COS 444
Internet Auctions:
Theory and Practice
Spring 2009
Ken Steiglitz
ken@cs.princeton.edu
week 6
1
Assignment 3: discussion
3.1 Techniques for average-of-losing-bids
auction
3.2 Best response
3.3 How can we encourage (subsidize?)
early entry?
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2
Conditional expectation
Intuitively clear, very useful in auction theory
If x is a random variable with cdf H, and A is an
event, define the conditional expectation of x
given A:
1
E[ x | A] 
xdH ( x)

prob{ A} A
Not defined when prob{ A } = 0.
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3
Two quick examples of conditional expectation
1) Suppose x is uniformly distributed on [0,1].
What is the expected value of x given that it is
less than a constant c ≤ 1?
2) Given two independent draws x1 and x2 , what
is the expected value of x1, given that x1≤ x2?
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4
Interpretation of FP equil.
Now take a look once more at the equilibrium
bidding function for a first-price IPV auction:
b fp

(v ) 
v
0
y dF ( y ) n1
F (v) n1
I claim this has the form of a conditional
expectation. What is the event A?
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Interpretation of FP equil.
Claim: event A = you win! = {Y1,(n-1) ≤ v }, where
Y1,(n-1) = highest of (n-1) independent draws
To check this
• prob { A}  F (v)n1
• Let y = Y1,(n-1) , the “next highest value”. The
cdf of y = Y1,(n-1) is F(v)n-1 so the integral in the
expected value of y given that you win is

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v
A
y dH ( y )   y dF ( y ) n1
0
6
Interpretation of FP equil.
Therefore,
b fp (v)  E[Y1,( n1) | Y1,( n1)  v]
 E[Y1,( n1) | v wins ]
That is, in equilibrium, bid the expected
next-highest value conditioned on your
winning. …Intuition?
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Stronger revenue equivalence
Let Psp(v) be the expected payment in equilibrium of
a bidder in a SP auction (and similarly for FP).
Psp (v)  prob {v wins} E [Y1,( n1) | v wins]
 prob {v wins }b fp (v)
 Pfp (v)
So SP and FP are revenue equivalent for each v !
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Graphical interpretation
• Once again, by parts:
Psp  Pfp  vF
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n 1
v
  F ( y ) n1 dy
0
9
Bidder preference revelation
Theorem: Suppose there exists a
symmetric Bayesian equilibrium in an IPV
auction, and assume high bidder wins.
Then this equilibrium bidding function is
monotonically nondecreasing.
*Thanks to Dilip Abreu for describing this elegant proof.
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Bidder preference revelation
• Proof: Bid as if your value is z when it’s
actually v. Let
w(z) = prob. of winning as fctn. of z
p(z) = exp. payment as fctn. of z
For convenience, let w=w(v), w΄=w(v΄),
p=p(v), p΄=p(v΄), for any v, v΄ .
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Bidder preference revelation
• From the definition of equilibrium, the
expected surplus satisfies:
v·w – p ≥ v·w΄ – p΄
v΄·w΄ – p΄ ≥ v΄·w – p
for every v,v΄ . Add:
(v – v΄ )·(w – w΄) ≥ 0 .
So v > v΄ → w ≥ w΄ → b(v) ≥ b(v΄). □
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12
Example of an IPV auction with
no symmetric Bayesian equil.:
third-price (see Krishna 02, p. 34)
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13
Riley & Samuelson 1981:
Optimal Auctions
• Elegant, landmark paper, constructs the
benchmark theory for optimal IPV
auctions with reserves
• Paradoxically, gets more powerful
results more easily by generalizing
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Riley & Samuelson’s class Ars
1.
2.
3.
4.
One seller, one indivisible object
Reserve b0 (open reserve, starting bid)
n bidders, with valuations vi i=1,…,n
Values iid according to cdf F, which is strictly
increasing, differentiable, with support [0,1] ( so f
>0)
5. There is a symmetric equilibrium bidding function
b(v) which is strictly increasing (we know by
preference revelation it must be nondecreasing)
6. Highest acceptable bid wins
7. Rules are anonymous
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Abstracting away…
• Bid as if value = z, and denote expected
payment of bidder by P(z). Then the
expected surplus is
v1F ( z )
n1
 P( z )
• For an equilibrium, this must be max at
z=v, so differentiate and set to 0:


d
n 1
x
F ( x)  P( x)  0
dx
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We need a boundary condition…
• Denote by v* the value at which it
becomes profitable to bid positively, called
the entry value:
P(v* )  v* F (v* )
n1
• Now integrate d.e. from v* to our value v1:
v1
P(v1 )  P(v* )   x dF ( x)
n 1
v*
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Once more, integrate by parts…
• And use the boundary condition:
P(v1 )  v1F (v1 )
n 1
v1
  F ( x) dx
n 1
v*
• A truly remarkable result! Why?
week 6
20
Once more, integrate by parts…
• And use the boundary condition:
P(v1 )  v1F (v1 )
n 1
v1
  F ( x) dx
n 1
v*
• A truly remarkable result! Why?
Where is the auction form? FP? SP?
Third-price? All-pay? …
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Revenue Equivalence Theorem
Theorem: In equilibrium the expected
revenue in an (optimal) Riley & Samuelson
auction depends only on the entry value v*
and not on the form of the auction. □
week 6
22
Marginal revenue, or virtual valuation
• Let’s put some work into this expected
revenue:
1
Rrs  n  [vF
v*
n1
v
  F ( x) dx]dF (v)
n1
v*
• And integrate by parts (of course, what
else?)…
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Marginal revenue, or virtual valuation
1  F (v )
n
Rrs   [v 
]
dF
(
v
)
v*
f (v )
1
1
Rrs   MR (v) dF (v)
n
v*
where
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1  F (v )
MR (v)  v 
f (v )
24
Interpretation of marginal revenue
Because F(v)n is the cdf of the highest,
winning value, we can interpret this as
saying:
The expected revenue of an (optimal) Riley
& Samuelson auction is the expected
marginal revenue of the winner.
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Hazard rate
Let the failure time of a device be distributed with
pdf f(t) and pdf F(t).
Define the “survival function” = R(t) = prob. of no
failure before time t = prob. of survival till time t.
Since F(t) = prob. of failure before t ,
R(t) = 1 – F(t).
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Hazard rate
The conditional prob. of failure in the interval
(t, t+Δt ] , given survival up to time t , is
R(t )  R(t  t )

R(t )
The “Hazard Rate” is the limit of this divided by
Δt as Δt → 0 :
 R
f
HR 

R
1 F
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Hazard rate
Thus, the marginal revenue, which is key to
finding the expected revenue in a RileySamuelson auction, is
1 F
1
MR  v 
v
f
HR
1/HR is the “Inverse Hazard Rate”
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Why call it “marginal revenue”?
Consider a monopolist seller who makes a takeit-or-leave-it offer to a single seller at a price p.
The buyer has value distribution F, so the prob.
of her accepting the offer is 1–F(p). Think of this
as the buyer’s demand curve. She buys, on the
average, quantity q = 1–F(p) at price p. Or, what
is the same thing, the seller offers price
p(q) = F-1(1– q) to sell quantity q.
after Krishna 02, BR 89
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Why call it “marginal revenue”?
The revenue function of the seller is therefore
q·p(q) = q F-1(1-q) , the revenue derived from
selling quantity q. The derivative of this wrt q is
by definition the marginal revenue of the
monopolist :
q
1
F (1  q) 
F ( F 1 (1  q))
F-1(1-q) = p, so this is
1  F ( p)
MR ( p)  p 
f ( p)
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□
30