PS3sol8.pdf-Problem Set 3 Problem 1 Part
PS3sol8.pdf-Problem Set 3 Problem 1 Part
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PS3sol8.pdf-Problem Set 3 Problem 1...
PS3sol8.pdf-Problem Set 3 Problem 1 Part
##### Page 1
Problem Set 3
Problem 1
Part (a)
Each bidder bids his own value.
Part (b)
To compute the expected revenue, we consider each case separately. Both values are above
the reserve price
p
with probability
(
10
-
p
/
10
)
2
. In this case, the seller gets the expected value
of the second highest draw from a uniform between
p
and
10
. Using the formula seen in class,
this can be computed as the second highest from a uniform between 0 and
10
-
p
plus
p
, i.e.
1
/
3
(10
-
p
)+
p
. When one value is above
p
and the other is below, the revenue is
p.
This
happens with probability
p
/
10
(1
-
p
/
10
)
. Therefore, the expected revenue is
2
·
p
/
10
(1
-
p
/
10
)
·
p
+(
10
-
p
/
10
)
2
(
10
-
p
/
3
+
p
)
Taking the derivative with respect to
p
and making it equal to 0, we ﬁnd
p
=5
. We then
replace this in the equation above and get that with
p
=5
, the expected revenue is
25
/
6
.
Problem 2
Part (a)
p
if his value is above
p
and bids
0
otherwise.
As a result, the expected
revenue of a price
p
is
p
(10
-
p
)
/
10
, and the optimal price is
p
=5
. Therefore, the expected
revenues is
2
.
5
.
Part (b)
Suppose the seller sets a
secret
reserve price
p
. All that the buyer knows is that the seller
has randomly chosen the reserve price from a uniform distribution on
[0
,
10]
. Otherwise, the
rules are the same as in Part (a). (The buyer makes a bid
b
. If
b
p
1

##### Page 2
good and pays
b
. If
b<p
, the buyer does not get the good and pays nothing.) What is the
buyer’s optimal behavior? What is the seller’s expected revenue?
v
, he gets
b
-
v
if his bid
b
is above
p
, which happens with probability
F
(
b
)=
b/
10
; otherwise with probability, he gets
0
. Therefore, the buyer’s expected utility
when bidding
b
is
(
v
-
b
)
b/
10
. The optimal bidding behavior is then
b
(
v
)=
v/
2
.
This problem is equivalent to a standard ﬁrst price auction with two players, only that when
the sellers “wins”, he does not get anything. Therefore, we can compute the expected revenue
for the seller as what a bidder in a ﬁrst price auction with two bidders would pay in expected
value. This will be the expected value of the highest bid among two. As values are between
0
and
10
, bids will be between
0
and
5
because the optimal bid is
v
/
2
. The expected value of
the highest bid is then
2
/
3
5=
10
/
3
. But only half of the times the buyer has a value higher
than the seller’s reserve price, and even when this happens the bidder only bids has its value
so it wins the auction half of the time. Therefore, the expected revenue is
1
/
2
1
/
2
10
/
3
=
5
/
6
.
Problem 3
We can use the slides on all pay auctions to solve this problem. First note that in this case
the “bid” is the squared e
ort, because this is what a worker has to pay in the auction. From
the slides, we know that the optimal bid in an all pay auction is
v
2
/
2
when the values are
uniformly distributed between
0
and
1
. Therefore, the optimal e
ort level would be
v
/
p
2
if
the values were between
0
and
1
. In this case, the values are distributed between
10
and
20
so the optimal bid has to be adjusted.
All the conditions for the revenue equivalence theorem hold in this case, so we can just look
for the expected payment for a bidder in a second price auction. The optimal bid in a second
price auction is the bidder’s value
p
v
, and the bidder wins when his value
v
i
is higher than
the other player’s value
v
j
, which happens with probability
v
i
-
10
10
. If the bidder wins, he will
have to pay the expected value of the other’s value, conditional on being lower than his own
value
v
(so the other’s value is between
10
and
v
). This is
v
+10
2
. Then the expected payment
for a bidder in a second price auction would be
v
+10
2
v
-
10
10
. As the revenue equivalence theorem
holds, the payo
in the all-pay auction will be the same as the expected payo
in the second
price auction:
(
v
-
10)(
v
+10)
20
. Bidders pay the square of their bid or e
ort, so the optimal e
ort
will be
e
=
q
(
v
-
10)(
v
+10)
20
.
2

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