Mixing
what is the Nash equilibrium in a Third price auction?
Consider an auction in which the winner is the highest bidder, and he pays the third highest bid. Suppose there are three players and each player $iin{1,2,3}$ has a valuation $v_i$ and bids $b_i$. Without loss of generality, assume $v_1>v_2>v_3$. In case of a tie, the bidder with the lowest index $i$ wins and pays $min_{i}(b_i)$. The payoff of the winner $i$ is given by $v_i-p_i$, where $p_i$ is the value he will have to pay. The other bidders get a payoff of zero.
(a) Show that truth-telling (i.e. $b_i=v_i$ $forall i$) is in general not a Nash equilibrium of this auction.
For this I have said:
Truth telling is not a dominant strategy with this auction. In order to explain this I will need you to suppose that I value an item at $100, and there are 2 other bids, $200 and $10. I should bid $201 and pay only $10 for the item (note that the bid is higher than my private valuation).
Consider three bidders. Suppose bidders 1 and 2 submit bids , respectively, and bidder 3's value is such that . If bidder 3 bids truthfully, her payoff is 0, because bidder 2 will win the object. However, if bidder 3 overbids, so that , then she would win the auction and get a positive payoff This shows that truthful bidding is not a dominant strategy: there exists a situation where playing truthful bidding is not a best response.
In the third price auction, the expected utility of for the bidder is: ( is your own bid, is the bid of the third place.)
(b) Show that everyone bidding the highest valuation (i.e. $b_i=v_1$ $forall i$) is a Nash equilibrium of
this auction.
I have no idea how to answer this one... someone please help
game-theory nash-equilibrium auction-theory
edited Nov 21 '18 at 9:23
smcc
4,297517
asked Nov 21 '18 at 1:00
Jose Luis Montalvo Ferreiro
61
add a comment |
Consider an auction in which the winner is the highest bidder, and he pays the third highest bid. Suppose there are three players and each player $iin{1,2,3}$ has a valuation $v_i$ and bids $b_i$. Without loss of generality, assume $v_1>v_2>v_3$. In case of a tie, the bidder with the lowest index $i$ wins and pays $min_{i}(b_i)$. The payoff of the winner $i$ is given by $v_i-p_i$, where $p_i$ is the value he will have to pay. The other bidders get a payoff of zero.
(a) Show that truth-telling (i.e. $b_i=v_i$ $forall i$) is in general not a Nash equilibrium of this auction.
For this I have said:
Truth telling is not a dominant strategy with this auction. In order to explain this I will need you to suppose that I value an item at $100, and there are 2 other bids, $200 and $10. I should bid $201 and pay only $10 for the item (note that the bid is higher than my private valuation).
Consider three bidders. Suppose bidders 1 and 2 submit bids , respectively, and bidder 3's value is such that . If bidder 3 bids truthfully, her payoff is 0, because bidder 2 will win the object. However, if bidder 3 overbids, so that , then she would win the auction and get a positive payoff This shows that truthful bidding is not a dominant strategy: there exists a situation where playing truthful bidding is not a best response.
In the third price auction, the expected utility of for the bidder is: ( is your own bid, is the bid of the third place.)
(b) Show that everyone bidding the highest valuation (i.e. $b_i=v_1$ $forall i$) is a Nash equilibrium of
this auction.
I have no idea how to answer this one... someone please help
game-theory nash-equilibrium auction-theory
edited Nov 21 '18 at 9:23
smcc
4,297517
asked Nov 21 '18 at 1:00
Jose Luis Montalvo Ferreiro
61
add a comment |
Consider an auction in which the winner is the highest bidder, and he pays the third highest bid. Suppose there are three players and each player $iin{1,2,3}$ has a valuation $v_i$ and bids $b_i$. Without loss of generality, assume $v_1>v_2>v_3$. In case of a tie, the bidder with the lowest index $i$ wins and pays $min_{i}(b_i)$. The payoff of the winner $i$ is given by $v_i-p_i$, where $p_i$ is the value he will have to pay. The other bidders get a payoff of zero.
