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ECON30290 – Auction Theory
Unit 2 – The Revenue Equivalence Theorem
Omer Edhan
Omer Edhan 1 / 29 1 / 29
Section 2.1: Revenue Equivalence
for 1st and 2nd Price Auctions
Omer Edhan 2 / 29 2 / 29
Section 2.1A: Payment and
Revenue in 1st Price Auction
Omer Edhan 3 / 29 3 / 29
1st Price – Payment and Revenue
Buyer i with valuation x expected payment:
mI (x) = Pr(Win)× Bid
= G (x)× E [Yi |Yi ≤ x ]
Seller’s expected revenue:
e I
=
n
∫ v
v
mI (x)
f (x)dx
Omer Edhan 4 / 29 4 / 29
1st Price – Payment and Revenue
Buyer i with valuation x expected payment:
mI (x) = Pr(Win)× Bid
= G (x)× E [Yi |Yi ≤ x ]
Seller’s expected revenue:
e I
=
n
∫ v
v
mI (x)
f (x)dx
Omer Edhan 4 / 29 4 / 29
1st Price – Payment and Revenue
Buyer i with valuation x expected payment:
mI (x) = Pr(Win)× Bid
= G (x)× E [Yi |Yi ≤ x ]
Seller’s expected revenue:
e I
=
n
∫ v
v
mI (x)
f (x)dx
Omer Edhan 4 / 29 4 / 29
1st Price – Payment and Revenue
Buyer i with valuation x expected payment:
mI (x) = Pr(Win)× Bid
= G (x)× E [Yi |Yi ≤ x ]
Seller’s expected revenue:
e I
=
n
∫ v
v
mI (x)
f (x)dx
Omer Edhan 4 / 29 4 / 29
1st Price – Payment and Revenue
Buyer i with valuation x expected payment:
mI (x) = Pr(Win)× Bid
= G (x)× E [Yi |Yi ≤ x ]
Seller’s expected revenue:
e I
=
n
∫ v
v
mI (x)
f (x)dx
Omer Edhan 4 / 29 4 / 29
1st Price – Payment and Revenue
Buyer i with valuation x expected payment:
mI (x) = Pr(Win)× Bid
= G (x)× E [Yi |Yi ≤ x ]
Seller’s expected revenue:
e I =
n
∫ v
v
mI (x)
f (x)dx
Omer Edhan 4 / 29 4 / 29
1st Price – Payment and Revenue
Buyer i with valuation x expected payment:
mI (x) = Pr(Win)× Bid
= G (x)× E [Yi |Yi ≤ x ]
Seller’s expected revenue:
e I =
n
∫ v
v
mI (x)f (x)dx
Omer Edhan 4 / 29 4 / 29
1st Price – Payment and Revenue
Buyer i with valuation x expected payment:
mI (x) = Pr(Win)× Bid
= G (x)× E [Yi |Yi ≤ x ]
Seller’s expected revenue:
e I = n
∫ v
v
mI (x)f (x)dx
Omer Edhan 4 / 29 4 / 29
Section 2.1B: Payment and
Revenue in 2nd Price Auction
Omer Edhan 5 / 29 5 / 29
2nd Price – Payments and Revenue
Under a symmetric equilibrium β(x) = x (bid the valuation).
Bidder 1 with valuation x has expected payment:
mII (x)
= Pr(Win)× E [2nd bid |x is highest]
= G (x)× E [Yi |Yi < x ] = mI (x) Seller’s expected revenue: e II = n ∫ v v mII (x)f (x)dx = n ∫ v v mI (x)f (x)dx = e I . Omer Edhan 6 / 29 6 / 29 2nd Price - Payments and Revenue Under a symmetric equilibrium β(x) = x (bid the valuation). Bidder 1 with valuation x has expected payment: mII (x) = Pr(Win)× E [2nd bid |x is highest] = G (x)× E [Yi |Yi < x ] = mI (x) Seller’s expected revenue: e II = n ∫ v v mII (x)f (x)dx = n ∫ v v mI (x)f (x)dx = e I . Omer Edhan 6 / 29 6 / 29 2nd Price - Payments and Revenue Under a symmetric equilibrium β(x) = x (bid the valuation). Bidder 1 with valuation x has expected payment: mII (x) = Pr(Win)× E [2nd bid |x is highest] = G (x)× E [Yi |Yi < x ] = mI (x) Seller’s expected revenue: e II = n ∫ v v mII (x)f (x)dx = n ∫ v v mI (x)f (x)dx = e I . Omer Edhan 6 / 29 6 / 29 2nd Price - Payments and Revenue Under a symmetric equilibrium β(x) = x (bid the valuation). Bidder 1 with valuation x has expected payment: mII (x) = Pr(Win)× E [2nd bid |x is highest] = G (x)× E [Yi |Yi < x ] = mI (x) Seller’s expected revenue: e II = n ∫ v v mII (x)f (x)dx = n ∫ v v mI (x)f (x)dx = e I . Omer Edhan 6 / 29 6 / 29 2nd Price - Payments and Revenue Under a symmetric equilibrium β(x) = x (bid the valuation). Bidder 1 with valuation x has expected payment: mII (x) = Pr(Win)× E [2nd bid |x is highest] = G (x)× E [Yi |Yi < x ] = mI (x) Seller’s expected revenue: e II = n ∫ v v mII (x)f (x)dx = n ∫ v v mI (x)f (x)dx = e I . Omer Edhan 6 / 29 6 / 29 2nd Price - Payments and Revenue Under a symmetric equilibrium β(x) = x (bid the valuation). Bidder 1 with valuation x has expected payment: mII (x) = Pr(Win)× E [2nd bid |x is highest] = G (x)× E [Yi |Yi < x ] = mI (x) Seller’s expected revenue: e II = n ∫ v v