9 E¢ ciency and the English Auction
Problem 9.1 (Two-bidder auctions) Suppose that there are two bidders with valua-
tions v1(x1; x2) = 2
3x1+1
3x2
v2(x1; x2) = 1
3x1+2
3x2
and all signals lie in [0;1].
a. Using the break-even conditions, find an e¢ cient ex post equilibrium of the
English (in this case, also the second-price) auction.
b. Show that the equilibrium strategies so determined survive the iterated elimina-
tion of weakly dominated strategies and are the only strategies to do so.
(Note: Iterated elimination of dominated strategies is carried out stepwise as follows.
In step 1, discard all weakly dominated strategies for both bidders. In step 2, discard
all weakly dominated strategies in the reduced game obtained after step 1. In step
3, discard all weakly dominated strategies in the reduced game obtained after step 2.
Continue in this fashion.)
Solution. Part a. Let 1(x)and 2(x)be the prices the bidders drop out at and
1and 2be the corresponding inverse functions. According to equation (9.3) in
section 9.2, the break-even conditions are
v1(1(p); 2(p)) = p
which can be rewritten as 2
31(p) + 1
32(p) = p
1
so
Hence the bidding functions are 1(x) = 2(x) = x.
Part b. Assume the conditional density functions fX1jX2fX2jX1are strictly
positive, and 1()and 1()are strictly increasing. If bidder 1 with signal x1bids
b1, his expected payo¤ is
2(b1)
and the marginal expected payo¤ is
@1
@b1
=2
3x1+1
31
2(b1)b1fX2jX11
2(b1)jx1
01(b1)
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Step 1. Because x22[0;1], we have 1
2(b1)2[0;1] :Therefore if b1>2
3x1+1
3,
the marginal expected payo¤ is
@1
@b1
=2
3x1+1
31
2(b1)b1fX2jX11
2(b1)jx1
01(b1)
2
2(b1)jx1
<0
where the first inequality comes from 1
2(b1)1. Hence bidder 1 could benefit by
reducing the bid, so he does not bid more than 2
3x1+1
3:Similarly, if b1<2
3x1;the
marginal expected payo¤ is
@1
@b1
=2
3x1+1
31
2(b1)b1fX2jX11
2(b1)jx1
01(b1)
2
2(b1)jx1
>0
where the first inequality comes from 1
2(b1)0;hence bidder 1 does not bid less
than 2
3x1:As a result, we have
Applying the same analysis to bidder 2 yields
Step 2. From step 1, we have 1(x1)22
3x1;2
3x1+1
3, therefore 1
2(b1)2
3
2b11
3;3
2b1:Repeating the same analysis in step 1 implies
1(x1)24
3x11
3;4
3x1
Step n: The strategies that survive the nsteps of iterated eliminations are
i(xi)2[nxi; nxin+ 1] if nis odd (14)
where
1=2
3
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Taking limit of both sides of the equation above yields lim
n!1n=2
31=lim
n!1n;therefore
lim
n!1n= 1. Therefore (14) and (15) imply that the set of bidder i’s strategies that
Problem 9.2 (Ordinality) Suppose is an increasing and di¤erentiable function
such that (0) = 0. If the valuations vsatisfy the average crossing condition, and
for all i,wi() = (vi()) then the valuations walso satisfy the average crossing
condition.
Solution. First, note that viis maximal at xif and only if wiis maximal at x. This
means that for all AN, the set of signals such that the values of all bidders in A
are maximal is the same for both vand w. Then, for any such vector of signals x
which satisfies that for all l2 A,
vl(x) = max
k2N vk(x)vx(16)
and for all i; j 2 A,i6=j,
@wA
@xj
(x) = 1
]AXi2A
@wi
@xj
(x)
=1
]AXi2A 0(vi(x)) @vi
@xj
(x)
@vi
@xj
where the third equality comes from (16) and the inequality comes from the average
crossing condition for v. Thus, the average crossing condition is also satisfied by w.
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