Is it true that $(A, mathcal F_A,mathbb P(.|A))$ is a probability space.
$begingroup$
Let $(Omega ,mathcal F,mathbb P)$ a probability space and $Ain mathcal F$ s.t. $mathbb P(A)>0$. In an exercise I have, I need to prove that $(A, mathcal F_A, mathbb P(.|A))$ is a probability space where $mathbb P(Cmid A)=frac{mathbb P(Acap C)}{mathbb P(A)}$ and $mathcal F_A={Bcap Amid Bin mathcal F}$. Is it really true ?
I agree that $mathcal F_A$ is a $sigma -$agebra on $A$.
I agree that $mathbb Q(C)=frac{mathbb P(C)}{mathbb P(A)}$ for $Cin mathcal F_A$ is a measure.
But for me $mathbb P(.|A)$ is a measure on $Omega $, and $(Omega ,mathcal F,mathbb P(.|A))$ is a probability measure, but $(A,mathcal F_A,mathbb P(.|A))$ is not. So, what is $mathbb P(.|A)$ for $(A,mathcal F_A)$ ? or what represent the probability space $(Omega ,mathcal F,mathbb P(.|A))$ for $(A,mathcal F_A, mathbb Q)$ ?
probability measure-theory
asked Jan 9 at 19:31
NewMathNewMath
4059
$endgroup$
add a comment |
$begingroup$
Let $(Omega ,mathcal F,mathbb P)$ a probability space and $Ain mathcal F$ s.t. $mathbb P(A)>0$. In an exercise I have, I need to prove that $(A, mathcal F_A, mathbb P(.|A))$ is a probability space where $mathbb P(Cmid A)=frac{mathbb P(Acap C)}{mathbb P(A)}$ and $mathcal F_A={Bcap Amid Bin mathcal F}$. Is it really true ?
I agree that $mathcal F_A$ is a $sigma -$agebra on $A$.
I agree that $mathbb Q(C)=frac{mathbb P(C)}{mathbb P(A)}$ for $Cin mathcal F_A$ is a measure.
But for me $mathbb P(.|A)$ is a measure on $Omega $, and $(Omega ,mathcal F,mathbb P(.|A))$ is a probability measure, but $(A,mathcal F_A,mathbb P(.|A))$ is not. So, what is $mathbb P(.|A)$ for $(A,mathcal F_A)$ ? or what represent the probability space $(Omega ,mathcal F,mathbb P(.|A))$ for $(A,mathcal F_A, mathbb Q)$ ?
probability measure-theory
asked Jan 9 at 19:31
NewMathNewMath
4059
$endgroup$
1
$begingroup$
Obviously they mean $P(;{cdot}mid A)$ restricted to $mathcal F_A$ -- which is exactly the thing you want to call $Q$ instead.
$endgroup$
– Henning Makholm
Jan 9 at 19:38
1
$begingroup$
Note also that you could equally say $mathcal F_A = { Binmathcal F mid Bsubseteq A}$, which gives the same result as your definition because $mathcal F$ is an algebra and therefore closed under intersection.
$endgroup$
– Henning Makholm
Jan 9 at 19:43
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@HenningMakholm: But $mathbb P(cdot |A)$ can eat in $mathcal F$, not only $mathcal F_A$... that a bit my problem...
$endgroup$
– NewMath
Jan 9 at 20:01
$begingroup$
Why is that a problem? Just ignore what it does to inputs outside of $mathcal F_A$.
$endgroup$
– Henning Makholm
Jan 9 at 22:57
$begingroup$
What you need to have a probability measure on $mathcal F_A$ is to explain how to find the probability of an arbitrary element of $mathcal F_A$. Writing "$P(;{cdot}mid A)$" supplies such an explanation clearly and succinctly. That there are also other things it might explain should not be a problem.
$endgroup$
– Henning Makholm
Jan 9 at 22:59
add a comment |
$begingroup$
Let $(Omega ,mathcal F,mathbb P)$ a probability space and $Ain mathcal F$ s.t. $mathbb P(A)>0$. In an exercise I have, I need to prove that $(A, mathcal F_A, mathbb P(.|A))$ is a probability space where $mathbb P(Cmid A)=frac{mathbb P(Acap C)}{mathbb P(A)}$ and $mathcal F_A={Bcap Amid Bin mathcal F}$. Is it really true ?