mII (x)f (x)dx = n ∫ v v mI (x)f (x)dx = e I . Omer Edhan 6 / 29 6 / 29 2nd Price - Payments and Revenue Under a symmetric equilibrium β(x) = x (bid the valuation). Bidder 1 with valuation x has expected payment: mII (x) = Pr(Win)× E [2nd bid |x is highest] = G (x)× E [Yi |Yi < x ] = mI (x) Seller’s expected revenue: e II = n ∫ v v mII (x)f (x)dx = n ∫ v v mI (x)f (x)dx = e I . Omer Edhan 6 / 29 6 / 29 2nd Price - Payments and Revenue Under a symmetric equilibrium β(x) = x (bid the valuation). Bidder 1 with valuation x has expected payment: mII (x) = Pr(Win)× E [2nd bid |x is highest] = G (x)× E [Yi |Yi < x ] = mI (x) Seller’s expected revenue: e II = n ∫ v v mII (x)f (x)dx = n ∫ v v mI (x)f (x)dx = e I . Omer Edhan 6 / 29 6 / 29 2nd Price - Payments and Revenue Under a symmetric equilibrium β(x) = x (bid the valuation). Bidder 1 with valuation x has expected payment: mII (x) = Pr(Win)× E [2nd bid |x is highest] = G (x)× E [Yi |Yi < x ] = mI (x) Seller’s expected revenue: e II = n ∫ v v mII (x)f (x)dx = n ∫ v v mI (x)f (x)dx = e I . Omer Edhan 6 / 29 6 / 29 2nd Price - Payments and Revenue Under a symmetric equilibrium β(x) = x (bid the valuation). Bidder 1 with valuation x has expected payment: mII (x) = Pr(Win)× E [2nd bid |x is highest] = G (x)× E [Yi |Yi < x ] = mI (x) Seller’s expected revenue: e II = n ∫ v v mII (x)f (x)dx = n ∫ v v mI (x)f (x)dx = e I . Omer Edhan 6 / 29 6 / 29 Revenue Equivalence Theorem (Revenue Equivalence) In the 1st price and 2nd price symmetric sealed bid auctions with private and independent values: Buyers with the same valuation have the same expected payments. The sellers have the same expected revenue. Omer Edhan 7 / 29 7 / 29 Section 2.2: Revenue Equivalence Theorem for Symmetric Auctions with Private and Independent Values Omer Edhan 8 / 29 8 / 29 Section 2.2A: The Average Payment Approach Omer Edhan 9 / 29 9 / 29 Notation Recall the following notation in an Auction A: mA(x) = expected payment of buyer with valuation x . eA = expected revenue of the seller. Yi ≡ max j 6=i Xj the maximal competing valuation for buyer i . G (x) - the CDF of Yi . g(x) ≡ G ′(x) the PDF of Yi . Omer Edhan 10 / 29 10 / 29 Notation Recall the following notation in an Auction A: mA(x) = expected payment of buyer with valuation x . eA = expected revenue of the seller. Yi ≡ max j 6=i Xj the maximal competing valuation for buyer i . G (x) - the CDF of Yi . g(x) ≡ G ′(x) the PDF of Yi . Omer Edhan 10 / 29 10 / 29 Notation Recall the following notation in an Auction A: mA(x) = expected payment of buyer with valuation x . eA = expected revenue of the seller. Yi ≡ max j 6=i Xj the maximal competing valuation for buyer i . G (x) - the CDF of Yi . g(x) ≡ G ′(x) the PDF of Yi . Omer Edhan 10 / 29 10 / 29 The Equilibrium Payment in 1st and 2nd Price Auctions The equilibrium payment, conditional on winning, turns to be m(x) = G (x)E [Yi |Yi ≤ x ] It depends on x being the highest valuation as a condition for winning. Idea: Prove that this remains the same as long as having the highest valuation is the condition for winning. Omer Edhan 11 / 29 11 / 29 The Equilibrium Payment in 1st and 2nd Price Auctions The equilibrium payment, conditional on winning, turns to be m(x) = G (x)E [Yi |Yi ≤ x ] It depends on x being the highest valuation as a condition for winning. Idea: Prove that this remains the same as long as having the highest valuation is the condition for winning. Omer Edhan 11 / 29 11 / 29 The Equilibrium Payment in 1st and 2nd Price Auctions The equilibrium payment, conditional on winning, turns to be m(x) = G (x)E [Yi |Yi ≤ x ] It depends on x being the highest valuation as a condition for winning. Idea: Prove that this remains the same as long as having the highest valuation is the condition for winning. Omer Edhan 11 / 29 11 / 29 The Average Payment Approach Assume the highest bidder wins the auction. Let m(z) = mA(z) be the expected payment for buyer bidding b = β(z). The buyer’s