(a) Show that truth-telling (i.e. $b_i=v_i$ $forall i$) is in general not a Nash equilibrium of this auction.
For this I have said:
Truth telling is not a dominant strategy with this auction. In order to explain this I will need you to suppose that I value an item at $100, and there are 2 other bids, $200 and $10. I should bid $201 and pay only $10 for the item (note that the bid is higher than my private valuation).
Consider three bidders. Suppose bidders 1 and 2 submit bids , respectively, and bidder 3's value is such that . If bidder 3 bids truthfully, her payoff is 0, because bidder 2 will win the object. However, if bidder 3 overbids, so that , then she would win the auction and get a positive payoff This shows that truthful bidding is not a dominant strategy: there exists a situation where playing truthful bidding is not a best response.
In the third price auction, the expected utility of for the bidder is: ( is your own bid, is the bid of the third place.)
(b) Show that everyone bidding the highest valuation (i.e. $b_i=v_1$ $forall i$) is a Nash equilibrium of
this auction.
I have no idea how to answer this one... someone please help
game-theory nash-equilibrium auction-theory
edited Nov 21 '18 at 9:23
smcc
4,297517
asked Nov 21 '18 at 1:00
Jose Luis Montalvo Ferreiro
61
Consider an auction in which the winner is the highest bidder, and he pays the third highest bid. Suppose there are three players and each player $iin{1,2,3}$ has a valuation $v_i$ and bids $b_i$. Without loss of generality, assume $v_1>v_2>v_3$. In case of a tie, the bidder with the lowest index $i$ wins and pays $min_{i}(b_i)$. The payoff of the winner $i$ is given by $v_i-p_i$, where $p_i$ is the value he will have to pay. The other bidders get a payoff of zero.
(a) Show that truth-telling (i.e. $b_i=v_i$ $forall i$) is in general not a Nash equilibrium of this auction.
For this I have said:
Truth telling is not a dominant strategy with this auction. In order to explain this I will need you to suppose that I value an item at $100, and there are 2 other bids, $200 and $10. I should bid $201 and pay only $10 for the item (note that the bid is higher than my private valuation).
Consider three bidders. Suppose bidders 1 and 2 submit bids , respectively, and bidder 3's value is such that . If bidder 3 bids truthfully, her payoff is 0, because bidder 2 will win the object. However, if bidder 3 overbids, so that , then she would win the auction and get a positive payoff This shows that truthful bidding is not a dominant strategy: there exists a situation where playing truthful bidding is not a best response.
In the third price auction, the expected utility of for the bidder is: ( is your own bid, is the bid of the third place.)
(b) Show that everyone bidding the highest valuation (i.e. $b_i=v_1$ $forall i$) is a Nash equilibrium of
this auction.
I have no idea how to answer this one... someone please help
game-theory nash-equilibrium auction-theory
game-theory nash-equilibrium auction-theory
edited Nov 21 '18 at 9:23
smcc
4,297517
asked Nov 21 '18 at 1:00
Jose Luis Montalvo Ferreiro
61
edited Nov 21 '18 at 9:23
smcc
4,297517
asked Nov 21 '18 at 1:00
Jose Luis Montalvo Ferreiro
61
edited Nov 21 '18 at 9:23
smcc
4,297517
edited Nov 21 '18 at 9:23
smcc
4,297517
edited Nov 21 '18 at 9:23
smcc
4,297517
4,297517
asked Nov 21 '18 at 1:00
Jose Luis Montalvo Ferreiro
61
asked Nov 21 '18 at 1:00
Jose Luis Montalvo Ferreiro
61
asked Nov 21 '18 at 1:00
Jose Luis Montalvo Ferreiro
61
61
add a comment |
add a comment |
1 Answer
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In (a) you are not asked whether bidding the valuation is a dominant strategy. Instead, you need to show that $(b_1,b_2,b_3)=(v_1,v_2,v_3)$ is not a Nash equilibrium. To do this, you just need to show that a player has a strictly profitable deviation. So, suppose $(b_1,b_2,b_3)=(v_1,v_2,v_3)$. Player 2's pay off is zero. If they increase their bid to something more than $v_1$, then they win the auction, pay the third highest bid $v_3$, and get a payoff of $v_2-v_3>0$. This is a strictly profitable deviation, so all players bidding their valuation is not a Nash equilibrium.
