In Bayesian games, is the actions set for a player fixed?
In a Bayesian game, is the actions set for a player fixed? Or does her actions set depend on her type?
In other words, can I model as a Bayesian game the following situation.
I have two players: master and worker. The worker is honest or malicious.
- If the worker is honest, she can only work honestly (she has only one action).
- If the worker is malicious, she can work honestly or maliciously (she has two actions).
game-theory
add a comment |
In a Bayesian game, is the actions set for a player fixed? Or does her actions set depend on her type?
In other words, can I model as a Bayesian game the following situation.
I have two players: master and worker. The worker is honest or malicious.
- If the worker is honest, she can only work honestly (she has only one action).
- If the worker is malicious, she can work honestly or maliciously (she has two actions).
game-theory
So you're saying the probability of being malicious when the type is honest is given by $p(action=malicious|type=honest)=0$? I think this might work for a Bayesian game.
– Jay Schyler Raadt
Nov 23 '18 at 4:36
add a comment |
In a Bayesian game, is the actions set for a player fixed? Or does her actions set depend on her type?
In other words, can I model as a Bayesian game the following situation.
I have two players: master and worker. The worker is honest or malicious.
- If the worker is honest, she can only work honestly (she has only one action).
- If the worker is malicious, she can work honestly or maliciously (she has two actions).
game-theory
In a Bayesian game, is the actions set for a player fixed? Or does her actions set depend on her type?
In other words, can I model as a Bayesian game the following situation.
I have two players: master and worker. The worker is honest or malicious.
- If the worker is honest, she can only work honestly (she has only one action).
- If the worker is malicious, she can work honestly or maliciously (she has two actions).
game-theory
game-theory
edited Nov 24 '18 at 13:32
mlc
4,87931332
4,87931332
asked Nov 22 '18 at 11:30
Tail of Godzilla
535
535
So you're saying the probability of being malicious when the type is honest is given by $p(action=malicious|type=honest)=0$? I think this might work for a Bayesian game.
– Jay Schyler Raadt
Nov 23 '18 at 4:36
add a comment |
So you're saying the probability of being malicious when the type is honest is given by $p(action=malicious|type=honest)=0$? I think this might work for a Bayesian game.
– Jay Schyler Raadt
Nov 23 '18 at 4:36
So you're saying the probability of being malicious when the type is honest is given by $p(action=malicious|type=honest)=0$? I think this might work for a Bayesian game.
– Jay Schyler Raadt
Nov 23 '18 at 4:36
So you're saying the probability of being malicious when the type is honest is given by $p(action=malicious|type=honest)=0$? I think this might work for a Bayesian game.
– Jay Schyler Raadt
Nov 23 '18 at 4:36
add a comment |
1 Answer
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Yes, you can assume different action sets for different types of a player.
To simplify notation, it is a common assumption in the literature to assume that all types of a player have the same set of actions.
If you like this approach for your example, assume that both types can work honestly or maliciously but the honest type has a payoff of $- infty$ (or some other very low value) if he works maliciously, so that he would never opt for this action. In other words, assume that working honestly strictly dominates working maliciously for the honest type.
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1 Answer
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1 Answer
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active
oldest
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oldest
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active
oldest
votes
Yes, you can assume different action sets for different types of a player.
To simplify notation, it is a common assumption in the literature to assume that all types of a player have the same set of actions.
If you like this approach for your example, assume that both types can work honestly or maliciously but the honest type has a payoff of $- infty$ (or some other very low value) if he works maliciously, so that he would never opt for this action. In other words, assume that working honestly strictly dominates working maliciously for the honest type.
add a comment |
Yes, you can assume different action sets for different types of a player.
To simplify notation, it is a common assumption in the literature to assume that all types of a player have the same set of actions.
If you like this approach for your example, assume that both types can work honestly or maliciously but the honest type has a payoff of $- infty$ (or some other very low value) if he works maliciously, so that he would never opt for this action. In other words, assume that working honestly strictly dominates working maliciously for the honest type.
add a comment |
Yes, you can assume different action sets for different types of a player.
To simplify notation, it is a common assumption in the literature to assume that all types of a player have the same set of actions.
If you like this approach for your example, assume that both types can work honestly or maliciously but the honest type has a payoff of $- infty$ (or some other very low value) if he works maliciously, so that he would never opt for this action. In other words, assume that working honestly strictly dominates working maliciously for the honest type.
Yes, you can assume different action sets for different types of a player.
To simplify notation, it is a common assumption in the literature to assume that all types of a player have the same set of actions.
If you like this approach for your example, assume that both types can work honestly or maliciously but the honest type has a payoff of $- infty$ (or some other very low value) if he works maliciously, so that he would never opt for this action. In other words, assume that working honestly strictly dominates working maliciously for the honest type.
answered Nov 24 '18 at 13:36
mlc
4,87931332
4,87931332
add a comment |
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So you're saying the probability of being malicious when the type is honest is given by $p(action=malicious|type=honest)=0$? I think this might work for a Bayesian game.
– Jay Schyler Raadt
Nov 23 '18 at 4:36