expected payoff is UA(z |x) = G (z)x mA(z). Maximize w.r.t. z . FOC: 0 = g(z)x m′(z) Omer Edhan 12 / 29 12 / 29 The Average Payment Approach Assume the highest bidder wins the auction. Let m(z) = mA(z) be the expected payment for buyer bidding b = β(z). The buyer’s expected payoff is UA(z |x) = G (z)x mA(z). Maximize w.r.t. z . FOC: 0 = g(z)x m′(z) Omer Edhan 12 / 29 12 / 29 The Average Payment Approach Assume the highest bidder wins the auction. Let m(z) = mA(z) be the expected payment for buyer bidding b = β(z). The buyer’s expected payoff is UA(z |x) = G (z)x mA(z). Maximize w.r.t. z . FOC: 0 = g(z)x m′(z) Omer Edhan 12 / 29 12 / 29 The Average Payment Approach Assume the highest bidder wins the auction. Let m(z) = mA(z) be the expected payment for buyer bidding b = β(z). The buyer’s expected payoff is UA(z |x) = G (z)x mA(z). Maximize w.r.t. z . FOC: 0 = g(z)x m′(z) Omer Edhan 12 / 29 12 / 29 The Average Payment Approach Assume the highest bidder wins the auction. Let m(z) = mA(z) be the expected payment for buyer bidding b = β(z). The buyer’s expected payoff is UA(z |x) = G (z)x mA(z). Maximize w.r.t. z . FOC: 0 = g(z)x m′(z) Omer Edhan 12 / 29 12 / 29 The Average Payment Approach Assume the highest bidder wins the auction. Let m(z) = mA(z) be the expected payment for buyer bidding b = β(z). The buyer’s expected payoff is UA(z |x) = G (z)x mA(z). Maximize w.r.t. z . FOC: 0 = g(z)x m′(z) Omer Edhan 12 / 29 12 / 29 The Average Payment Approach In equilibrium z = x . Thus 0 = g(x)x m′(x) x m′(x) = xg(x) Omer Edhan 12 / 29 12 / 29 The Average Payment Approach In equilibrium z = x . Thus 0 = g(x)x m′(x) x m′(x) = xg(x) Omer Edhan 12 / 29 12 / 29 The Average Payment Approach In equilibrium z = x . Thus 0 = g(x)x m′(x) x m′(x) = xg(x) Omer Edhan 12 / 29 12 / 29 The Average Payment Approach In equilibrium z = x . Thus 0 = g(x)x m′(x) x m′(x) = xg(x) ∫ x x0 mA(x) = mA(x0) + ∫ x x0 yg(y)dy Omer Edhan 12 / 29 12 / 29 The Average Payment Approach In equilibrium z = x . Thus 0 = g(x)x m′(x) x m′(x) = xg(x) mA(x) = mA(x0) + ∫ x x0 yg(y)dy Omer Edhan 12 / 29 12 / 29 Section 2.2B: The Derivation of the Revenue Equivalence Theorem Omer Edhan 13 / 29 13 / 29 Revenue Equivalence Theorem Theorem (Revenue Equivalence Theorem (RET)) Consider two symmetric private values auctions A1,A2 in which 1. The highest bidder wins. 2. There is a symmetric increasing equilibrium. 3. In equilibrium mA1(x0) = m A2(x0) for some value x0. Then the buyer’s expected payment and the seller’s expected revenue is the same in both auctions. Proof. mA1(x) = mA1(x0) + ∫ x 0 yg(y)dy = mA2(x0) + ∫ x 0 yg(y)dy = mA2(x). Omer Edhan 14 / 29 14 / 29 Revenue Equivalence Theorem Theorem (Revenue Equivalence Theorem (RET)) Consider two symmetric private values auctions A1,A2 in which 1. The highest bidder wins. 2. There is a symmetric increasing equilibrium. 3. In equilibrium mA1(x0) = m A2(x0) for some value x0. Then the buyer’s expected payment and the seller’s expected revenue is the same in both auctions. Proof. mA1(x) = mA1(x0) + ∫ x 0 yg(y)dy = mA2(x0) + ∫ x 0 yg(y)dy = mA2(x). Omer Edhan 14 / 29 14 / 29 Revenue Equivalence Theorem Theorem (Revenue Equivalence Theorem (RET)) Consider two symmetric private values auctions A1,A2 in which 1. The highest bidder wins. 2. There is a symmetric increasing equilibrium. 3. In equilibrium mA1(x0) = m A2(x0) for some value x0. Then the buyer’s expected payment and the seller’s expected revenue is the same in both auctions. Proof. mA1(x) = mA1(x0) + ∫ x 0 yg(y)dy = mA2(x0) + ∫ x 0 yg(y)dy = mA2(x). Omer Edhan 14 / 29 14 / 29 Revenue Equivalence Theorem Theorem (Revenue Equivalence Theorem (RET)) Consider two symmetric private values auctions A1,A2 in which 1. The highest bidder wins. 2. There is a symmetric increasing equilibrium. 