Subgame Perfect Equilibria in a one-stage game
$begingroup$
Consider the following one stage game, with two players A and B.
There is a pie which is to be divided between the two players. A can offer B any fraction of the cake, which B can accept or reject. If B accepts the offer (say $1-x$), then A gets $x$ and B gets $1-x$. If B rejects the offer, both players get $0$.
As far as the Subgame Perfect Equilibria are concerned, ($1$,accept always) is definitely one such equilibrium. What about ($1-epsilon$,reject only when $x=1$), where $epsilon$ is a very small number?
game-theory economics nash-equilibrium
edited Dec 23 '18 at 8:49
mlc
4,89431333
asked Dec 22 '18 at 20:10
StudentStudent
5811
$endgroup$
add a comment |
$begingroup$
Consider the following one stage game, with two players A and B.
There is a pie which is to be divided between the two players. A can offer B any fraction of the cake, which B can accept or reject. If B accepts the offer (say $1-x$), then A gets $x$ and B gets $1-x$. If B rejects the offer, both players get $0$.
As far as the Subgame Perfect Equilibria are concerned, ($1$,accept always) is definitely one such equilibrium. What about ($1-epsilon$,reject only when $x=1$), where $epsilon$ is a very small number?
game-theory economics nash-equilibrium
edited Dec 23 '18 at 8:49
mlc
4,89431333
asked Dec 22 '18 at 20:10
StudentStudent
5811
$endgroup$
add a comment |
$begingroup$
Consider the following one stage game, with two players A and B.
There is a pie which is to be divided between the two players. A can offer B any fraction of the cake, which B can accept or reject. If B accepts the offer (say $1-x$), then A gets $x$ and B gets $1-x$. If B rejects the offer, both players get $0$.
As far as the Subgame Perfect Equilibria are concerned, ($1$,accept always) is definitely one such equilibrium. What about ($1-epsilon$,reject only when $x=1$), where $epsilon$ is a very small number?
game-theory economics nash-equilibrium
edited Dec 23 '18 at 8:49
mlc
4,89431333
asked Dec 22 '18 at 20:10
StudentStudent
5811
$endgroup$
Consider the following one stage game, with two players A and B.
There is a pie which is to be divided between the two players. A can offer B any fraction of the cake, which B can accept or reject. If B accepts the offer (say $1-x$), then A gets $x$ and B gets $1-x$. If B rejects the offer, both players get $0$.
As far as the Subgame Perfect Equilibria are concerned, ($1$,accept always) is definitely one such equilibrium. What about ($1-epsilon$,reject only when $x=1$), where $epsilon$ is a very small number?
game-theory economics nash-equilibrium
game-theory economics nash-equilibrium
edited Dec 23 '18 at 8:49
mlc
4,89431333
asked Dec 22 '18 at 20:10
StudentStudent
5811
edited Dec 23 '18 at 8:49
mlc
4,89431333
asked Dec 22 '18 at 20:10
StudentStudent
5811
edited Dec 23 '18 at 8:49
mlc
4,89431333
edited Dec 23 '18 at 8:49
mlc
4,89431333
edited Dec 23 '18 at 8:49
mlc
4,89431333
4,89431333
asked Dec 22 '18 at 20:10
StudentStudent
5811
asked Dec 22 '18 at 20:10
StudentStudent
5811
asked Dec 22 '18 at 20:10
StudentStudent
5811
5811
add a comment |
add a comment |
1 Answer
1
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votes
$begingroup$
This is not even a Nash equilibrium, because the proposer has profitable deviations: e.g., A can propose $1 - (epsilon/2)$ and B would accept it. This gives A a higher payoff: $1 - (epsilon/2)$ instead of $1 - epsilon$.
answered Dec 23 '18 at 8:48
mlcmlc
4,89431333
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
This is not even a Nash equilibrium, because the proposer has profitable deviations: e.g., A can propose $1 - (epsilon/2)$ and B would accept it. This gives A a higher payoff: $1 - (epsilon/2)$ instead of $1 - epsilon$.
answered Dec 23 '18 at 8:48
mlcmlc
4,89431333
$endgroup$
add a comment |
$begingroup$
This is not even a Nash equilibrium, because the proposer has profitable deviations: e.g., A can propose $1 - (epsilon/2)$ and B would accept it. This gives A a higher payoff: $1 - (epsilon/2)$ instead of $1 - epsilon$.
answered Dec 23 '18 at 8:48
mlcmlc
4,89431333
$endgroup$
add a comment |
$begingroup$
This is not even a Nash equilibrium, because the proposer has profitable deviations: e.g., A can propose $1 - (epsilon/2)$ and B would accept it. This gives A a higher payoff: $1 - (epsilon/2)$ instead of $1 - epsilon$.
answered Dec 23 '18 at 8:48
mlcmlc
4,89431333
$endgroup$
This is not even a Nash equilibrium, because the proposer has profitable deviations: e.g., A can propose $1 - (epsilon/2)$ and B would accept it. This gives A a higher payoff: $1 - (epsilon/2)$ instead of $1 - epsilon$.
answered Dec 23 '18 at 8:48
mlcmlc
4,89431333
answered Dec 23 '18 at 8:48
mlcmlc
4,89431333
answered Dec 23 '18 at 8:48
mlcmlc
4,89431333
answered Dec 23 '18 at 8:48
mlcmlc
4,89431333
4,89431333
add a comment |
add a comment |
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