For (b), to show $(b_1,b_2,b_3)=(v_1,v_1,v_1)$ is a Nash equilibrium, you need to show that no player has a strictly profitable deviation from bidding $v_1$ (when the other players are bidding $v_1$). When players all bid $v_1$, they all get zero payoff (players 2 and 3 because they do not win the auction, and player 1 because they win the auction and pay $v_1$).
- If player 1 raises their bid, they still win the auction and pay $v_1$ and get zero payoff. If player 1 lowers their bid, they lose the auction and get zero payoff.
- If player 2 or 3 raise their bid, they win the auction and pay $v_1(>v_2>v_3)$ so make a loss. If player 2 or 3 lowers their bid, they lose the auction and get zero payoff.
Thus, no player has a strictly profitable deviation, and hence it is a Nash equilibrium for all players to bid the highest valuation $v_1$.
answered Nov 21 '18 at 9:17
smcc
4,297517
add a comment |
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In (a) you are not asked whether bidding the valuation is a dominant strategy. Instead, you need to show that $(b_1,b_2,b_3)=(v_1,v_2,v_3)$ is not a Nash equilibrium. To do this, you just need to show that a player has a strictly profitable deviation. So, suppose $(b_1,b_2,b_3)=(v_1,v_2,v_3)$. Player 2's pay off is zero. If they increase their bid to something more than $v_1$, then they win the auction, pay the third highest bid $v_3$, and get a payoff of $v_2-v_3>0$. This is a strictly profitable deviation, so all players bidding their valuation is not a Nash equilibrium.
For (b), to show $(b_1,b_2,b_3)=(v_1,v_1,v_1)$ is a Nash equilibrium, you need to show that no player has a strictly profitable deviation from bidding $v_1$ (when the other players are bidding $v_1$). When players all bid $v_1$, they all get zero payoff (players 2 and 3 because they do not win the auction, and player 1 because they win the auction and pay $v_1$).
- If player 1 raises their bid, they still win the auction and pay $v_1$ and get zero payoff. If player 1 lowers their bid, they lose the auction and get zero payoff.
- If player 2 or 3 raise their bid, they win the auction and pay $v_1(>v_2>v_3)$ so make a loss. If player 2 or 3 lowers their bid, they lose the auction and get zero payoff.
Thus, no player has a strictly profitable deviation, and hence it is a Nash equilibrium for all players to bid the highest valuation $v_1$.
answered Nov 21 '18 at 9:17
smcc
4,297517
add a comment |
In (a) you are not asked whether bidding the valuation is a dominant strategy. Instead, you need to show that $(b_1,b_2,b_3)=(v_1,v_2,v_3)$ is not a Nash equilibrium. To do this, you just need to show that a player has a strictly profitable deviation. So, suppose $(b_1,b_2,b_3)=(v_1,v_2,v_3)$. Player 2's pay off is zero. If they increase their bid to something more than $v_1$, then they win the auction, pay the third highest bid $v_3$, and get a payoff of $v_2-v_3>0$. This is a strictly profitable deviation, so all players bidding their valuation is not a Nash equilibrium.
For (b), to show $(b_1,b_2,b_3)=(v_1,v_1,v_1)$ is a Nash equilibrium, you need to show that no player has a strictly profitable deviation from bidding $v_1$ (when the other players are bidding $v_1$). When players all bid $v_1$, they all get zero payoff (players 2 and 3 because they do not win the auction, and player 1 because they win the auction and pay $v_1$).