3. In equilibrium mA1(x0) = m A2(x0) for some value x0. Then the buyer’s expected payment and the seller’s expected revenue is the same in both auctions. Proof. mA1(x) = mA1(x0) + ∫ x 0 yg(y)dy = mA2(x0) + ∫ x 0 yg(y)dy = mA2(x). Omer Edhan 14 / 29 14 / 29 Example n buyers, F (x) = x2, x ∈ [0, 1]. Omer Edhan 15 / 29 15 / 29 Example F (x) = x2, x ∈ [0, 1] G (x) = F (x)n 1 = x2n 2. In any “highest bidder wins” private value symmetric auction A with mA(0) = 0: mA(x) = ∫ x 0 yg(y)dy = ∫ x 0 (2n 2)y2n 2dy = 2n 2 2n 1x 2n 1. The expected revenue is eA = nE [mA] = n ∫ 1 0 mA(x)f (x)dx = n ∫ 1 0 4n 4 2n 1x 2ndx = 4(n 1)n (2n 1)(2n + 1) = 4n(n 1) 4n2 1 Omer Edhan 15 / 29 15 / 29 Example F (x) = x2, x ∈ [0, 1] G (x) = F (x)n 1 = x2n 2. In any “highest bidder wins” private value symmetric auction A with mA(0) = 0: mA(x) = ∫ x 0 yg(y)dy = ∫ x 0 (2n 2)y2n 2dy = 2n 2 2n 1x 2n 1. The expected revenue is eA = nE [mA] = n ∫ 1 0 mA(x)f (x)dx = n ∫ 1 0 4n 4 2n 1x 2ndx = 4(n 1)n (2n 1)(2n + 1) = 4n(n 1) 4n2 1 Omer Edhan 15 / 29 15 / 29 Example F (x) = x2, x ∈ [0, 1] G (x) = F (x)n 1 = x2n 2. In any “highest bidder wins” private value symmetric auction A with mA(0) = 0: mA(x) = ∫ x 0 yg(y)dy = ∫ x 0 (2n 2)y2n 2dy = 2n 2 2n 1x 2n 1. The expected revenue is eA = nE [mA] = n ∫ 1 0 mA(x)f (x)dx = n ∫ 1 0 4n 4 2n 1x 2ndx = 4(n 1)n (2n 1)(2n + 1) = 4n(n 1) 4n2 1 Omer Edhan 15 / 29 15 / 29 Example F (x) = x2, x ∈ [0, 1] G (x) = F (x)n 1 = x2n 2. In any “highest bidder wins” private value symmetric auction A with mA(0) = 0: mA(x) = ∫ x 0 yg(y)dy = ∫ x 0 (2n 2)y2n 2dy = 2n 2 2n 1x 2n 1. The expected revenue is eA = nE [mA] = n ∫ 1 0 mA(x)f (x)dx = n ∫ 1 0 4n 4 2n 1x 2ndx = 4(n 1)n (2n 1)(2n + 1) = 4n(n 1) 4n2 1 Omer Edhan 15 / 29 15 / 29 Example F (x) = x2, x ∈ [0, 1] G (x) = F (x)n 1 = x2n 2. In any “highest bidder wins” private value symmetric auction A with mA(0) = 0: mA(x) = ∫ x 0 yg(y)dy = ∫ x 0 (2n 2)y2n 2dy = 2n 2 2n 1x 2n 1. The expected revenue is eA = nE [mA] = n ∫ 1 0 mA(x)f (x)dx = n ∫ 1 0 4n 4 2n 1x 2ndx = 4(n 1)n (2n 1)(2n + 1) = 4n(n 1) 4n2 1 Omer Edhan 15 / 29 15 / 29 Example F (x) = x2, x ∈ [0, 1] G (x) = F (x)n 1 = x2n 2. In any “highest bidder wins” private value symmetric auction A with mA(0) = 0: mA(x) = ∫ x 0 yg(y)dy = ∫ x 0 (2n 2)y2n 2dy = 2n 2 2n 1x 2n 1. The expected revenue is eA = nE [mA] = n ∫ 1 0 mA(x)f (x)dx = n ∫ 1 0 4n 4 2n 1x 2ndx = 4(n 1)n (2n 1)(2n + 1) = 4n(n 1) 4n2 1 Omer Edhan 15 / 29 15 / 29 Example F (x) = x2, x ∈ [0, 1] G (x) = F (x)n 1 = x2n 2. In any “highest bidder wins” private value symmetric auction A with mA(0) = 0: mA(x) = ∫ x 0 yg(y)dy = ∫ x 0 (2n 2)y2n 2dy = 2n 2 2n 1x 2n 1. The expected revenue is eA = nE [mA] = n ∫ 1 0 mA(x)f (x)dx = n ∫ 1 0 4n 4 2n 1x 2ndx = 4(n 1)n (2n 1)(2n + 1) = 4n(n 1) 4n2 1 Omer Edhan 15 / 29 15 / 29 Example F (x) = x2, x ∈ [0, 1] G (x) = F (x)n 1 = x2n 2. In any “highest bidder wins” private value symmetric auction A with mA(0) = 0: mA(x) = ∫ x 0 yg(y)dy = ∫ x 0 (2n 2)y2n 2dy = 2n 2 2n 1x 2n 1. The expected revenue is eA = nE [mA] = n ∫ 1 0 mA(x)f (x)dx = n ∫ 1 0 4n 4 2n 1x 2ndx = 4(n 1)n (2n 1)(2n + 1) = 4n(n 1) 4n2 1 Omer Edhan 15 / 29 15 / 29 Example F (x) = x2, x ∈ [0, 1] G (x) = F (x)n 1 = x2n 2. In any “highest bidder wins” private value symmetric auction A with mA(0) = 0: mA(x) = ∫ x 0 yg(y)dy = ∫ x 0 (2n 2)y2n 2dy = 2n 2 2n 1x 2n 1. The expected revenue is eA = nE [mA] = n ∫ 1 0 mA(x)f (x)dx = n ∫ 1 0 4n 4 2n 1x 2ndx = 4(n 1)n (2n 1)(2n + 1) = 4n(n 1) 4n2 1 Omer Edhan 15 / 29 15 / 29 Section 2.3: Applications Omer Edhan 16 / 29 16 / 29 Section 2.3A: All-Pay Auctions Omer Edhan 17 / 29 17 / 29 All Pay Auctions All the buyers pay their bid. Consider a “highest bidder wins” all-pay auction. Under the assumptions of RET, and mAP(0) = 0: βAP(x) = mAP(x) = ∫ x 0 yG ′(y)dy Is it really an equilibrium Omer Edhan 18 / 29 18 / 29 All Pay Auctions All the buyers pay their bid. Consider a “highest bidder wins” all-pay auction. Under the assumptions of RET, and mAP(0) = 0: βAP(x) = mAP(x) = ∫ x 0 yG ′(y)dy I

s it really an equilibrium
Omer Edhan 18 / 29 18 / 29
All Pay Auctions
All the buyers pay their bid.
