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Bid Rigging An Analysis of Corruption in Auctions Yvan Lengwiler (Univ of Basel, Switzerland) Elmar Wolfstetter (Humboldt Univ, Berlin)

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Corruption: What and How? A buyer/seller delegates to an agent auctioneer. Auctioneer twists the auction rules in favor of some bidder(s) in exchange for bribes. Three ways: in a multidimensional (scoring) auction by manipulating the quality assessment, › we don't consider this by orchestrating bids (auctioneer works to make collusion among bidders possible), › I'll say some words on this that is our model by rigging the submitted bids. › that is our model

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Example of orchestrating bids The Economist, "Just how rotten?", Oct 23, Mr. Spitzer against Marsh & McLennan and other major insurance brokers.

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Example of orchestrating bids Insurance brokers have directed business to companies willing to pay "contingent fees." Brokers have forged bids to simulate competition. > $800 million just in Very large gain from corruption because the auctioneer controls the whole vector of bids. But the strategy is risky: many parties involved. Maybe, a more prudent corrupt auctioneer should behave in a more discrete fashion?

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Bid rigging After inspecting all bids, auctioneer approaches just one (possibly two) bidders. Allows this bidder to alter bid in exchange for a bribe. Advantage: small number of people with hard evidence on illegal activity. Select that bidder(s) that allow(s) the greatest gain from corruption, given the submitted bids. Example: Construction of Berlin airport. Example: Siemens bribed a Singapore government official to get access to competitors bids.

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The model: preliminaries One seller of a single good, n ≥ 2 potential buyers. Symmetric independent private values. Everyone is risk neutral. Seller delegates the auction to an auctioneer, who runs a sealed-bid first- or second-price auction. Valuations are denoted with v 1, v 2, … Submitted bids are b 1, b 2, … Wlog b i ≥ b i+1 Valuations are drawn from distribution F with support [0,1]. Y 1 and Y 2 are the highest- and second highest order statistics. Y 1 / G(x) := F(x) n-1. Joint density of Y 1 and Y 2 is (for z ≤ y)

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Revenue Equivalence Theorem Let P be the allocation rule: bidder with valuation v wins the auction with probability P(v). Let u(v) be the expected equilibrium payoff of a bidder with valuation v. Myerson (1981) has shown: u(v) = u(0) + s 0 v P(y) dy. Efficient auction allocates good to the bidder with the highest valuation, so allocation rule is P = G = F n-1. G(v) is probability that v is larger than all the other n-1 draws. RE generalized form: u(v) = u(0) + s 0 v G(y) dy. RE strict form: in addition, u(0) = 0.

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The model: corruption in FPA Consider first-price auction (FPA). Auctioneer views all bids. He contacts the highest bidder and offers him to revise his bid from b 1 to b 2 (plus ), in exchange for a bribe. › TYPE I After viewing the bids, the auctioneer contacts the highest loser and offers him to revise his bid from b 2 to b 1 (plus ), in exchange for a bribe. › TYPE II The gain from type I corruption is b 1 – b 2. The gain from type II corruption is v 2 – b 1. If equilibrium bid function is monotone, auctioneer can infer valuation v 2 = -1 (b 2 ). Auctioneer chooses type I or type II, whichever is more profitable.

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The model: corruption in SPA Type I: Auctioneer contacts two highest bidders, allows highest losing bidder to withdraw his bid, in exchange for a side payment and a bribe paid for by the winner. Type II: Auctioneer contacts highest losing bidder and allows him to match the highest submitted bid. in exchange for a bribe. Gain from type I corruption is b 2 – b 3. Gain from type II corruption is v 2 – b 1. We will see that in the SPA, type II corruption never occurs. Type I corruption, however, does occur in equilibrium.

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The model: cake-sharing and efficiency Exogenous shares for the involved bidder(s) [ ] and for the auctioneer [1- or 1-2 , depending on whether 1 or 2 bidders are involved]. NOTE: type I corruption does not jeopardize efficiency because it is still the bidder with the highest valuation who wins. It only affects the distribution of payoffs. › generalized RE holds This is not true for type II, because here it is not the highest valuation bidder who is allocated the good. › RE fails

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Second-price auction Start with the simpler SPA. We look for symmetric separating (monotone) equilibrium, . Strategy: assume that auctioneer contemplates only type I corruption, compute equilibrium bid function , then verify that type II is never profitable in equilibrium. Gain from corruption: b 2 – b 3. Cake sharing: the two involved bidders each receive a share , the auctioneer gets a share 1 – 2 . U(v,x) = expected payoff of bidder with valuation v who submits a bid (x), assuming that everyone else plays strategy . Symmetric equilibrium requires v 2 argmax x U(v,x).

