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
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
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