Consider a “highest bidder wins” all-pay auction.
Under the assumptions of RET, and mAP(0) = 0:
βAP(x)
= mAP(x) =
∫ x
0
yG ′(y)dy
Is it really an equilibrium
Omer Edhan 18 / 29 18 / 29
All Pay Auctions
All the buyers pay their bid.
Consider a “highest bidder wins” all-pay auction.
Under the assumptions of RET, and mAP(0) = 0:
βAP(x)
= mAP(x) =
∫ x
0
yG ′(y)dy
Is it really an equilibrium
Omer Edhan 18 / 29 18 / 29
All Pay Auctions
All the buyers pay their bid.
Consider a “highest bidder wins” all-pay auction.
Under the assumptions of RET, and mAP(0) = 0:
βAP(x)
= mAP(x) =
∫ x
0
yG ′(y)dy
Is it really an equilibrium
Omer Edhan 18 / 29 18 / 29
All Pay Auctions
All the buyers pay their bid.
Consider a “highest bidder wins” all-pay auction.
Under the assumptions of RET, and mAP(0) = 0:
βAP(x) = mAP(x)
=
∫ x
0
yG ′(y)dy
Is it really an equilibrium
Omer Edhan 18 / 29 18 / 29
All Pay Auctions
All the buyers pay their bid.
Consider a “highest bidder wins” all-pay auction.
Under the assumptions of RET, and mAP(0) = 0:
βAP(x) = mAP(x) =
∫ x
0
yG ′(y)dy
Is it really an equilibrium
Omer Edhan 18 / 29 18 / 29
All Pay Auctions
All the buyers pay their bid.
Consider a “highest bidder wins” all-pay auction.
Under the assumptions of RET, and mAP(0) = 0:
βAP(x) = mAP(x) =
∫ x
0
yG ′(y)dy
Is it really an equilibrium
Omer Edhan 18 / 29 18 / 29
All Pay Auctions – Symmetric Equilibrium
Consider β = βAP
A bidder with valuation x bids b. Let z = β 1(b).
The bidder’s payoff
G (β 1(b))x b = G (z)x β(z)
= G (z)x
∫ z
0
yG ′(y)dy
=∫
by parts
G (z)(x z) +
∫ z
0
G (y)dy .
Choose z = x to maximise the expression βAP is Equilibrium.
Omer Edhan 19 / 29 19 / 29
All Pay Auctions – Symmetric Equilibrium
Consider β = βAP
A bidder with valuation x bids b. Let z = β 1(b).
The bidder’s payoff
G (β 1(b))x b = G (z)x β(z)
= G (z)x
∫ z
0
yG ′(y)dy
=∫
by parts
G (z)(x z) +
∫ z
0
G (y)dy .
Choose z = x to maximise the expression βAP is Equilibrium.
Omer Edhan 19 / 29 19 / 29
All Pay Auctions – Symmetric Equilibrium
Consider β = βAP
A bidder with valuation x bids b. Let z = β 1(b). The bidder’s payoff
G (β 1(b))x b = G (z)x β(z)
= G (z)x
∫ z
0
yG ′(y)dy
=∫
by parts
G (z)(x z) +
∫ z
0
G (y)dy .
Choose z = x to maximise the expression βAP is Equilibrium.
Omer Edhan 19 / 29 19 / 29
All Pay Auctions – Symmetric Equilibrium
Consider β = βAP
A bidder with valuation x bids b. Let z = β 1(b). The bidder’s payoff
G (β 1(b))x b = G (z)x β(z)
= G (z)x
∫ z
0
yG ′(y)dy
=∫
by parts
G (z)(x z) +
∫ z
0
G (y)dy .
Choose z = x to maximise the expression βAP is Equilibrium.
Omer Edhan 19 / 29 19 / 29
All Pay Auctions – Symmetric Equilibrium
Consider β = βAP
A bidder with valuation x bids b. Let z = β 1(b). The bidder’s payoff
G (β 1(b))x b = G (z)x β(z)
= G (z)x
∫ z
0
yG ′(y)dy
=∫
by parts
G (z)(x z) +
∫ z
0
G (y)dy .
Choose z = x to maximise the expression βAP is Equilibrium.
Omer Edhan 19 / 29 19 / 29
All Pay Auctions – Symmetric Equilibrium
Consider β = βAP
A bidder with valuation x bids b. Let z = β 1(b). The bidder’s payoff
G (β 1(b))x b = G (z)x β(z)
= G (z)x
∫ z
0
yG ′(y)dy
=∫
by parts
G (z)(x z) +
∫ z
0
G (y)dy .
Choose z = x to maximise the expression βAP is Equilibrium.
Omer Edhan 19 / 29 19 / 29
Section 2.3B: Political Contests I –
Lobbying
Omer Edhan 20 / 29 20 / 29
Model
Omer Edhan 21 / 29 21 / 29
Model
Consider the following model for Lobbies:
1. A politician determines which lobbyist receives a prize.
2. Lobbyist i valuation is xi > 0, chosen i.i.d. with CDF F .
3. Lobbyists pay the politician, up front, for her “services”.
Omer Edhan 21 / 29 21 / 29
Model
Consider the following model for Lobbies:
1. A politician determines which lobbyist receives a prize.
2. Lobbyist i valuation is xi > 0, chosen i.i.d. with CDF F .
3. Lobbyists pay the politician, up front, for her “services”.
Omer Edhan 21 / 29 21 / 29
Model
Consider the following model for Lobbies:
1. A politician determines which lobbyist receives a prize.
2. Lobbyist i valuation is xi > 0, chosen i.i.d. with CDF F .
3. Lobbyists pay the politician, up front, for her “services”.
Omer Edhan 21 / 29 21 / 29
More Formally
“Bribes” = bi , N(b) = {i : bi = maxj bj}.