I agree that $mathcal F_A$ is a $sigma -$agebra on $A$.
I agree that $mathbb Q(C)=frac{mathbb P(C)}{mathbb P(A)}$ for $Cin mathcal F_A$ is a measure.
But for me $mathbb P(.|A)$ is a measure on $Omega $, and $(Omega ,mathcal F,mathbb P(.|A))$ is a probability measure, but $(A,mathcal F_A,mathbb P(.|A))$ is not. So, what is $mathbb P(.|A)$ for $(A,mathcal F_A)$ ? or what represent the probability space $(Omega ,mathcal F,mathbb P(.|A))$ for $(A,mathcal F_A, mathbb Q)$ ?
probability measure-theory
asked Jan 9 at 19:31
NewMathNewMath
4059
$endgroup$
Let $(Omega ,mathcal F,mathbb P)$ a probability space and $Ain mathcal F$ s.t. $mathbb P(A)>0$. In an exercise I have, I need to prove that $(A, mathcal F_A, mathbb P(.|A))$ is a probability space where $mathbb P(Cmid A)=frac{mathbb P(Acap C)}{mathbb P(A)}$ and $mathcal F_A={Bcap Amid Bin mathcal F}$. Is it really true ?
I agree that $mathcal F_A$ is a $sigma -$agebra on $A$.
I agree that $mathbb Q(C)=frac{mathbb P(C)}{mathbb P(A)}$ for $Cin mathcal F_A$ is a measure.
But for me $mathbb P(.|A)$ is a measure on $Omega $, and $(Omega ,mathcal F,mathbb P(.|A))$ is a probability measure, but $(A,mathcal F_A,mathbb P(.|A))$ is not. So, what is $mathbb P(.|A)$ for $(A,mathcal F_A)$ ? or what represent the probability space $(Omega ,mathcal F,mathbb P(.|A))$ for $(A,mathcal F_A, mathbb Q)$ ?
probability measure-theory
probability measure-theory
asked Jan 9 at 19:31
NewMathNewMath
4059
asked Jan 9 at 19:31
NewMathNewMath
4059
asked Jan 9 at 19:31
NewMathNewMath
4059
asked Jan 9 at 19:31
NewMathNewMath
4059
asked Jan 9 at 19:31
NewMathNewMath
4059
4059
1
$begingroup$
Obviously they mean $P(;{cdot}mid A)$ restricted to $mathcal F_A$ -- which is exactly the thing you want to call $Q$ instead.
$endgroup$
– Henning Makholm
Jan 9 at 19:38
1
$begingroup$
Note also that you could equally say $mathcal F_A = { Binmathcal F mid Bsubseteq A}$, which gives the same result as your definition because $mathcal F$ is an algebra and therefore closed under intersection.
$endgroup$
– Henning Makholm
Jan 9 at 19:43
$begingroup$
@HenningMakholm: But $mathbb P(cdot |A)$ can eat in $mathcal F$, not only $mathcal F_A$... that a bit my problem...
$endgroup$
– NewMath
Jan 9 at 20:01
$begingroup$
Why is that a problem? Just ignore what it does to inputs outside of $mathcal F_A$.
$endgroup$
– Henning Makholm
Jan 9 at 22:57
$begingroup$
What you need to have a probability measure on $mathcal F_A$ is to explain how to find the probability of an arbitrary element of $mathcal F_A$. Writing "$P(;{cdot}mid A)$" supplies such an explanation clearly and succinctly. That there are also other things it might explain should not be a problem.
$endgroup$
– Henning Makholm
Jan 9 at 22:59
add a comment |
1
$begingroup$
Obviously they mean $P(;{cdot}mid A)$ restricted to $mathcal F_A$ -- which is exactly the thing you want to call $Q$ instead.
