi(vi) – ∫0vi xi(z) dz = infz {z: xi(z) = 1}

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Recall that the surplus is defined by ∑i vixi – c(x).

Consider an allocation rule that maximizes the surplus, i.e. given b, output allocation x maximizing ∑i bixi – c(x).

Theorem. The surplus maximizing allocation is monotone.

Proof. Let S(v,x) = ∑i vixi – c(x). Let v’ be the vector of values where v’i = vi + δ, and v’j = vjfor j ≠ i. ThenS(v’,x) = ∑i v’ixi – c(x) = δ∙xi + ∑i vixi – c(x) = δ∙xi+S(v,x)

Consider any x maximizing S(v,x) with the largest xi, then x maximizes S(v’,x) as well.

Thus, the value of xi is monotone.

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Assume that the value vi of agent i is drawn independently at random from Fi. Let F = F1 x F2 x … x Fn

The cumulative distribution function for viis Fi(z) = Pr[vi < z]

The probability density function for vi is fi(z) = F’i(z)

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The virtual valuation of agent i with value vi is

Given value vi, virtual valuation Φi(vi), and allocation x, the virtual surplus is

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Theorem. (Myerson’81) The expected profit of a mechanism is equal to its expected virtual surplus.

Myerson’s optimal mechanism outputs x to maximize the virtual surplus.

Given bids b, compute the “virtual bids” b’i = Φi(bi).

Compute the surplus maximizing allocation x’ and corresponding payment p’.

Output x = x’ and p, where pi= bi ∙xi(bi) – ∫0bi xi(z) dz

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Since maximizing-surplus allocation rule is monotone, Myerson’s optimal mechanism is truthful if and only if Φi(vi) is monotone in vi for any i.

The condition that Φi(vi) is monotone is consistent with the monotone hazard rate assumption, which is a common assumption in economics. (The hazard rate is define as f(z) / (1-F(z)).)

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Lemma. The expected payment of an agent, as a function of their bid bi and with all other bids fixed b-i, satisfies

Proof. (drop the subscript i) The bid b is selected at random from distribution F with density function f. Hence,

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Focusing on the 2nd term and switching the order of integration, we have

Rename z to b and factor x(b)f(b) out

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Proof of Myerson’s Theorem. Since the values of all agents are independently distributed andby linearity of expectation, we have

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Single item auction:

In single-item auctions, the surplus maximizing allocation gives the item to the agent with the highest value, unless the highest value is less than 0. (Usually the values of all agents are assumed to be at least 0.)

Myerson’s mechanism computes virtual surplus maximizing allocation, given virtual valuations, which can be negative. The mechanism, thus, allocates the item to the agent with the largest positive virtual valuation.

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Consider there are only two agents.

Agent 1 wins when Φ1(b1) ≥ max{Φ2(b2), 0}.

This is a deterministic allocation rule. Thus the payment is p1 = inf {b: Φ1(b) > Φ2(b2) and Φ1(b) > 0}

Suppose that F1 = F2 = F, then Φ1(z) = Φ2(z) = Φ(z)

If agent 1 wins, his payment isp1 = inf {b: Φ(b) > Φ(b2) and Φ(b) > 0}which isp1 = inf {b: b > b2 and b > Φ-1(0)}

If agent 2 winsp2 = inf {b: b > b1 and b > Φ-1(0)}

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Theorem. The optimal single-item auction for agents with values drawn independently from F is precisely the Vickrey auction with reservation price Φ-1(0).

For example, when F = [0,1], then

the cumulative distribution function F(z) = z

the density function f(z) = F’(z) = 1

Thus Φ(z) = 2z – 1 and the virtual valuations are drawn uniformly from [-1,1]

Hence, Φ-1(0) = 1/2 and the optimal mechanism for two agents with uniform value on [0,1] is the Vickrey auction with reservation price 1/2.

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Characterization of truthful mechanisms.

Steps to design an optimal mechanism to maximize the expected profit by Myerson’s optimal mechanism:

valuation distribution of each agent

virtual valuation and surplus

allocation rule maximizing virtual surplus

payment rule (according to the characterization theorem)

simplify allocation and payment rule.

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Assume that there are n agents, and the true value vi of each agent i is drawn independently from the same distribution F(v) on [vmin,vmax] (i.e. F(vmin)=0 and F(vmax)=1) with density function f(v). Assume that f(v) > 0 for any v∈[vmin,vmax].

In a Bayesian Nash equilibrium, agent i submits his bid (a function of vi) according to the equilibrium

assume that the probability that i wins is qi(vi)

assume that the expected payment of agent i is pi(vi)

The utility function of agent i is ui(vi) = vi∙qi(vi) – pi(vi)

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Lemma. For any value v and v’, ui(v) ≥ ui(v’) + (v – v’)∙qi(v’)

Proof. Let v be the true value of agent i, thenui(v’) + (v – v’)∙qi(v’) = v’qi(v’) – pi(v’) + (v – v’)∙qi(v’)= vqi(v’) – pi(v’)≤ ui(v)

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Given the lemma, ui(v) ≥ ui(v’) + (v – v’)∙qi(v’)

when v is the true value, we have ui(v) ≥ ui(v+dv) + (– dv)∙qi(v+dv)

when v+dv is the true value, we have ui(v+dv) ≥ ui(v) + (dv)∙qi(v)

Hence,

Taking the limit dv 0, we have

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expected

utility

the slope of the utility function is qi(v)

ui(vmin)

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Now consider any two mechanisms that have the same ui(vmin) and the same qi(v) functions for all v and agent i.

Asthey have the same utility function ui(v).

Hence, for any given value vi, agent i makes the same expected payment in these two mechanisms.

This implies the expected payment of agent i, across all possible values vi, is also the same.

Therefore, the two mechanisms have the same expected revenue.

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Consider any mechanisms which always

give the item to the highest-bid agent.

give an agent of lowest possible value no chance of any positive utility, i.e. ui(vmin) = 0.

Then qi(v) = (F(v))n-1, and all these mechanisms have the same expected revenue.

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Revenue Equivalence Theorem. Assume that

there are k items and n agents, where each agent desires one item.

the values are independently drawn from a common distribution [vmin,vmax], which is strictly increasing

Then any mechanism in which

the items always go to the k agents with the highest values,

any agent with value vmin expects 0 utility,

yields the same expected revenue, and each agent i has the same expected payment.

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The condition that f(v) > 0 for any v∈[vmin,vmax] is crucial for the theorem.

For example, there are two agents and their values are independently and equally either 0 or 1.

Consider the following mechanism: offer a price p to two agents simultaneously

if only one accepts, then sell him at p

if both accept, then choose a winner at random at price p

if both reject, then choose a winner at random at price 0

Then ui(0) = 0 and the higher-bid always wins. But the revenue strictly increases in p, so Revenue Equivalence fails.

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