Mixing
Two formulations of continuity in Expected Utility Theory
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In von Neumann-Morgenstern's expected utility theory, there are two ways that the continuity axioms are usually stated.
Some authors state the axiom as follows:
C1. The preference relation $succsim$ is continuous if for all lotteries $L,L',L''$, the two sets below are both closed:
begin{align}
S&={alphain[0,1]:alpha L+(1-alpha)L''succsim L'}\
T&={alphain[0,1]:L'succsim alpha L+(1-alpha)L''}
end{align}
Other authors use a somewhat different formulation:
C2. The preference relation $succsim$ is continuous if for all lotteries $L,L',L''$ with $Lsuccsim L'succsim L''$, there exists an $alphain[0,1]$ such that
begin{equation}
L'sim alpha L+(1-alpha)L''
end{equation}
Are the two formulations equivalent? It's easy to see how C1 implies C2 (just take $alphain Scap T$). But I'm not sure how C2 implies C1.
The proofs using C2 usually first establish that, by invoking the independence axiom (for all lotteries $L,L',L''$ and any $alphain[0,1]$, $Lsuccsim L''Leftrightarrowalpha L+(1-alpha)L'succsim alpha L''+(1-alpha)L'$), we have
begin{equation}
beta L+(1-beta)L'succ alpha L+(1-alpha)L'
end{equation}
whenever $Lsucc L'$ and $1>beta>alpha>0$. Hence, C2 and independence together imply C1. But does C2 imply C1 without independence?
real-analysis continuity
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Herr K.
5681516
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In von Neumann-Morgenstern's expected utility theory, there are two ways that the continuity axioms are usually stated.
Some authors state the axiom as follows:
C1. The preference relation $succsim$ is continuous if for all lotteries $L,L',L''$, the two sets below are both closed:
begin{align}
S&={alphain[0,1]:alpha L+(1-alpha)L''succsim L'}\
T&={alphain[0,1]:L'succsim alpha L+(1-alpha)L''}
end{align}
Other authors use a somewhat different formulation:
C2. The preference relation $succsim$ is continuous if for all lotteries $L,L',L''$ with $Lsuccsim L'succsim L''$, there exists an $alphain[0,1]$ such that
begin{equation}
L'sim alpha L+(1-alpha)L''
end{equation}
Are the two formulations equivalent? It's easy to see how C1 implies C2 (just take $alphain Scap T$). But I'm not sure how C2 implies C1.
The proofs using C2 usually first establish that, by invoking the independence axiom (for all lotteries $L,L',L''$ and any $alphain[0,1]$, $Lsuccsim L''Leftrightarrowalpha L+(1-alpha)L'succsim alpha L''+(1-alpha)L'$), we have
begin{equation}
beta L+(1-beta)L'succ alpha L+(1-alpha)L'
end{equation}
whenever $Lsucc L'$ and $1>beta>alpha>0$. Hence, C2 and independence together imply C1. But does C2 imply C1 without independence?
real-analysis continuity
asked yesterday
Herr K.
5681516
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In von Neumann-Morgenstern's expected utility theory, there are two ways that the continuity axioms are usually stated.
Some authors state the axiom as follows:
C1. The preference relation $succsim$ is continuous if for all lotteries $L,L',L''$, the two sets below are both closed:
begin{align}
S&={alphain[0,1]:alpha L+(1-alpha)L''succsim L'}\
T&={alphain[0,1]:L'succsim alpha L+(1-alpha)L''}
end{align}
Other authors use a somewhat different formulation:
C2. The preference relation $succsim$ is continuous if for all lotteries $L,L',L''$ with $Lsuccsim L'succsim L''$, there exists an $alphain[0,1]$ such that
begin{equation}
L'sim alpha L+(1-alpha)L''
end{equation}
Are the two formulations equivalent? It's easy to see how C1 implies C2 (just take $alphain Scap T$). But I'm not sure how C2 implies C1.
The proofs using C2 usually first establish that, by invoking the independence axiom (for all lotteries $L,L',L''$ and any $alphain[0,1]$, $Lsuccsim L''Leftrightarrowalpha L+(1-alpha)L'succsim alpha L''+(1-alpha)L'$), we have
begin{equation}
beta L+(1-beta)L'succ alpha L+(1-alpha)L'
end{equation}
whenever $Lsucc L'$ and $1>beta>alpha>0$. Hence, C2 and independence together imply C1. But does C2 imply C1 without independence?
real-analysis continuity
asked yesterday
Herr K.
5681516
In von Neumann-Morgenstern's expected utility theory, there are two ways that the continuity axioms are usually stated.
Some authors state the axiom as follows:
C1. The preference relation $succsim$ is continuous if for all lotteries $L,L',L''$, the two sets below are both closed:
begin{align}
S&={alphain[0,1]:alpha L+(1-alpha)L''succsim L'}\
T&={alphain[0,1]:L'succsim alpha L+(1-alpha)L''}
end{align}
Other authors use a somewhat different formulation:
C2. The preference relation $succsim$ is continuous if for all lotteries $L,L',L''$ with $Lsuccsim L'succsim L''$, there exists an $alphain[0,1]$ such that
begin{equation}
L'sim alpha L+(1-alpha)L''
end{equation}
Are the two formulations equivalent? It's easy to see how C1 implies C2 (just take $alphain Scap T$). But I'm not sure how C2 implies C1.
The proofs using C2 usually first establish that, by invoking the independence axiom (for all lotteries $L,L',L''$ and any $alphain[0,1]$, $Lsuccsim L''Leftrightarrowalpha L+(1-alpha)L'succsim alpha L''+(1-alpha)L'$), we have
begin{equation}
beta L+(1-beta)L'succ alpha L+(1-alpha)L'
end{equation}
whenever $Lsucc L'$ and $1>beta>alpha>0$. Hence, C2 and independence together imply C1. But does C2 imply C1 without independence?
real-analysis continuity
real-analysis continuity
asked yesterday
Herr K.
5681516
asked yesterday
Herr K.
5681516
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asked yesterday
Herr K.
5681516
asked yesterday
Herr K.
5681516
asked yesterday
Herr K.
5681516
5681516
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