Can you transform a continuous probability space into an equiprobable probability space?From conditional...
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Can you transform a continuous probability space into an equiprobable probability space?
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Can we have a probability space $(Omega, mathcal F, P)$ where $Omega$ is uncountable but $P$ is rational-valued (i.e., the range is a subset of $mathbb Q$)?
Why I’m asking: I was considering transforming a probability space $(Omega, mathcal F, P)$ into a ‘corresponding’ $(Omega_2, mathcal F_2, P_2)$ where each ${omega_2}$ for $omega_2 in Omega_2$ is equiprobable. When $P$ is rational-valued, I can find the GCD $G$, and stick $G div P({omega})$ corresponding elements in $Omega_2$ for every $omega in Omega$.
probability-theory
asked Mar 12 at 3:58
Yatharth AgarwalYatharth Agarwal
542418
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add a comment |
$begingroup$
Can we have a probability space $(Omega, mathcal F, P)$ where $Omega$ is uncountable but $P$ is rational-valued (i.e., the range is a subset of $mathbb Q$)?
Why I’m asking: I was considering transforming a probability space $(Omega, mathcal F, P)$ into a ‘corresponding’ $(Omega_2, mathcal F_2, P_2)$ where each ${omega_2}$ for $omega_2 in Omega_2$ is equiprobable. When $P$ is rational-valued, I can find the GCD $G$, and stick $G div P({omega})$ corresponding elements in $Omega_2$ for every $omega in Omega$.
probability-theory
asked Mar 12 at 3:58
Yatharth AgarwalYatharth Agarwal
542418
$endgroup$
add a comment |
$begingroup$
Can we have a probability space $(Omega, mathcal F, P)$ where $Omega$ is uncountable but $P$ is rational-valued (i.e., the range is a subset of $mathbb Q$)?
Why I’m asking: I was considering transforming a probability space $(Omega, mathcal F, P)$ into a ‘corresponding’ $(Omega_2, mathcal F_2, P_2)$ where each ${omega_2}$ for $omega_2 in Omega_2$ is equiprobable. When $P$ is rational-valued, I can find the GCD $G$, and stick $G div P({omega})$ corresponding elements in $Omega_2$ for every $omega in Omega$.
probability-theory
asked Mar 12 at 3:58
Yatharth AgarwalYatharth Agarwal
542418
$endgroup$
Can we have a probability space $(Omega, mathcal F, P)$ where $Omega$ is uncountable but $P$ is rational-valued (i.e., the range is a subset of $mathbb Q$)?
Why I’m asking: I was considering transforming a probability space $(Omega, mathcal F, P)$ into a ‘corresponding’ $(Omega_2, mathcal F_2, P_2)$ where each ${omega_2}$ for $omega_2 in Omega_2$ is equiprobable. When $P$ is rational-valued, I can find the GCD $G$, and stick $G div P({omega})$ corresponding elements in $Omega_2$ for every $omega in Omega$.
probability-theory
probability-theory
asked Mar 12 at 3:58
Yatharth AgarwalYatharth Agarwal
542418
asked Mar 12 at 3:58
Yatharth AgarwalYatharth Agarwal
542418
asked Mar 12 at 3:58
Yatharth AgarwalYatharth Agarwal
542418
asked Mar 12 at 3:58
Yatharth AgarwalYatharth Agarwal
542418
asked Mar 12 at 3:58
Yatharth AgarwalYatharth Agarwal
542418
542418
add a comment |
add a comment |
1 Answer
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No, the sum of uncountably many positive reals (and therefore also uncountably many positive rationals) is infinite. Therefore, unless your $P$ maps all but countably many ${omega}$ to $0$ (not what you intended), $P$ cannot sum to the finite value of 1 as required to be a probability space.
answered Mar 12 at 3:58
Yatharth AgarwalYatharth Agarwal
542418
$endgroup$
add a comment |
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$begingroup$
No, the sum of uncountably many positive reals (and therefore also uncountably many positive rationals) is infinite. Therefore, unless your $P$ maps all but countably many ${omega}$ to $0$ (not what you intended), $P$ cannot sum to the finite value of 1 as required to be a probability space.
answered Mar 12 at 3:58
Yatharth AgarwalYatharth Agarwal
542418
$endgroup$
add a comment |
$begingroup$
No, the sum of uncountably many positive reals (and therefore also uncountably many positive rationals) is infinite. Therefore, unless your $P$ maps all but countably many ${omega}$ to $0$ (not what you intended), $P$ cannot sum to the finite value of 1 as required to be a probability space.
answered Mar 12 at 3:58
Yatharth AgarwalYatharth Agarwal
542418
$endgroup$
add a comment |
$begingroup$
No, the sum of uncountably many positive reals (and therefore also uncountably many positive rationals) is infinite. Therefore, unless your $P$ maps all but countably many ${omega}$ to $0$ (not what you intended), $P$ cannot sum to the finite value of 1 as required to be a probability space.
answered Mar 12 at 3:58
Yatharth AgarwalYatharth Agarwal
542418
$endgroup$
No, the sum of uncountably many positive reals (and therefore also uncountably many positive rationals) is infinite. Therefore, unless your $P$ maps all but countably many ${omega}$ to $0$ (not what you intended), $P$ cannot sum to the finite value of 1 as required to be a probability space.
answered Mar 12 at 3:58
Yatharth AgarwalYatharth Agarwal
542418
answered Mar 12 at 3:58
Yatharth AgarwalYatharth Agarwal
542418
answered Mar 12 at 3:58
Yatharth AgarwalYatharth Agarwal
542418
answered Mar 12 at 3:58
Yatharth AgarwalYatharth Agarwal
542418
542418
add a comment |
add a comment |
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