Answer.
Bidder
v
won’t participate in the auction if
v<
0
.
2
.
In
this case, this bidder’s expected payment is 0
.
Suppose that
v
≥
0
.
2
.
Bidder
v
pays the larger value of 0
.
2o
r
v
(1
,
2)
(max
{
0
.
2
,v
(1
,
2)
}
) conditional on winning, i.e.
v
(1
,
2)
≤
v.
So
bidder
v
’s expected payment is
E
[max
{
0
.
2
,v
(1
,
2)
}
v
(1
,
2)
≤
v
]Pr
³
v
(1
,
2)
≤
v
´
The probability to win is Pr
³
v
(1
,
2)
≤
v
´
=
G
1
,
2
(
v
)=
F
(
v
)
2
=
v
2
.
Conditional on winning (
v
(1
,
2)
≤
v
),
v
(1
,
2)
is distributed ac
cording to the probability density function
g
1
,
2
(
x
)
G
1
,
2
(
v
)
(=
2
x
v
2
)on[0
,v
].
1
We can compute the conditional expectation of max
{
0
.
2
,v
(1
,
2)
}
with respect to this pdf.
Hence bidder
v
’s expected payment is
E
[max
{
0
.
2
,v
(1
,
2)
}
v
(1
,
2)
≤
v
]Pr
³
v
(1
,
2)
≤
v
´
=
Z
v
0
max
{
0
.
2
,x
}
g
1
,
2
(
x
)
G
1
,
2
(
v
)
dx
·
G
1
,
2
(
v
)
=
Z
v
0
max
{
0
.
2
,x
}
g
1
,
2
(
x
)
dx
=
Z
0
.
2
0
0
.
2
g
1
,
2
(
x
)
dx
+
Z
v
0
.
2
xg
1
,
2
(
x
)
dx
=
Z
0
.
2
0
0
.
4x
dx
+
Z
v
0
.
2
2
x
2
dx
=
2
3
v
3
+
0
.
008
3
(d) Compute the seller’s expected revenue when the reserve price is
0
.
2
.
Answer.
The seller’s total expected revenue is
nE
∙
2
3
v
3
+
0
.
008
3
¸
1
This is a slight generalization of the conditional expectation formula. Let
X
≥
0bea
random variable with CDF
F.
Let
q
be any function. We can compute the expected value
of
q
(
X
) conditional on
X
≤
k
(
k
is some
f
xed number) as follows:
E
[
q
(
X
)

X
≤
k
]=
Z
k
0
q
(
x
)
f
(
x
)
F
(
k
)
dx.
The formula we know:
E
[
X

X
≤
k
]=
R
k
0
x
f
(
x
)
F
(
k
)
dx
is a special case of this.