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Second-price auction Compute FOC, set x = v, use definition of f Y 2 Y 1, and rearrange… consider n > 2: expected share of corruption cake if highest bidder expected share of corruption cake if second-highest bidder

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Second-price auction can be further simplified to... Observe that K(0) = 1 and K(1) = 0. Integration by parts then yields the solution... Note that (v) > v for all v < 1, so there is overbidding. (Even though payoff function is different for n = 2, it turns out that the equilibrium bid function is identical, and independent of n.) This yields

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Second-price auction Why overbidding? First, similar to a third price auction (or rather 2½ price auction) because winner expects to pay less than 2 nd highest bid through corruption. Second, by bidding high, expect larger side payment in case you come in second. Strict RE holds for n>2, but fails for n=2, because then u(0)>0. uniform distribution n arbitrary, = 0.5

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Second-price auction Equilibrium bid function if auctioneer contemplates type I only exhibits overbidding. As a consequence, this remains an equilibrium even if we allow the auctioneer to consider both types of corruption, because… …whenever auctioneer approaches losing bidder and invites him to match the highest bid, that bidder will decline since his bid already exceeds his valuation. The we have found is also an equilibrium of the full game.

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First-price auction: restricted game Follow the same strategy as before: let auctioneer contemplate only type I corruption; check if resulting (symmetric) equilibrium is also an equilibrium of the full game. Gain from type I corruption: b 1 – b 2. Highest bidder receives a share , the auctioneer receives a share 1 – .

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FPA: restricted game Note: if = 0, then payoff function is same as in ordinary FPA. If = 1, then payoff function is same as in ordinary SPA. In general, this is like a 1½ price auction. The solution is … This is like the solution of a standard FPA with (n – ) / (1 – ) bidders.

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FPA: restricted game Is this also an equilibrium of the full game? Note that is continuous and exhibits bid shading (the opposite of overbidding). Thus, there are draws of the two highest valuations so that the two highest bids are arbitrarily close to each other. Gain from type I is then very small. Gain from type II would be much greater in these cases. This is not an equilibrium of the full game. uniform distribution n = 2, = 0.5

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First-price auction: full game We look for a symmetric equilibrium which is strictly monotone and exhibits bid shading. (Both assumptions will verify.) Auctioneer receives the bids b 1, b 2, b 3, …, b n. He could propose type I corruption to the highest bidder: gain to be shared is b 1 – b 2. Alternatively, he could propose type II corruption to the highest loser. The gain from corruption is v 2 – b 1 in that case. Because the equilibrium is separating, the auctioneer can infer v 2 = -1 (b 2 ). The fact that auctioneer must infer the second highest valuation introduces a signalling aspect into the problem. Off-equilibrium beliefs: if b 2 > (1), assume b 2 = 1; any other bid not in the range of , assume b 2 = 0.

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First-price auction: full game As before, consider bidder with valuation v submitting a bid (x). Assume everyone else plays strategy (strictly monotone). Suppose (x) is the second-highest bid. Valuation of highest rival bidder is y ; highest competing bid is (y). For type II corruption to occur, two things must be true: auctioneer must prefer type II over type I, the second highest bidder must accept to participate in this corrupt scheme.

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First-price auction: full game Auctioneer proposes type II instead of type I if x – (y) ≥ (y) – (x). Auctioneer proposes type II instead of type I if Highest losing bidder accepts this proposal if v – (y) – (1 – )(x – (y)) ≥ 0, i.e. if Solve this for y : define

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First-price auction: full game Both restrictions must be satisfied for type II corruption to occur, In equilibrium (i.e., when setting x = v is optimal for the bidder), the accept-restriction is never binding though, because (by the bid shading assumption) unless x is considerably larger than v.

