Two bidders compete in a sealed-bid auction for a single indivisible object. For each bidder the valuation of the object is uniformly distributed on [0,1]. The valuations of the two idders are independent. Each bidder knows her own valuation, but not the valuation of he other bidder. The bidders simultaneously submit bids b € [0, ∞), and whoever submits he higher bid wins the object and pays the average of the two bids. In case of a tie, each idder wins the object with probability 1/2 and again the winner pays the average of the two ids. Find a Nash equilibrium of this game in which each bidder's strategy is linear in her aluation. Solution: Look for a symmetric equilibrium where each player bids according to b(v) = av+c.
Two bidders compete in a sealed-bid auction for a single indivisible object. For each bidder the valuation of the object is uniformly distributed on [0,1]. The valuations of the two idders are independent. Each bidder knows her own valuation, but not the valuation of he other bidder. The bidders simultaneously submit bids b € [0, ∞), and whoever submits he higher bid wins the object and pays the average of the two bids. In case of a tie, each idder wins the object with probability 1/2 and again the winner pays the average of the two ids. Find a Nash equilibrium of this game in which each bidder's strategy is linear in her aluation. Solution: Look for a symmetric equilibrium where each player bids according to b(v) = av+c.
Chapter1: Making Economics Decisions
Section: Chapter Questions
Problem 1QTC
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max┬(v^'∈[0,1]) Pr(b_2<b(v^' ))(v_1-1/2 (b(v^' )+E[b(v_2 )∣v_2<v^' ]))". "
![Two bidders compete in a sealed-bid auction for a single indivisible object. For each bidder
i, the valuation of the object is uniformly distributed on [0,1]. The valuations of the two
bidders are independent. Each bidder knows her own valuation, but not the valuation of
the other bidder. The bidders simultaneously submit bids b € [0, 0), and whoever submits
the higher bid wins the object and pays the average of the two bids. In case of a tie, each
bidder wins the object with probability 1/2 and again the winner pays the average of the two
bids. Find a Nash equilibrium of this game in which each bidder's strategy is linear in her
valuation.
Solution: Look for a symmetric equilibrium where each player bids according to b(v) = av+c.
First note that in equilibrium, b(0) = 0 since otherwise types with valuations just above 0 win
with positive probability and pay more than their valuation, and hence would prefer to lower
their bids. Hence e=0. For every valuation vi for player 1, v₁ = must solve
max_ Pr(b2 < b(v')) (v₁ − ½ (b(v¹) + E [b(v2) | v2 < v′])) .
We have Pr(b₂ <b(v¹)) = Pr(v₂ < v¹) = v'. Moreover, since b(v₂) is linear and b(0) = 0,
E[b(v₂) | 1¹₂ < v²] = b(r)/2. Substituting and differentiating leads to the first-order condition
- ³²b(v) — ²/vb'(v) = 0,
which is solved for all v if a = 2/3. There is a symmetric NE in which both players use the
strategy b(v) = 2v/3.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F609dbd69-2d00-4841-aec2-a392b4858745%2F6c46deca-de56-4731-bde5-bd5edb66a31a%2F4gbxmwh_processed.png&w=3840&q=75)
Transcribed Image Text:Two bidders compete in a sealed-bid auction for a single indivisible object. For each bidder
i, the valuation of the object is uniformly distributed on [0,1]. The valuations of the two
bidders are independent. Each bidder knows her own valuation, but not the valuation of
the other bidder. The bidders simultaneously submit bids b € [0, 0), and whoever submits
the higher bid wins the object and pays the average of the two bids. In case of a tie, each
bidder wins the object with probability 1/2 and again the winner pays the average of the two
bids. Find a Nash equilibrium of this game in which each bidder's strategy is linear in her
valuation.
Solution: Look for a symmetric equilibrium where each player bids according to b(v) = av+c.
First note that in equilibrium, b(0) = 0 since otherwise types with valuations just above 0 win
with positive probability and pay more than their valuation, and hence would prefer to lower
their bids. Hence e=0. For every valuation vi for player 1, v₁ = must solve
max_ Pr(b2 < b(v')) (v₁ − ½ (b(v¹) + E [b(v2) | v2 < v′])) .
We have Pr(b₂ <b(v¹)) = Pr(v₂ < v¹) = v'. Moreover, since b(v₂) is linear and b(0) = 0,
E[b(v₂) | 1¹₂ < v²] = b(r)/2. Substituting and differentiating leads to the first-order condition
- ³²b(v) — ²/vb'(v) = 0,
which is solved for all v if a = 2/3. There is a symmetric NE in which both players use the
strategy b(v) = 2v/3.
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