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where E ( π w ) is the winner's expected profit. This profit is:
E ( π w )= E ( f 1 )
E ( s 1 )
(5)
On the basis of the above equilibrium for a single object, we determine equilibrium
for sequential auctions for the m objects defined in Section 2 as follows.
Multiple objects . We first introduce some notation and them derive the equilibrium.
We will denote the first order statistic of the surplus for the j th (for j =1 ,...,m )
auction as f n−j +1
j and the second order statistic as s n−j + j . Also, we denote a bidder's
cumulative ex-ante expected profit from auctions j to m (where 1
m )as α j .
Finally, we denote the winner's expected profit for the j th auction as E ( π wj ).Given
this, the following theorem characterises the equilibrium for m> 1 objects.
Theorem 1. Fo r 1
j
m ,let β j and α j be defined as:
j
m and j
y
y− 1
β j =[
(1
1 / ( n
y + k + 1))][1 / ( n
j +1)]
(6)
k = j
m
( β j [ E ( f n−j +1 )
E ( s n−j +1 )+ α j +1 ])
α j =
(7)
y = j
where α m +1 =0 . Then the n -tuple of strategies ( B (
·
) ,...,B (
·
)) with B (
·
) defined in
Equation 8 constitutes an equilibrium for the j th (for j =1 ,..., ( m
1) ) auction at a
stage where k bidders have dropped out:
B 0 ( x ij )= E ( v ij
c ij |
S ij = x ij )
α j +1
B k ( x ij ; b 1 ,...,b k )= n
j +1
k
E ( v ij |
S ij = x ij )
E ( c ij |
S ij = x ij )
n
z +1
k− 1
1
+
E ( v ij |
B y ( S ij ; b 1 ,...,b y )= b y +1 )
n
j +1
y =0
α j +1
(8)
For the last auction, the equilibrium is as given in Equation 3 with n replaced with
( n
m +1) .
For the above equilibrium, the winner for the j th (for j =1 ,...,m ) auction is the bid-
der with the highest surplus for that auction (see proof of Theorem 3 in the appendix for
details). The following two theorems characterise the expected revenue and the winner's
expected profit.
Theorem 2. Fo r t h e j th (for j =1 ,...,m
1 ) auction, the winner's expected profit
(denoted E ( π wj ) )is:
E ( π wj )= E ( f n−j +1
E ( s n−j +1
j
)
)+ α j +1
(9)
j
and for the last auction, the winner's expected profit is:
E ( π wm )= E ( f n−m +1
) − E ( s n−m +1
m
)
(10)
m
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