Equivalent definitions of perfect equilibrium of a normal form game
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Let $Gamma$ be a normal form game. An equilibrium $s$ of $Gamma$ is a perfect equilibrium of $Gamma$ if $s$ is a limit point of a sequence $s ( eta ) _eta downarrow 0$ with $s(eta)in E(Gamma,eta)$ for all $eta $ i.e. $s$ is perfect if there exist sequences $s(t)_t inmathbbN$ and $eta(t)_tin mathbbN$ with
$s(t)in E(Gamma,eta(t))$ for all $tinmathbbN
$, and such that $s(t)$ converges to $s$ and $eta(t)$ converges to zero, as $t$ tends to infinity.
My question is: how do I get and precisely understand the switch from $$s(eta) text to s(t)$$ ?
sequences-and-series game-theory nash-equilibrium
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Let $Gamma$ be a normal form game. An equilibrium $s$ of $Gamma$ is a perfect equilibrium of $Gamma$ if $s$ is a limit point of a sequence $s ( eta ) _eta downarrow 0$ with $s(eta)in E(Gamma,eta)$ for all $eta $ i.e. $s$ is perfect if there exist sequences $s(t)_t inmathbbN$ and $eta(t)_tin mathbbN$ with
$s(t)in E(Gamma,eta(t))$ for all $tinmathbbN
$, and such that $s(t)$ converges to $s$ and $eta(t)$ converges to zero, as $t$ tends to infinity.
My question is: how do I get and precisely understand the switch from $$s(eta) text to s(t)$$ ?
sequences-and-series game-theory nash-equilibrium
add a comment |Â
up vote
1
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up vote
1
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Let $Gamma$ be a normal form game. An equilibrium $s$ of $Gamma$ is a perfect equilibrium of $Gamma$ if $s$ is a limit point of a sequence $s ( eta ) _eta downarrow 0$ with $s(eta)in E(Gamma,eta)$ for all $eta $ i.e. $s$ is perfect if there exist sequences $s(t)_t inmathbbN$ and $eta(t)_tin mathbbN$ with
$s(t)in E(Gamma,eta(t))$ for all $tinmathbbN
$, and such that $s(t)$ converges to $s$ and $eta(t)$ converges to zero, as $t$ tends to infinity.
My question is: how do I get and precisely understand the switch from $$s(eta) text to s(t)$$ ?
sequences-and-series game-theory nash-equilibrium
Let $Gamma$ be a normal form game. An equilibrium $s$ of $Gamma$ is a perfect equilibrium of $Gamma$ if $s$ is a limit point of a sequence $s ( eta ) _eta downarrow 0$ with $s(eta)in E(Gamma,eta)$ for all $eta $ i.e. $s$ is perfect if there exist sequences $s(t)_t inmathbbN$ and $eta(t)_tin mathbbN$ with
$s(t)in E(Gamma,eta(t))$ for all $tinmathbbN
$, and such that $s(t)$ converges to $s$ and $eta(t)$ converges to zero, as $t$ tends to infinity.
My question is: how do I get and precisely understand the switch from $$s(eta) text to s(t)$$ ?
sequences-and-series game-theory nash-equilibrium
asked Aug 7 at 20:08
user122424
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9521614
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If I understand your question correctly, this is just notational density. You use $eta$ as the perturbation (minimal tremble) of the stratgies. That is $E(Gamma, eta)$ is the set of equilibria of the game $Gamma$ such that all strategies for all agents puts weight at least $eta$ on each action (pure strategy). We call an equilibrium proper if it can be approximated arbitrarily well by such restricted games.
Your sequence $eta(t)$ is just $eta_1, eta_2, ...$, is a sequential relaxation of the perturbation of the game (in the sense of action weights). What is needed then is a corresponding sequence of equilibria $sbig(eta(t)big) in Ebig(Gamma, eta(t)big)$ that similarly converge to $s$. Think of it as approximation: if as a sequence of perturbations subsides to zero, we would wish that there is a sequence of equilibria of the perturbed games that converges (in the space of strategies) to the equilibrium of interest, $s$.
In action, consider the following game.
$$ beginarrayc hline
& L & R \ hline
T & 1,1 & 0,0 \ hline
B & 0,0 & 0,0 \ hline
endarray $$
Clearly $B,R$ is an equilibrium. It is not, however, perfect. This is because $R$ is a best response for the column player if and only if the row player's strategy plays $B$ with probability one. In any tremble, row puts at most probability $1-eta$ on $B$, hence every equilibrium of every perturbed game with $eta > 0$ puts weight $eta$ on $R$ for column, hence as $etato 0$, any sequence of equilibria of perturbed games must converge toward $T,L$, the unique perfect equilibrium of this game.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If I understand your question correctly, this is just notational density. You use $eta$ as the perturbation (minimal tremble) of the stratgies. That is $E(Gamma, eta)$ is the set of equilibria of the game $Gamma$ such that all strategies for all agents puts weight at least $eta$ on each action (pure strategy). We call an equilibrium proper if it can be approximated arbitrarily well by such restricted games.
Your sequence $eta(t)$ is just $eta_1, eta_2, ...$, is a sequential relaxation of the perturbation of the game (in the sense of action weights). What is needed then is a corresponding sequence of equilibria $sbig(eta(t)big) in Ebig(Gamma, eta(t)big)$ that similarly converge to $s$. Think of it as approximation: if as a sequence of perturbations subsides to zero, we would wish that there is a sequence of equilibria of the perturbed games that converges (in the space of strategies) to the equilibrium of interest, $s$.
