Vitaminc,
Hi again. I assume we're still working out way through Robert Gibbon's
Game Theory book :)
Anyway here's the approach that I would take with such a question.
Firstly - You want to get a good diagram of how the game progresses.
That's going to be tough here (I think I outdid myself with my last
answer and the ascii art diagrams... but let me try one more)
t=0 B's valuation is determined - 1<=x<=100
/ \
/ \
/ \
t=1 B determines their optimal bid
(with full information of his valuation and knowledge of A's
valuation)
/ \
/ \
/ \
t=2 (A determines their optimal bid with only knowledge of vA=50)
(Nodes are circled to show that they are indistinguishable to
A)
/ \
/ \
/ \
t=3 Bids are revealed and valuations seen.
We now figure out the utility for each agent :
Let vA = 50 be the valuation of the object to A.
Let vB = x be the valuation of the object to B.
Let bidA = A's Bid.
Let bidB = B's Bid.
See below for the following sections :
PART A - SECOND PRICE AUCTION - Sale price is lower bid price
PART B - FIRST PRICE AUCTION - Sale price is high bid price
------
PART A - SECOND PRICE AUCTION - Sale price is lower bid price
Given vA, vB, bidA and bidB : A receives :
vA - bidB if bidA > bidB
0 if bidA <= bidB
Given vA, vB, bidA and bidB : B receives :
0 if bidA > bidB
vB - bidA if bidA <= bidB
Now we consider the optimal bid of B, given that A bids bidA and that
B observes a valuation of vB :
Suppose bidB > vB.
Then expected payoff = 0 for bidA > bidB > vB
= vB - BidA < 0 for bidB > bidA > vB
= 0 for bidA = vB
= vB - BidA > 0 for vB > bidA
Suppose bidB < vB.
Then expected payoff = 0 for bidA > vB > bidB
= 0 for bidA = vB
= 0 for vB > bidA > bidB
= vB - BidA > 0 for vB > bidB = bidA
= vB - BidA > 0 for vB > bidB > bidA
Suppose bidB = vB.
Then expected payoff = 0 for bidA > vB
= 0 for bidA = vB
= vB - BidA > 0 for bidA < bidB
The third strategy strictly beats the first one and weakly beats the
second one. So assume that bidB = vB.
Given that this is the expected behavior of B, we now turn to the
behavior of A. First we condition on vB = x (and later take
expectations)
A gets :
(vA - bidB) * 1{bidA > bidB|x}
= (vA - x) * [1{bidA > x|x}]
Where 1{} is an indicator function that takes the value 1 if the bid
is successful given x.
A's expected payoff is :
Sum of (vA - x) * 1{bidA > x|x} * P(vB = x) over x = 1 to 100.
= Sum of (vA - x) * 1{bidA > x|x} / 100 over x = 1 to 100.
= 0.01 * Sum of (vA - x) from x = 0 to bidA
(using the fact that sum x from 1 to n is .5n(n+1)
= 0.01 * [bidA * vA - .5*bidA*(bidA+1)]
Optimum occurs when vA = bidA + .5
As this occurs at bidA = 49.5, we must evaluate the utility for bidA =
49 and bidA = 50.
When bidA = 50, expected payoff is 0.01 * [50*50 - .5*50*51] = 12.25
When bidA = 49, expected payoff is 0.01 * [49*50 - .5*49*50] = 12.25
So the pure strategy nash equilibria are :
1)
bidA = 50
bidB = vB
2)
bidA = 49
bidB = vB
------
PART B - FIRST PRICE AUCTION - Sale price is high bid price
Given vA, vB, bidA and bidB : A receives :
vA - bidA if bidA > bidB
0 if bidA <= bidB
Given vA, vB, bidA and bidB : B receives :
0 if bidA > bidB
vB - bidB if bidA <= bidB
Now we consider the optimal bid of B, given that A bids bidA and that
B observes a valuation of vB :
If BidA > vB then optimal bid is vB < bidA (Receives 0 >=
alternative)
If BidA = vB then optimal bid is vB <= bidA (Receives 0 >=
alternative)
If BidA < vB then optimal bid is bidA + 1 (Receives vB - BidA + 1
>= 0 (alternative))
In other words, B will bid to win if BidA < vB. If BidA < vB then B
will bid to lose. If BidA = vB then B is indifferent between winning
and losing. (Identical so far to Part A)
Given that this is the expected behavior of B, we now turn to the
behavior of A. First we condition on vB = x (and later take
expectations)
A gets :
(vA - bidA) * 1{bidA > bidB|x}
= (vA - bidA) * [1{bidA > x|x} + 1{bidA = x and B bids|x}]
Where 1{} is an indicator function that takes the value 1 if the bid
is successful given x.
A's expected payoff is :
Sum of (vA - bidA) * 1{bidA > bidB|x} * P(vB = x) over x = 1 to 100.
= Sum of (vA - bidA) * 1{bidA > bidB|x} / 100 over x = 1 to 100.
= 0.01 * Sum of (vA - bidA) from x = 0 to bidA
= 0.01 * bidA*[vA - bidA]
Optimum occurs when vA = 2*bidA
So bidA = 25
So the pure strategy nash equilibria are :
1)
bidA = 25
bidB = 26 if vB >= 25, <=vB otherwise
2)
bidA = 25
bidB = 26 if vB > 25, <=vB otherwise
I hope this helps. Good luck and feel free to ask if anything doesn't
make immediate sense.
Regards
calebu2-ga |
,
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