1
problem set 12
from Binmore’s
Fun and Games.
p.564 Ex. 41
p.565 Ex. 42
The Optimality of Auctions
An example:
• A seller sells an object whose value to him is
zero, he faces two buyers.
• The seller does not know the value of the object
to the buyers.
• Each of the buyers has the valuation 3 or 4 with
probability p, 1-p (respc.)
?
• The seller wishes to design a mechanism that
will yield the highest possible expected payoff.
3
First best
4 – p2
Posting price 3
3
Posting price 4
4(1 – p2)
1st price auction
????
2nd price auction
3+(1 – p)2
Modified 2nd price
auction
4–p
The
results
so
far
4
The Optimality of Auctions
First price auction
Player L will bid 3.
It is not optimal for H to play a pure strategy.
We look for an equilibrium in which H mixes with
probability distribution F(  ) on  3, K  .
When bidding b he expects to earn :
 4 - b   p +  1 - p  F  b 
This should be constant over the support,
with F  3  = 0, F  K  = 1
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The Optimality of Auctions
First price auction
When bidding b he expects to earn :
 4 - b   p +  1 - p  F  b 
This should be constant over the support,
with F  3  = 0, F  K  = 1
 4 - b   p +  1 - p  F  b   Const.
 4 - 3   p +  1 - p  F  3   p  Const.
p b-3
F b =
1- p 4 -b
K =4- p
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The Optimality of Auctions
First price auction
The expected payoff to the seller is :
3 + 1 - p
2
(same as 2
nd
price auction)
see ex. 11.10.39 in Binmore p.564
7
First best
4 – p2
Posting price 3
3
Posting
Posting price 4
price 4
4(1 – p2)
1st price auction
3+(1 – p)2
2nd price auction
3+(1 – p)2
Modified 2nd price
auction
4–p
4(1 – p2)
max
0
¼
Is there
a better
mechanism,
one that
yields a
higher
payoff
to the
seller?
4–p
1
p
8
The Optimality of Auctions
Player 2
types:
A general mechanism G
H , L
An Equilibrium:
4 , 3
tH
tL
sH
Player 1
types:
$
H
L
sL
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A Direct mechanism
Player 2
types:
{H,L}
The strategy set of player 2: {H,L}
H, L
H
The strategy set of player 1:
tH
When the players choose
the strategies X,Y
The outcome is
4 , 3
tL
sH
G(sX, tY)
Player 1
types:
$
H
L
sLL
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Player 2
types:
A Direct mechanism
H , L
Truth telling is an equilibrium
of the direct mechanism.
The payoff in this equilibrium
is the same as the payoff of
the quilibrium of G
Player 1
types:
H
L
4 , 3
tH
tL
sH
sL
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The Revelation Principle
For every mechanism G, and an equilibrium
E of it, there is a direct mechanism*D
in which truth telling is an equilibrium.
The truth telling equilibrium implements the
outcome of E.
*(the strategy set of each player is his set of types)
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A general direct mechanism D
Each player is allowed to announce H or
L.
D( , ): D(H , H), D(H , L), D(L , H) ,D(L , L)
describes the probability of obtaining the object and the
expected payoff
We are interested in describing a truth telling equilibrium
The mechanism can be described by 4 numbers:
h,l,H,L
Assuming the other player tells the truth:
A player who announces L
H
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wins the object with probability lh, ,and
andexpects
expectsto
topay
payLH
..
The seller chooses h,l,H,L to maximize his expected profits:
2[(1-p)H + pL]
subject to:
Individual rationality (participation) constraints:
4h - H ≥ 0
IRH
3l - L ≥ 0
IRL
Incentive Compatibility constraints
4h - H ≥ 4l - L
ICH
3l - L ≥ 3h -H
ICL
Assume we are at the optimal h,l,H,L,
which of the constraints are equalities?
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If IRL is a strict inequality
3l – L > 0
max 2[(1-p)H + pL]
s.t.
4h - H ≥ 0
IRH
Then by ICH :
3l - L ≥ 0
IRL
4h - H ≥ 4l - L ≥ 3l –L > 0
4h - H ≥ 4l - L
ICH
3l - L ≥ 3h -H
ICL
4h –H > 0
Now, increasing both H and L by the (same, small) constant
will keep IRH , IRL and will not change ICH , ICL.
However, this means that we could not have been at a
maximum of the objective function.
An increase in H and L improves the seller’s payoff
IRL must be an equality.
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If ICH is a strict inequality
4h – H > 4l - L
max 2[(1-p)H + pL]
s.t.
4h - H ≥ 0
IRH
Then by ICH and IRL :
3l - L = 0
IRL
4h - H > 4l - L ≥ 3l –L = 0
4h - H ≥ 4l - L
ICH
3l - L ≥ 3h -H
ICL
4h –H > 0
IRH is a strict inequality.
Now, increasing H by a small constant
will keep ICH , IRH and will not change IRL , ICL.
However, this means that we could not have been at a
maximum of the objective function.
An increase in H improves the seller’s payoff
ICH must be an equality.
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max 2[(1-p)H + pL]
Moreover, If ICH and IRL are
equalities, then IRH is satisfied.
And if h ≥ l then also ICL is
satisfied.
s.t.
4h - H ≥ 0
3l - L = 0
IRL
4h - H = 4l - L ≥ 3l –L = 0
4h - H = 4l - L
ICH
3l - L ≥ 3h -H
ICL
4h –H ≥ 0
IRH
IRH is satisfied
4h - H = 4l - L
3h - H = 4l – L - h = 3l – L + l - h ≤ 3l – L
ICL is satisfied
Note, that if ICL is satisfied
then h ≥ l
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max 2[(1-p)H + pL]
We therefore need to consider
s.t.
only ICH and IRL (and ensure
4h--LH=≥0 0
IRH
that the outcome satisfies h ≥ l ) 3l
IR
L
3l - L = 0
IR
4h - H = 4l - L
IC L
H
L = 3l
4h - H = 4l - L
ICH
3l - L ≥ 3h -H
ICL
H = 4h – l
max 2  1 - p  4h - l  + p  3l  
h,l
max 2  4h  1 - p  + l  4p - 1  
h,l
equivalent to maximizing:
1

h1 - p + l  p - 
4

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But symmetry imposes
additional constraints

