Order Statistics and Revenue Equivalence
In the first section, we derive the first- and second-order statistics
for the uniform distribution on [0, 1], quickly introducing cdfs and
pdfs for those who don’t know them along the way. In the second
section, we derive the full kth order statistic. Finally, we use the results to prove that the expected revenue from the first-price auction is
identical to that of the second-price auction.
Definition 0.1. The kth order statistic, denoted X(k) of a statistical
sample is the kth largest draw (of n) from a sample.
For example, the first-order statistic is the maximum of n draws;
the nth order statistic is the minimum of n draws.
1
First-Order Statistic
1.1 Motivation
We want to compute the expected revenue of running the first-price
auction with n bidders, each of whose valuations are i.i.d. uniform
on [0, 1]. This is simply the highest bidder’s bid, or
bi =
n−1
vi ,
n
where vi is her valuation. But what is her valuation? Since the equilibrium bid function is increasing with increasing valuation, the expected valuation of the winner is simply the expected maximum of n
i.i.d. draws from the uniform distribution on [0, 1]. We compute this
value now.
1.2 Probability Review
To accommodate students with different probability backgrounds, we
first formalize exactly what we mean by expected value.
In the discrete world, the expected value of a random variable X is
given by the following:
n
E[ X ] =
∑ x i · P( X = x i ).
i =1
Consider, for example, the roll of a die. The expected value is given
by the following:
6
1
1
6
1 6·7
7
= , or 3.5.
2
2
∑i· 6 = 6 ∑i = 6 ·
i =1
i =1
order statistics and revenue equivalence
For a continuous analogue, we would like to write something like:
E[ X ] =
Z ∞
−∞
x · P( X = x ) dx,
however the probability of X equalling one precise value is always 0.
Therefore, we use instead:
E[ X ] =
Z ∞
−∞
x f X ( x ) dx,
where f X ( x ) is called the probability density function (PDF) of X with
the defining property that
P( a ≤ X ≤ b ) =
1.3
Z b
a
f X ( x ) dx.
Computation
What we want specifically is the expected value of X(1) , the first
order statistic, where X is drawn from a uniform distribution on
[0, 1]. That is,
E [ X (1) ] =
Z 1
0
x f X(1) ( x ) dx.
We first need to compute the PDF, f X(1) ( x ). We can easily compute
P( X(1) < x ), for some x ∈ [0, 1]; this is simply the probability that all
n draws are less than x, or
P ( X (1) < x ) = x n .
But we can also express this probability in terms of the PDF:
P ( X (1) < x ) =
Z x
−∞
f X(1) (t) dt = FX(1) ( x ).
This precise function has a name: the cumulative distribution function.
We use a capital F because it follows from the Second Fundamental
Theorem of Calculus that
f X (1) ( x ) =
Therefore,
f X (1) ( x ) =
d
FX ( x ).
dx (1)
d n
x = nx n−1
dx
and so the expected value is
E [ X (1) ] =
Z 1
0
x f X ( x ) dx =
Z 1
0
nx n dx =
n
.
n+1
2
order statistics and revenue equivalence
1.4 Second Order Statistic
We do the same computation, but much quicker this time.
P( X(2) < x ) = x n + nx n−1 (1 − x ).
f X(2) ( x ) = nx n−1 + n(n − 1) x n−2 (1 − x ) − nx n−1 = n(n − 1) x n−2 (1 − x ).
E [ X (2) ] =
Z 1
0
x f X(2) ( x )dx
Z 1
x n−1 − x n dx
0
1
1
−
= n ( n − 1)
n n+1
1
= n ( n − 1)
n ( n + 1)
n−1
=
.
n+1
= n ( n − 1)
2
kth Order Statistic
2.1 Beta Function
In this derivation, we make use of the beta function. The beta function B( x, y) is defined by the following integral:
B( x, y) =
Z 1
0
t x−1 (1 − t)y−1 dy.
Additionally, when x and y are positive integers, it is equal to the
following expression:
B( x, y) =
( x − 1) ! ( y − 1) !
.
( x + y − 1) !
We will make use of both definitions.