Politician payoff W = E
n∑
i=1
bi .
Lobbyist i payoff pii =
{
xi
N(b) bi , bi = maxj bj
bi , otherwise
.
This is a “highest bidder wins” all pay auction!
Omer Edhan 22 / 29 22 / 29
More Formally
“Bribes” = bi , N(b) = {i : bi = maxj bj}.
Politician payoff W = E
n∑
i=1
bi .
Lobbyist i payoff pii =
{
xi
N(b) bi , bi = maxj bj
bi , otherwise
.
This is a “highest bidder wins” all pay auction!
Omer Edhan 22 / 29 22 / 29
More Formally
“Bribes” = bi , N(b) = {i : bi = maxj bj}.
Politician payoff W = E
n∑
i=1
bi .
Lobbyist i payoff pii =
{
xi
N(b) bi , bi = maxj bj
bi , otherwise
.
This is a “highest bidder wins” all pay auction!
Omer Edhan 22 / 29 22 / 29
More Formally
“Bribes” = bi , N(b) = {i : bi = maxj bj}.
Politician payoff W = E
n∑
i=1
bi .
Lobbyist i payoff pii =
{
xi
N(b) bi , bi = maxj bj
bi , otherwise
.
This is a “highest bidder wins” all pay auction!
Omer Edhan 22 / 29 22 / 29
Symmetric Equilibrium
RET implies the existence of symmetric equilibrium:
βAP(x) =
∫ x
0
yg(y)dy
with mAP(0) = 0.
Omer Edhan 23 / 29 23 / 29
Section 2.3C: Political Contests II
– Elections
Omer Edhan 24 / 29 24 / 29
Model
Consider elections with two candidates.
The candidate with highest budget campaign b wins.
Budget arrives from resource x .
Resources are i.i.d. with CDF F .
Resource x leads to payoff h(x) conditional on winning, h(0) = 0.
Losing leads to payoff 0.
Omer Edhan 25 / 29 25 / 29
Model
Consider elections with two candidates.
The candidate with highest budget campaign b wins.
Budget arrives from resource x .
Resources are i.i.d. with CDF F .
Resource x leads to payoff h(x) conditional on winning, h(0) = 0.
Losing leads to payoff 0.
Omer Edhan 25 / 29 25 / 29
Model
Consider elections with two candidates.
The candidate with highest budget campaign b wins.
Budget arrives from resource x .
Resources are i.i.d. with CDF F .
Resource x leads to payoff h(x) conditional on winning, h(0) = 0.
Losing leads to payoff 0.
Omer Edhan 25 / 29 25 / 29
Model
Consider elections with two candidates.
The candidate with highest budget campaign b wins.
Budget arrives from resource x .
Resources are i.i.d. with CDF F .
Resource x leads to payoff h(x) conditional on winning, h(0) = 0.
Losing leads to payoff 0.
Omer Edhan 25 / 29 25 / 29
Model
Consider elections with two candidates.
The candidate with highest budget campaign b wins.
Budget arrives from resource x .
Resources are i.i.d. with CDF F .
Resource x leads to payoff h(x) conditional on winning, h(0) = 0.
Losing leads to payoff 0.
Omer Edhan 25 / 29 25 / 29
Analysis
A “highest bidder wins” all pay auction.
For budget b, resource x , and z = β 1(b) the payoff is
G (z)h(x) mA(x) = G (z)h(x)
∫ z
0
h(y)G ′(y)dy
= G (z)(h(x) h(z)) +
∫ z
0
G (y)h′(y)dy .
If h is increasing and h′ is continuous, maximising over z we have x = z .
Thus
β(x) =
∫ x
0
h(y)G ′(y)dy
=
z=h(y)
∫ h 1(x)
0
z
G ′ h 1(z)
h′(z)
dz y = h 1(z) dy = dz
h′(z)
Omer Edhan 26 / 29 26 / 29
Analysis
A “highest bidder wins” all pay auction.
For budget b, resource x , and z = β 1(b) the payoff is
G (z)h(x) mA(x) = G (z)h(x)
∫ z
0
h(y)G ′(y)dy
= G (z)(h(x) h(z)) +
∫ z
0
G (y)h′(y)dy .
If h is increasing and h′ is continuous, maximising over z we have x = z .
Thus
β(x) =
∫ x
0
h(y)G ′(y)dy
=
z=h(y)
∫ h 1(x)
0
z
G ′ h 1(z)
h′(z)
dz y = h 1(z) dy = dz
h′(z)
Omer Edhan 26 / 29 26 / 29
Analysis
A “highest bidder wins” all pay auction.
For budget b, resource x , and z = β 1(b) the payoff is
G (z)h(x) mA(x) = G (z)h(x)
∫ z
0
h(y)G ′(y)dy
= G (z)(h(x) h(z)) +
∫ z
0
G (y)h′(y)dy .
If h is increasing and h′ is continuous, maximising over z we have x = z .
Thus
β(x) =
∫ x
0
h(y)G ′(y)dy
=
z=h(y)
∫ h 1(x)
0
z
G ′ h 1(z)
h′(z)
dz y = h 1(z) dy = dz
h′(z)
Omer Edhan 26 / 29 26 / 29
Analysis
A “highest bidder wins” all pay auction.