1 

max  h  1 - p  + l  p -  
h,l
4 


The probability that a given
player wins is ≤ ½
(1-p)h + pl ≤ ½
The winning probabilities h,l
cannot be too large
h ≤ p+ ½(1-p) =½(1+p)
l ≤ 1 - p+ ½p =1 -½p
L
H wins
wins against
against H
L and
and wins
wins with
with prob.
prob. ½ against
against L
H
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
1 

max  h  1 - p  + l  p -  
h,l
4 


(1-p)h + pl ≤ ½
h ≤ ½(1 + p)
l ≤ 1 -½p
l
1 -½p
slope: -(1-p)/p
(1-p)h + pl = ½
½(1 + p)
h
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
1 

max  h  1 - p  + l  p -  
h,l
4 


For
For pp << ¼
¼ the slope of
the
maximum
is atConst
h(1-p)
+ l(p- ¼)=
lis=(1-p)/(¼-p)
0, h = ½(1> +0 p)
(1-p)h + pl ≤ ½
h ≤ ½(1 + p)
l ≤ 1 -½p
l
1 -½p
slope: -(1-p)/p
(1-p)h + pl = ½
½(1 + p)
h
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
1 

For
For pp >> ¼
¼ the slope of
max  h  1 - p  + l  p -  
h,l
4 


the
maximum
is
at
h(1-p) + l(p- ¼)= Const
(1-p)h + pl ≤ ½
lis=-(1
½p,- p)h/(p
= -½(1
+
p)
¼) < -(1 - p)/p
h ≤ ½(1 + p)
l ≤ 1 -½p
l
1 -½p
slope: -(1-p)/p
(1-p)h + pl = ½
½p
½(1 + p)
h
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For p > ¼

1 

max  h  1 - p  + l  p -  
h,l
4 


the maximum is at
l = ½p, h = ½(1 + p)
(1-p)h + pl ≤ ½
h ≤ ½(1 + p)
For p < ¼
l ≤ 1 -½p
the maximum is at
l = 0,
l h = ½(1 + p)
1 -½p
(1-p)h + pl = ½
½p
½(1 + p)
h
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For p > ¼

1 

max  h  1 - p  + l  p -  
h,l
4 


the maximum is at
l = ½p, h = ½(1 + p)
(1-p)h + pl ≤ ½
h ≤ ½(1 + p)
For p < ¼
l ≤ 1 -½p
the maximum is at
l = 0,
l h = ½(1 + p)
1 -½p
(1-p)h + pl = ½
½p
½(1 + p)
h
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For p > ¼
the maximum is at
l = ½p, h = ½(1 + p)

1 

max  h  1 - p  + l  p -  
h,l
4 


(1-p)h + pl ≤ ½
h ≤ ½(1 + p)
l ≤ 1 -½p
What is the seller’s payoff at this point ??
H = 4h – l = 2 + 3p/2
L=3l
= 3p/2
max 2  H  1 - p  + Lp  = 4 - p
Interpretation
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For p > ¼
the maximum is at
l = ½p, h = ½(1 + p)

1 

max  h  1 - p  + l  p -  
h,l
4 


(1-p)h + pl ≤ ½
h ≤ ½(1 + p)
H = 2 + 3p/2 L= 3p/2
l ≤ 1 -½p
max 2  H  1 - p  + Lp  = 4 - p
L wins against L with prob. ½:
½p
and pays 3 when he wins (expected payoff 3p/2)
H wins against L, and against H with prob. ½:
p + ½(1 - p) = ½(1 + p)
He pays 4 when he wins against H, and 3½ against L
(expected payoff: 3½·p +4·½(1 - p) = 2+3p/2 )
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For p > ¼
the maximum is at
l = ½p, h = ½(1 + p)
H = 2 + 3p/2 L= 3p/2
max 2  H  1 - p  + Lp  = 4 - p

1 

max  h  1 - p  + l  p -  
h,l
4 


(1-p)h + pl ≤ ½
h ≤ ½(1 + p)
l ≤ 1 -½p
Indeed, he could pay x when he wins against H,
and y against L, with
px +½(1 - p)y = 2+3p/2.
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For p < ¼
the maximum is at
l = 0, h = ½(1 + p)

1 

max  h  1 - p  + l  p -  
h,l
4 


(1-p)h + pl ≤ ½
h ≤ ½(1 + p)
l ≤ 1 -½p
What is the seller’s payoff at this point ??
H = 4h – l = 2(1 + p)
L=3l
= 0
max 2  H  1 - p  + Lp  = 4(1 – p2)
L never wins. H wins against L,
and against H
with prob. ½: p+ ½(1 - p) = ½(1 + p)
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and pays 4 when he wins (expected payoff 2(1 + p))