2.2 Derivation
We begin by computing the probability that the kth order statistic
lies in some small interval [ x, x + ∆x ] ⊂ [0, 1]. For convenience, we
consider each draw as a valuation by some agent; we regard these
agents as distinct individuals for counting purposes.
Where X is a random variable drawn uniformly from [0, 1],
P( X(k) ∈ [ x, x + ∆x ]) = P( X < x )
n−k
· P( X ∈ [ x, x + ∆x ]) · P( X > x + ∆x )
The first three terms are the probabilities of
k −1
n−1
·n
+ O(∆x2 )
k−1
3
order statistics and revenue equivalence
• n − k draws less than x,
• exactly one draw between x and x + ∆x,
• and k − 1 draws greater than x + ∆x,
respectively. However, this gives the probability of one specific arrangement of this form, so we multiply by the number of possible arrangements. There are n possible agents who could have a valuation
−1
between x and x + ∆, after which there are (nk−
1 ) possible groups of
agents who have valuations greater than x, after which the remaining
n − k agents are fixed so we don’t need to include that term. There
is a chance that there are multiple valuations between x and x + ∆x,
but each will contain a ∆xi term with i ≥ 2. Since we ultimately send
∆x → 0, we don’t bother writing these terms out.
The assumption of a uniform distribution gives us the following
simplification:
n−1
P( X(k) ∈ [ x, x + ∆x ]) = x n−k · ∆x · (1 − x − ∆x )k−1 · n
+ O(∆x2 )
k−1
n!
=
x n−k (1 − x − ∆x )k−1 ∆x + O(∆x2 ).
( k − 1) ! ( n − k ) !
Recall that discrete expectation is given by
m
∑ xi P(X(k) ∈ [xi , xi+1 ]),
i =1
where xi+1 = xi + ∆x (assume even-spaced intervals of length ∆x).
To get continuous expectation we take the limit as m → ∞. As we
do this, we can throw out ∆x terms because they become arbitrarily
small.
m
X(k) = lim
m→∞
= lim
m→∞
∑ xi P(X(k) ∈ [xi , xi + ∆x])
i =1
m
n!
∑ (k − 1)!(n − k)! xin−k+1 (1 − xi − ∆x)k−1 ∆x + O(∆x2 )
i =1
Z 1
n!
x n−k+1 (1 − x )k−1 dx
( k − 1) ! ( n − k ) !
Z 1
n!
=
x n−k+1 (1 − x )k−1 dx.
( k − 1) ! ( n − k ) ! 0
=
0
Finally, we make use of the beta function to produce the desired
result.
n!
X( k ) =
B(n − k + 2, k)
( k − 1) ! ( n − k ) !
n!
( n − k + 1) ! ( k − 1) !
=
( k − 1) ! ( n − k ) !
( n + 1) !
n+1−k
=
.
n+1
4
order statistics and revenue equivalence
3
Revenue Equivalence
Combining the above with the kth order statistic result,
X( k ) =
n+1−k
,
n+1
we can easily show that the expected revenue of running the firstprice auction is identical to that of the second-price auction.
Theorem 3.1. If bidder’s valuations are uniform and i.i.d., the expected
revenue of the first-price auction is identical to that of the second-price
auction, assuming bidders behave according to their respective dominant or
equilibrium strategies.
Proof. The support of the uniform distribution does not matter, so
we choose [0, 1] for convenience. Let R1 and R2 denote the expected
revenue of the first- and second-price auctions, respectively.
First we compute R1 . Let i∗ be the bidder with the highest valn−1
uation. Since the equilibrium bid function bi =
vi is strictly
n
increasing w.r.t. valuation, the winner is simply the bidder with the
highest type. Therefore, the expected revenue is equal to i∗ ’s bid:
R1 =
n−1
E [ v i ∗ ].
n
E[vi∗ ] is the expected value of maximum of n i.i.d. random variables
n
uniformly distributed on [0, 1], which is
. Thus,
n+1
R1 =
n−1 n
n−1
=
.
n n+1
n+1
In the second-price auction, the bidder with the highest valuation
wins, paying exactly the second highest valuation. Therefore, the
expected revenue is equal to second order statistic for a uniform
n−1
distribution on [0, 1], which is
. Thus,
n+1
R2 =
R1 = R2 .
n−1
.
n+1
5