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First-price auction: full game Similarly, our bidder (with valuation x) is offered type I corruption if x > y and y is not offered type II corruption, (x) – (y) > y – (x), so y < (x) where (x) + ( (x)) < 2 (x). y x)x) *(v,x)*(v,x) x x wins with type I y wins with type II x wins with type II y wins with type I

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First-price auction: full game We can now state the payoff function of bidder with valuation v, bidding (x), and assuming that everyone plays strategy , We differentiate with respect to x, set x equal to v (equilibrium requirement) which yields, with some effort, a delayed differential equation.

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FPA: the monster

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FPA: existence? This is a delayed differential equation of a more complicated kind. The delays are not constant. One of the delays ( ) is itself an implicit function. There is (most probably) no closed-form solution for this… …except in special cases: if = 1 or when n ! 1, then (v) = v. Theory guarantees existence of some solution (for arbitrary and n), but not necessarily a monotone solution. We have no existence proof. In fact, we have a necessary condition for existence that is not always met. Assume uniform distribution. Consider 1st order Taylor approx around v = 0. Let s := ´(0), then (v) ¼ 2s/(1+s) and (s) ¼ (1+s)/2s for v ¼ 0.

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FPA: existence? s solves the polynomial We need a positive root because ´(0) must be > 0.

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FPA: existence? For too small , there is no positive s that solves the polynomial non-existence. This is only a necessary condition for existence, not a sufficient one. s

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Numerical procedure Assume uniform distribution. Turn the infinite dimensional root-finding problem into a finite dimensional problem by discretizing the valuation space, {0, 1/g, 2/g, …, (g –1)/g, 1} and looking for a piecewise linear solution. Sequence of steps for finding a root: 1.Choose an initial function as close as possible to the solution. 2.Compute vector of changes that potentially reduces RMSE. 3.Optimize step size (how far to go in the chosen direction).

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Numerical procedure Choice of starting point: Use linear bid function (e.g. the solution of the game with type I corruption only). Alternatively, do a grid search (is feasible only with very small g). Have tried different root-finding methods: steepest descent, Gauss-Newton, hybrid method (switch between the two), Levenberg-Marquardt (dynamic transition between steepest descent and Gauss-Newton). Have tried two line-search methods for optimizing stepsize: If RMSE is single-peaked along the chosen direction, golden section search is fast. Otherwise, do "brute force" (like grid search in one dimension).

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Numerical procedure Following combination works wonderfully for our problem. Reasonably fine grid (g = 200, so this is a 200-dimensional problem). Start from linear bid function. Use Levenberg-Marquardt to find direction in which to search … … then optimize step size with golden-section search. Implementation in C # (code available for download).

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# iterSSE # iterSSE n = E-22n = E E E E E E E E E E E E E-01 n = E-07n = E E E E E E E E E E E E E E E-05

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Numerical results: bid functions n = 2n = 5

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Numerical results: inefficiency n = 2, = v1v1 v2v2 bidder 2 wins bidder 1 wins inefficiency v1v1 n = 2, = 0.8 There is a wedge indicating cases with inefficient allocation (that is, whenever type II corruption occurs). The inefficiency-region becomes smaller with and n.

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Numerical results: allocation rule ALLOCATION RULE (n = 2)EXPECTED PAYOFF (gain compared to standard auction) Allocation rule is distorted compared to efficient rule. Note: bidder with valuation 1.0 does not win with for sure. All bidders gain in expectation. Surprisingly: They gain more if their share of the corruption cake is smaller!

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Welfare and distribution welfaresellerauctioneer welfaresellerauctioneer n = n = n = n =

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The bottom line Bid rigging is a more cautious scheme for corruption than bid orchestration. Bid rigging in FPA jeopardizes efficiency. Bid rigging in SPA does not jeopardize efficiency, but still hurts the seller to the benefit of the auctioneer only (except if n = 2, then bidders also benefit from it). Bid rigging in FPA benefits the auctioneer and also all bidders. It may be difficult to detect since no-one with hard evidence has a financial interest in exposing it. Bidders benefit more from bid rigging in FPA the smaller their share of the gain from corruption is (the weaker their bargaining position vis-à-vis the auctioneer). However, if the bidders’ bargaining power becomes too weak, an equilibrium fails to exist.

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