- If player 1 raises their bid, they still win the auction and pay $v_1$ and get zero payoff. If player 1 lowers their bid, they lose the auction and get zero payoff.
- If player 2 or 3 raise their bid, they win the auction and pay $v_1(>v_2>v_3)$ so make a loss. If player 2 or 3 lowers their bid, they lose the auction and get zero payoff.
Thus, no player has a strictly profitable deviation, and hence it is a Nash equilibrium for all players to bid the highest valuation $v_1$.
answered Nov 21 '18 at 9:17
smcc
4,297517
add a comment |
In (a) you are not asked whether bidding the valuation is a dominant strategy. Instead, you need to show that $(b_1,b_2,b_3)=(v_1,v_2,v_3)$ is not a Nash equilibrium. To do this, you just need to show that a player has a strictly profitable deviation. So, suppose $(b_1,b_2,b_3)=(v_1,v_2,v_3)$. Player 2's pay off is zero. If they increase their bid to something more than $v_1$, then they win the auction, pay the third highest bid $v_3$, and get a payoff of $v_2-v_3>0$. This is a strictly profitable deviation, so all players bidding their valuation is not a Nash equilibrium.
For (b), to show $(b_1,b_2,b_3)=(v_1,v_1,v_1)$ is a Nash equilibrium, you need to show that no player has a strictly profitable deviation from bidding $v_1$ (when the other players are bidding $v_1$). When players all bid $v_1$, they all get zero payoff (players 2 and 3 because they do not win the auction, and player 1 because they win the auction and pay $v_1$).
- If player 1 raises their bid, they still win the auction and pay $v_1$ and get zero payoff. If player 1 lowers their bid, they lose the auction and get zero payoff.
- If player 2 or 3 raise their bid, they win the auction and pay $v_1(>v_2>v_3)$ so make a loss. If player 2 or 3 lowers their bid, they lose the auction and get zero payoff.
Thus, no player has a strictly profitable deviation, and hence it is a Nash equilibrium for all players to bid the highest valuation $v_1$.
answered Nov 21 '18 at 9:17
smcc
4,297517
In (a) you are not asked whether bidding the valuation is a dominant strategy. Instead, you need to show that $(b_1,b_2,b_3)=(v_1,v_2,v_3)$ is not a Nash equilibrium. To do this, you just need to show that a player has a strictly profitable deviation. So, suppose $(b_1,b_2,b_3)=(v_1,v_2,v_3)$. Player 2's pay off is zero. If they increase their bid to something more than $v_1$, then they win the auction, pay the third highest bid $v_3$, and get a payoff of $v_2-v_3>0$. This is a strictly profitable deviation, so all players bidding their valuation is not a Nash equilibrium.
For (b), to show $(b_1,b_2,b_3)=(v_1,v_1,v_1)$ is a Nash equilibrium, you need to show that no player has a strictly profitable deviation from bidding $v_1$ (when the other players are bidding $v_1$). When players all bid $v_1$, they all get zero payoff (players 2 and 3 because they do not win the auction, and player 1 because they win the auction and pay $v_1$).
- If player 1 raises their bid, they still win the auction and pay $v_1$ and get zero payoff. If player 1 lowers their bid, they lose the auction and get zero payoff.
- If player 2 or 3 raise their bid, they win the auction and pay $v_1(>v_2>v_3)$ so make a loss. If player 2 or 3 lowers their bid, they lose the auction and get zero payoff.
Thus, no player has a strictly profitable deviation, and hence it is a Nash equilibrium for all players to bid the highest valuation $v_1$.
answered Nov 21 '18 at 9:17
smcc
4,297517
edited Nov 23 '18 at 7:45
answered Nov 21 '18 at 9:17
smcc
4,297517
answered Nov 21 '18 at 9:17
smcc
4,297517
answered Nov 21 '18 at 9:17
smcc
4,297517
4,297517
add a comment |
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