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Equation 8. We know from Equation 4, that the expected revenue for the single
object case is the second order statistic of the surplus. The difference between the
equilibrium bids for the single object case and the ( m
1)th auction of the m
objects case is α m (see Equations 3 and 8). Hence, the expected revenue for the
( m
1)th auction is E ( s n−m +2
m− 1
α m . This implies that the winner's expected
)
profit for the ( m
1)th auction is:
E ( π w ( m− 1) )= E ( f n−m +2
E ( s n−m +2
m− 1
)+ α m
)
(17)
m− 1
-First ( m
m ,
let β j denote the ex-ante probability that a bidder wins the y th auction in the series
from j th to the m th auction. This probability is
2) auctions. We now find α 1 ,...,α m .For1
j
m and j
y
y
1
β j =[
(1
1 / ( n
y + k + 1))][1 / ( n
j +1)]
(18)
k = j
Also, let α j denote a bidder's cumulative ex-ante expected profit for all the auctions
from the j th to the m th. This profit is:
m
( β j [ E ( f n−j +1 )
E ( s n−j +1 )+ α j +1 ])
α j =
(19)
y = j
where α m +1 =0. Generalising Equation 17 to the first ( m
1) auctions, we get
the winner's expected profit ( E ( π wj ))as:
E ( π wj )= E ( f n−j +1
E ( s n−j +1
j
)
)+ α j +1
(20)
j
In other words, a bidder's optimal bid for the j th auction is obtained by discounting
the single object equilibrium bid by α j +1 . Hence, we get the equilibrium bids in
Equation 8.
B
Proof of Theorem 3
Proof. For the j th (for j =1 ,...,m ) auction, the total expected surplus that gets split
between the auctioneer and the winning bidder is E ( V )
s = f n−j + j ).From
this surplus, the winning bidder gets a share of E ( π wj ). Hence, the seller's revenue is
ER j = E ( V )
E ( c
|
s = f n−j +1
j
E ( c
|
)
E ( π wj ).
C
Proof of Theorem 2
Proof. The m th auction is identical to the single object case. Hence, the expected profit
for this auction is:
E ( s n−m + m )
Since the difference between the expected revenue for the single object case and the j th
(for j =1 ,...,m− 1)auctionofthe m objects case is α j +1 , the winner's expected
profit for the j th auction is E ( π wj )= E ( f n−j +1
E ( π wm )= E ( f n−m +1
)
m
E ( s n−j +1
j
)
)+ α j +1 .
j
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