In action, consider the following game.
$$ beginarrayc hline
& L & R \ hline
T & 1,1 & 0,0 \ hline
B & 0,0 & 0,0 \ hline
endarray $$
Clearly $B,R$ is an equilibrium. It is not, however, perfect. This is because $R$ is a best response for the column player if and only if the row player's strategy plays $B$ with probability one. In any tremble, row puts at most probability $1-eta$ on $B$, hence every equilibrium of every perturbed game with $eta > 0$ puts weight $eta$ on $R$ for column, hence as $etato 0$, any sequence of equilibria of perturbed games must converge toward $T,L$, the unique perfect equilibrium of this game.
add a comment |Â
up vote
1
down vote
accepted
If I understand your question correctly, this is just notational density. You use $eta$ as the perturbation (minimal tremble) of the stratgies. That is $E(Gamma, eta)$ is the set of equilibria of the game $Gamma$ such that all strategies for all agents puts weight at least $eta$ on each action (pure strategy). We call an equilibrium proper if it can be approximated arbitrarily well by such restricted games.
Your sequence $eta(t)$ is just $eta_1, eta_2, ...$, is a sequential relaxation of the perturbation of the game (in the sense of action weights). What is needed then is a corresponding sequence of equilibria $sbig(eta(t)big) in Ebig(Gamma, eta(t)big)$ that similarly converge to $s$. Think of it as approximation: if as a sequence of perturbations subsides to zero, we would wish that there is a sequence of equilibria of the perturbed games that converges (in the space of strategies) to the equilibrium of interest, $s$.
In action, consider the following game.
$$ beginarrayc hline
& L & R \ hline
T & 1,1 & 0,0 \ hline
B & 0,0 & 0,0 \ hline
endarray $$
Clearly $B,R$ is an equilibrium. It is not, however, perfect. This is because $R$ is a best response for the column player if and only if the row player's strategy plays $B$ with probability one. In any tremble, row puts at most probability $1-eta$ on $B$, hence every equilibrium of every perturbed game with $eta > 0$ puts weight $eta$ on $R$ for column, hence as $etato 0$, any sequence of equilibria of perturbed games must converge toward $T,L$, the unique perfect equilibrium of this game.
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If I understand your question correctly, this is just notational density. You use $eta$ as the perturbation (minimal tremble) of the stratgies. That is $E(Gamma, eta)$ is the set of equilibria of the game $Gamma$ such that all strategies for all agents puts weight at least $eta$ on each action (pure strategy). We call an equilibrium proper if it can be approximated arbitrarily well by such restricted games.
Your sequence $eta(t)$ is just $eta_1, eta_2, ...$, is a sequential relaxation of the perturbation of the game (in the sense of action weights). What is needed then is a corresponding sequence of equilibria $sbig(eta(t)big) in Ebig(Gamma, eta(t)big)$ that similarly converge to $s$. Think of it as approximation: if as a sequence of perturbations subsides to zero, we would wish that there is a sequence of equilibria of the perturbed games that converges (in the space of strategies) to the equilibrium of interest, $s$.
In action, consider the following game.
$$ beginarrayc hline
& L & R \ hline
T & 1,1 & 0,0 \ hline
B & 0,0 & 0,0 \ hline
endarray $$
Clearly $B,R$ is an equilibrium. It is not, however, perfect. This is because $R$ is a best response for the column player if and only if the row player's strategy plays $B$ with probability one. In any tremble, row puts at most probability $1-eta$ on $B$, hence every equilibrium of every perturbed game with $eta > 0$ puts weight $eta$ on $R$ for column, hence as $etato 0$, any sequence of equilibria of perturbed games must converge toward $T,L$, the unique perfect equilibrium of this game.
If I understand your question correctly, this is just notational density. You use $eta$ as the perturbation (minimal tremble) of the stratgies. That is $E(Gamma, eta)$ is the set of equilibria of the game $Gamma$ such that all strategies for all agents puts weight at least $eta$ on each action (pure strategy). We call an equilibrium proper if it can be approximated arbitrarily well by such restricted games.
Your sequence $eta(t)$ is just $eta_1, eta_2, ...$, is a sequential relaxation of the perturbation of the game (in the sense of action weights). What is needed then is a corresponding sequence of equilibria $sbig(eta(t)big) in Ebig(Gamma, eta(t)big)$ that similarly converge to $s$. Think of it as approximation: if as a sequence of perturbations subsides to zero, we would wish that there is a sequence of equilibria of the perturbed games that converges (in the space of strategies) to the equilibrium of interest, $s$.
In action, consider the following game.
$$ beginarrayc hline
& L & R \ hline
T & 1,1 & 0,0 \ hline
B & 0,0 & 0,0 \ hline
endarray $$
Clearly $B,R$ is an equilibrium. It is not, however, perfect. This is because $R$ is a best response for the column player if and only if the row player's strategy plays $B$ with probability one. In any tremble, row puts at most probability $1-eta$ on $B$, hence every equilibrium of every perturbed game with $eta > 0$ puts weight $eta$ on $R$ for column, hence as $etato 0$, any sequence of equilibria of perturbed games must converge toward $T,L$, the unique perfect equilibrium of this game.
answered Aug 7 at 20:37
Pete Caradonna
1,3191620
1,3191620
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