For budget b, resource x , and z = β 1(b) the payoff is
G (z)h(x) mA(x)
= G (z)h(x)
∫ z
0
h(y)G ′(y)dy
= G (z)(h(x) h(z)) +
∫ z
0
G (y)h′(y)dy .
If h is increasing and h′ is continuous, maximising over z we have x = z .
Thus
β(x) =
∫ x
0
h(y)G ′(y)dy
=
z=h(y)
∫ h 1(x)
0
z
G ′ h 1(z)
h′(z)
dz y = h 1(z) dy = dz
h′(z)
Omer Edhan 26 / 29 26 / 29
Analysis
A “highest bidder wins” all pay auction.
For budget b, resource x , and z = β 1(b) the payoff is
G (z)h(x) mA(x) = G (z)h(x)
∫ z
0
h(y)G ′(y)dy
= G (z)(h(x) h(z)) +
∫ z
0
G (y)h′(y)dy .
If h is increasing and h′ is continuous, maximising over z we have x = z .
Thus
β(x) =
∫ x
0
h(y)G ′(y)dy
=
z=h(y)
∫ h 1(x)
0
z
G ′ h 1(z)
h′(z)
dz y = h 1(z) dy = dz
h′(z)
Omer Edhan 26 / 29 26 / 29
Analysis
A “highest bidder wins” all pay auction.
For budget b, resource x , and z = β 1(b) the payoff is
G (z)h(x) mA(x) = G (z)h(x)
∫ z
0
h(y)G ′(y)dy
= G (z)(h(x) h(z)) +
∫ z
0
G (y)h′(y)dy .
If h is increasing and h′ is continuous, maximising over z we have x = z .
Thus
β(x) =
∫ x
0
h(y)G ′(y)dy
=
z=h(y)
∫ h 1(x)
0
z
G ′ h 1(z)
h′(z)
dz y = h 1(z) dy = dz
h′(z)
Omer Edhan 26 / 29 26 / 29
Analysis
A “highest bidder wins” all pay auction.
For budget b, resource x , and z = β 1(b) the payoff is
G (z)h(x) mA(x) = G (z)h(x)
∫ z
0
h(y)G ′(y)dy
= G (z)(h(x) h(z)) +
∫ z
0
G (y)h′(y)dy .
If h is increasing and h′ is continuous, maximising over z we have x = z .
Thus
β(x) =
∫ x
0
h(y)G ′(y)dy
=
z=h(y)
∫ h 1(x)
0
z
G ′ h 1(z)
h′(z)
dz y = h 1(z) dy = dz
h′(z)
Omer Edhan 26 / 29 26 / 29
Analysis
A “highest bidder wins” all pay auction.
For budget b, resource x , and z = β 1(b) the payoff is
G (z)h(x) mA(x) = G (z)h(x)
∫ z
0
h(y)G ′(y)dy
= G (z)(h(x) h(z)) +
∫ z
0
G (y)h′(y)dy .
If h is increasing and h′ is continuous, maximising over z we have x = z .
Thus
β(x)
=
∫ x
0
h(y)G ′(y)dy
=
z=h(y)
∫ h 1(x)
0
z
G ′ h 1(z)
h′(z)
dz y = h 1(z) dy = dz
h′(z)
Omer Edhan 26 / 29 26 / 29
Analysis
A “highest bidder wins” all pay auction.
For budget b, resource x , and z = β 1(b) the payoff is
G (z)h(x) mA(x) = G (z)h(x)
∫ z
0
h(y)G ′(y)dy
= G (z)(h(x) h(z)) +
∫ z
0
G (y)h′(y)dy .
If h is increasing and h′ is continuous, maximising over z we have x = z .
Thus
β(x) =
∫ x
0
h(y)G ′(y)dy
=
z=h(y)
∫ h 1(x)
0
z
G ′ h 1(z)
h′(z)
dz y = h 1(z) dy = dz
h′(z)
Omer Edhan 26 / 29 26 / 29
Analysis
A “highest bidder wins” all pay auction.
For budget b, resource x , and z = β 1(b) the payoff is
G (z)h(x) mA(x) = G (z)h(x)
∫ z
0
h(y)G ′(y)dy
= G (z)(h(x) h(z)) +
∫ z
0
G (y)h′(y)dy .
If h is increasing and h′ is continuous, maximising over z we have x = z .
Thus
β(x) =
∫ x
0
h(y)G ′(y)dy
=
z=h(y)
∫ h 1(x)
0
z
G ′ h 1(z)
h′(z)
dz
y = h 1(z) dy = dz
h′(z)
Omer Edhan 26 / 29 26 / 29
Analysis
A “highest bidder wins” all pay auction.
For budget b, resource x , and z = β 1(b) the payoff is
G (z)h(x) mA(x) = G (z)h(x)
∫ z
0
h(y)G ′(y)dy
= G (z)(h(x) h(z)) +
∫ z
0
G (y)h′(y)dy .
If h is increasing and h′ is continuous, maximising over z we have x = z .
Thus
β(x) =
∫ x
0
h(y)G ′(y)dy
=
z=h(y)
∫ h 1(x)
0
z
G ′ h 1(z)
h′(z)
dz y = h 1(z) dy = dz
h′(z)
Omer Edhan 26 / 29 26 / 29
Section 2.3D: Uncertain Number
of Buyers
Omer Edhan 27 / 29 27 / 29
Model
A known set of potential buyers N = {1, …, n}.
Actual buyers are a sample A N.
Buyer i Information: CDF F for values and a distribution
pim = Pr(|A| = m + 1) for sample size.
Symmetric Information: If i ∈ A then pim = pm.
Independence: Buyers values are chosen independently from the
sample.
Omer Edhan 28 / 29 28 / 29
Model
A known set of potential buyers N = {1, …, n}.
