Econ 145: Problem Set IV
Suggested Answer
March 13, 2010
1. (A.151, SSS) You are one of
n
bidders who draw i.i.d. valuations from
F
(
x
). The probability that your
draw
x
is the highest is
F
(
x
)
n
−
1
.
Then what is the probability that your draw is
k
th highest? What
are these probabilities when each draw follow the uniform distribution on [0
,
1]?
Answer:
Denote
{
v
1
,v
2
,
···
n
−
1
n
}
are descendingly ordered draws among
n
bidders. I.e.
v
1
denotes the highest
draw and
v
n
denotes the lowest draw. Assuming your draw is
k
th highest and denote your draw as
x
(i.e.
x
=
v
k
). This means there are (
k
−
1) draws which are above than yours, and there are (
n
−
k
)draws
wh
icharebe
lowthanyours
.
v
n
≤
v
n
−
1
≤
≤
v
k
+1

{z
}
(
n
−
k
) draws below your draw
≤
x
{z}
your draw =
k
th highest dras =
v
k
≤
v
k
−
1
≤
≤
v
2
≤
v
1

{z
}
(
k
−
1) drasws above your draw
First, we consider following two probabilities without considering combinations.
(i) Probability of (
k
−
1) draws above your draw
x
Pr
[(
k
−
1) draws are above your draw
x
]=
[(
v
k
−
1
>x
)and
and (
v
2
)and(
v
1
)]
=
[(
v
k
−
1
)]
·····
[(
v
2
)]
·
[(
v
1
)]
=
{
1
−
F
(
x
)
}
k
−
1
Note that the second equality comes from the independence of draw.
(ii) Probability of (
n
−
k
) draws are below your draw
x
[(
n
−
k
) draws are below your draw
x
[(
v
n
≤
x
v
n
−
1
≤
x
and (
v
k
−
1
≤
x
)]
=
[(
v
n
≤
x
)]
·
[(
v
n
−
1
≤
x
)]
[(
v
k
−
1
≤
x
)]
=
{
F
(
x
)
}
n
−
k
Again, the second equality comes from the independence of draw.
Second, we consider the combinations. We know that there are (
k
−
1) draws above your draw
x
.out of
totally (
n
−
1) draws (note: excluding your draw). To choose such (
k
−
1) draws out of (
n
−
1), there are
μ
n
−
1
k
−
1
¶
combinations to do so
1
. Once you calculate “above” combinations, ”below” combinations are automati
cally determined. Since we have to consider all combinations, the probability of your draw is
k
th highest
1
Some math textbook use the notation of
n
−
1
C
k
−
1
to express combinations
1
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Pr
[your draw
x
is
k
th highest] =
μ
n
−
1
k
−
1
¶
·{
1
−
F
(
x
)
}
k
−
1
1
−
F
(
x
)
}
n
−
k
See footnote
2
.
2. Consider a second price auction with
n
bidders.
(a) Compute the seller’s expected revenue (in symmetric Bayesian Nash equilibrium) when each bidder’s
value is independently and uniformly distributed on [0
,
2]
.
Note: CDF of Order Statistic
There are
n
bidders. Assume valuation
v
is drawn independently from the Cumulative Distribution
Function (henceforce CDF)
F
(
x
).
(1)
G
1
,n
(
x
): CDF of highest valuation among
n
draws
G
1
,n
(
v
)=
{
F
(
x
)
}
n
=
⎧
⎪
⎨
⎪
⎩
F
(
x
)

{z
}
prob that a draw is below
x
⎫
⎪
⎬
⎪
⎭
n
The interpretation is (very) easy. To make the highest draw be less than or equal to
x
,a
l
l
n
inde
pendent draws have to be smaller than
x
.
(2)
G
2
,n
(
x
): CDF of highest valuation among
n
draws
G
2
,n
(
x
{
F
(
x
)
}
n
+
n
(1
−
F
(
x
))
{
F
(
x
)
}
n
−
1
=
{
F
(
x
)
}
n

{z
}
(i)
+
n
(1
−
F
(
x
))
{
F
(
x
)
}
n
−
1

{z
}
(ii)
=
{
F
(
x
)
}
n
+
n
{z}
n
possibilities
,
·
(1
−
F
(
x
))

{z
}
one draw is above
x,
F
(
x
)
}
n
−
1

{z
}
(
n
−
1) draws are below
x
The interpretation of this CDF is little bit tricky.
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 Winter '10
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 Auction, First Price Auction, Vwill, dvwill bequili, equilibrium bidding function

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