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Bid Rigging An Analysis of Corruption in Auctions

Bid Rigging An Analysis of Corruption in Auctions. Yvan Lengwiler (Univ of Basel, Switzerland) Elmar Wolfstetter (Humboldt Univ, Berlin). Corruption: What and How?. A buyer/seller delegates to an agent auctioneer.

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Bid Rigging An Analysis of Corruption in Auctions

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  1. Bid RiggingAn Analysis of Corruption in Auctions Yvan Lengwiler (Univ of Basel, Switzerland) Elmar Wolfstetter (Humboldt Univ, Berlin)

  2. 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 • by rigging the submitted bids. ›that is our model

  3. The Economist, "Just how rotten?", Oct 23, 2004. Example of orchestrating bids Mr. Spitzer against Marsh & McLennan and other major insurance brokers.

  4. 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 2003. • 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?

  5. 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.

  6. 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 v1, v2, … • Submitted bids are b1, b2, … Wlog bi≥ bi+1 • Valuations are drawn from distribution F with support [0,1]. Y1 and Y2 are the highest- and second highest order statistics. • Y1/G(x) := F(x)n-1. Joint density of Y1 and Y2 is (for z ≤ y)

  7. 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) + s0vP(y) dy. • Efficient auction allocates good to the bidder with the highest valuation, so allocation rule is P = G = Fn-1. G(v) is probability that v is larger than all the other n-1 draws. • RE generalized form: u(v) = u(0) + s0vG(y) dy. • RE strict form: in addition, u(0) = 0.

  8. 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 b1 to b2 (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 b2 to b1 (plus ), in exchange for a bribe. ›TYPE II • The gain from type I corruption is b1 – b2. • The gain from type II corruption is v2 – b1. • If equilibrium bid function  is monotone, auctioneer can infer valuation v2 = -1(b2). • Auctioneer chooses type I or type II, whichever is more profitable.

  9. 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 b2 – b3. Gain from type II corruption is v2 – b1. • We will see that in the SPA, type II corruption never occurs. Type I corruption, however, does occur in equilibrium.

  10. 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

  11. 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: b2 – b3. • 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 v2 argmaxxU(v,x).

  12. Second-price auction consider n > 2: Compute FOC, set x = v, use definition of fY2Y1, and rearrange… expected share of corruption cake if highest bidder expected share of corruption cake if second-highest bidder

  13. Second-price auction This yields 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.)

  14. Second-price auction • Why overbidding? • First, similar to a third price auction (or rather 2½ price auction) because winner expects to pay less than 2nd 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

  15. 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.

  16. 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: b1 – b2. Highest bidder receives a share , the auctioneer receives a share 1– .

  17. 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.

  18. 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

  19. 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 b1, b2, b3, …, bn. • He could propose type I corruption to the highest bidder: gain to be shared is b1 – b2. • Alternatively, he could propose type II corruption to the highest loser. The gain from corruption is v2 – b1 in that case. • Because the equilibrium is separating, the auctioneer can infer v2 = -1(b2). • The fact that auctioneer must infer the second highest valuation introduces a signalling aspect into the problem. • Off-equilibrium beliefs: if b2 > (1), assume b2 = 1; any other bid not in the range of , assume b2 = 0.

  20. 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.

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

  22. 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. First-price auction: full game Both restrictions must be satisfied for type II corruption to occur,

  23. x wins with type I y wins with type II x wins with type II y wins with type I y *(v,x) (x) x 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).

  24. 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.

  25. FPA: the monster

  26. 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 v¼ 0.

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

  28. s 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.

  29. 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: • Choose an initial function as close as possible to the solution. • Compute vector of changes that potentially reduces RMSE. • Optimize step size (how far to go in the chosen direction).

  30. 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).

  31. 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).

  32. Numerical results: bid functions n = 2 n = 5

  33. Numerical results: inefficiency n = 2,  = 0.6101 n = 2,  = 0.8 inefficiency bidder 2 wins v2 bidder 1 wins v1 v1 There is a wedge indicating cases with inefficient allocation (that is, whenever type II corruption occurs). The inefficiency-region becomes smaller with  and n.

  34. 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!

  35. Welfare and distribution

  36. 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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