Actual buyers are a sample A N.
Buyer i Information: CDF F for values and a distribution
pim = Pr(|A| = m + 1) for sample size.
Symmetric Information: If i ∈ A then pim = pm.
Independence: Buyers values are chosen independently from the
sample.
Omer Edhan 28 / 29 28 / 29
Analysis
Conditional on |A| = m + 1, a buyer bidding β(z) wins with probability
G (z |m) = F (z)m
Her probability to win the auction is
G (z) =
n 1∑
m=0
pmG (z |m) =
n 1∑
m=0
pmF (z)
m
If her value x then her payoff is:
pi(z |x) = G (z)x mA(z).
The RET follows as before.
Omer Edhan 29 / 29 29 / 29
Analysis
Conditional on |A| = m + 1, a buyer bidding β(z) wins with probability
G (z |m) = F (z)m
Her probability to win the auction is
G (z) =
n 1∑
m=0
pmG (z |m) =
n 1∑
m=0
pmF (z)
m
If her value x then her payoff is:
pi(z |x) = G (z)x mA(z).
The RET follows as before.
Omer Edhan 29 / 29 29 / 29
Analysis
Conditional on |A| = m + 1, a buyer bidding β(z) wins with probability
G (z |m) = F (z)m
Her probability to win the auction is
G (z)
=
n 1∑
m=0
pmG (z |m) =
n 1∑
m=0
pmF (z)
m
If her value x then her payoff is:
pi(z |x) = G (z)x mA(z).
The RET follows as before.
Omer Edhan 29 / 29 29 / 29
Analysis
Conditional on |A| = m + 1, a buyer bidding β(z) wins with probability
G (z |m) = F (z)m
Her probability to win the auction is
G (z) =
n 1∑
m=0
pmG (z |m)
=
n 1∑
m=0
pmF (z)
m
If her value x then her payoff is:
pi(z |x) = G (z)x mA(z).
The RET follows as before.
Omer Edhan 29 / 29 29 / 29
Analysis
Conditional on |A| = m + 1, a buyer bidding β(z) wins with probability
G (z |m) = F (z)m
Her probability to win the auction is
G (z) =
n 1∑
m=0
pmG (z |m) =
n 1∑
m=0
pmF (z)
m
If her value x then her payoff is:
pi(z |x) = G (z)x mA(z).
The RET follows as before.
Omer Edhan 29 / 29 29 / 29
Analysis
Conditional on |A| = m + 1, a buyer bidding β(z) wins with probability
G (z |m) = F (z)m
Her probability to win the auction is
G (z) =
n 1∑
m=0
pmG (z |m) =
n 1∑
m=0
pmF (z)
m
If her value x then her payoff is:
pi(z |x) = G (z)x mA(z).
The RET follows as before.
Omer Edhan 29 / 29 29 / 29
Analysis
Conditional on |A| = m + 1, a buyer bidding β(z) wins with probability
G (z |m) = F (z)m
Her probability to win the auction is
G (z) =
n 1∑
m=0
pmG (z |m) =
n 1∑
m=0
pmF (z)
m
If her value x then her payoff is:
pi(z |x) = G (z)x mA(z).
The RET follows as before.
Omer Edhan 29 / 29 29 / 29
Analysis
Conditional on |A| = m + 1, a buyer bidding β(z) wins with probability
G (z |m) = F (z)m
Her probability to win the auction is
G (z) =
n 1∑
m=0
pmG (z |m) =
n 1∑
m=0
pmF (z)
m
If her value x then her payoff is:
pi(z |x) = G (z)x mA(z).
The RET follows as before.
Omer Edhan 29 / 29 29 / 29
Analysis
In 2nd price auction:
mII (x) =
n 1∑
m=0
pmF (z)
mE [Y (m)|Y (m) < x ]. Average payment = expectation over payments of 2nd price auction with m buyers, m ～ pm. In 1st price auction: mI (x) = G (x)βI (x). As mI (x) = mII (x) by the RET (please verify!): βI (x) = n 1∑ m=0 pmF (z) m G (x) E [Y (m)|Y (m) < x ] Omer Edhan 29 / 29 29 / 29 Analysis In 2nd price auction: mII (x) = n 1∑ m=0 pmF (z) mE [Y (m)|Y (m) < x ]. Average payment = expectation over payments of 2nd price auction with m buyers, m ～ pm. In 1st price auction: mI (x) = G (x)βI (x). As mI (x) = mII (x) by the RET (please verify!): βI (x) = n 1∑ m=0 pmF (z) m G (x) E [Y (m)|Y (m) < x ] Omer Edhan 29 / 29 29 / 29 Analysis In 2nd price auction: mII (x) = n 1∑ m=0 pmF (z) mE [Y (m)|Y (m) < x ]. Average payment = expectation over payments of 2nd price auction with m buyers, m ～ pm. In 1st price auction: mI (x) = G (x)βI (x). As mI (x) = mII (x) by the RET (please verify!): βI (x) = n 1∑ m=0 pmF (z) m G (x) E [Y (m)|Y (m) < x ] Omer Edhan 29 / 29 29 / 29 Analysis In 2nd price auction: mII (x) = n 1∑ m=0 pmF (z) mE [Y (m)|Y (m) < x ]. Average payment = expectation over payments of 2nd price auction with m buyers, m ～ pm. In 1st price auction: mI (x) = G (x)βI (x). As mI (x) = mII (x) by the RET (please verify!): βI (x) = n 1∑ m=0 pmF (z) m G (x) E [Y (m)|Y (m) < x ] Omer Edhan 29 / 29 29 / 29