$endgroup$
– Henning Makholm
Jan 9 at 19:38
1
$begingroup$
Note also that you could equally say $mathcal F_A = { Binmathcal F mid Bsubseteq A}$, which gives the same result as your definition because $mathcal F$ is an algebra and therefore closed under intersection.
$endgroup$
– Henning Makholm
Jan 9 at 19:43
$begingroup$
@HenningMakholm: But $mathbb P(cdot |A)$ can eat in $mathcal F$, not only $mathcal F_A$... that a bit my problem...
$endgroup$
– NewMath
Jan 9 at 20:01
$begingroup$
Why is that a problem? Just ignore what it does to inputs outside of $mathcal F_A$.
$endgroup$
– Henning Makholm
Jan 9 at 22:57
$begingroup$
What you need to have a probability measure on $mathcal F_A$ is to explain how to find the probability of an arbitrary element of $mathcal F_A$. Writing "$P(;{cdot}mid A)$" supplies such an explanation clearly and succinctly. That there are also other things it might explain should not be a problem.
$endgroup$
– Henning Makholm
Jan 9 at 22:59
1
1
$begingroup$
Obviously they mean $P(;{cdot}mid A)$ restricted to $mathcal F_A$ -- which is exactly the thing you want to call $Q$ instead.
$endgroup$
– Henning Makholm
Jan 9 at 19:38
$begingroup$
Obviously they mean $P(;{cdot}mid A)$ restricted to $mathcal F_A$ -- which is exactly the thing you want to call $Q$ instead.
$endgroup$
– Henning Makholm
Jan 9 at 19:38
1
1
$begingroup$
Note also that you could equally say $mathcal F_A = { Binmathcal F mid Bsubseteq A}$, which gives the same result as your definition because $mathcal F$ is an algebra and therefore closed under intersection.
$endgroup$
– Henning Makholm
Jan 9 at 19:43
$begingroup$
Note also that you could equally say $mathcal F_A = { Binmathcal F mid Bsubseteq A}$, which gives the same result as your definition because $mathcal F$ is an algebra and therefore closed under intersection.
$endgroup$
– Henning Makholm
Jan 9 at 19:43
$begingroup$
@HenningMakholm: But $mathbb P(cdot |A)$ can eat in $mathcal F$, not only $mathcal F_A$... that a bit my problem...
$endgroup$
– NewMath
Jan 9 at 20:01
$begingroup$
@HenningMakholm: But $mathbb P(cdot |A)$ can eat in $mathcal F$, not only $mathcal F_A$... that a bit my problem...
$endgroup$
– NewMath
Jan 9 at 20:01
$begingroup$
Why is that a problem? Just ignore what it does to inputs outside of $mathcal F_A$.
$endgroup$
– Henning Makholm
Jan 9 at 22:57
$begingroup$
Why is that a problem? Just ignore what it does to inputs outside of $mathcal F_A$.
$endgroup$
– Henning Makholm
Jan 9 at 22:57
$begingroup$
What you need to have a probability measure on $mathcal F_A$ is to explain how to find the probability of an arbitrary element of $mathcal F_A$. Writing "$P(;{cdot}mid A)$" supplies such an explanation clearly and succinctly. That there are also other things it might explain should not be a problem.
$endgroup$
– Henning Makholm
Jan 9 at 22:59
$begingroup$
What you need to have a probability measure on $mathcal F_A$ is to explain how to find the probability of an arbitrary element of $mathcal F_A$. Writing "$P(;{cdot}mid A)$" supplies such an explanation clearly and succinctly. That there are also other things it might explain should not be a problem.
$endgroup$
– Henning Makholm
Jan 9 at 22:59
add a comment |
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$begingroup$
Obviously they mean $P(;{cdot}mid A)$ restricted to $mathcal F_A$ -- which is exactly the thing you want to call $Q$ instead.
$endgroup$
– Henning Makholm
Jan 9 at 19:38
1
$begingroup$
Note also that you could equally say $mathcal F_A = { Binmathcal F mid Bsubseteq A}$, which gives the same result as your definition because $mathcal F$ is an algebra and therefore closed under intersection.
$endgroup$
– Henning Makholm
Jan 9 at 19:43
$begingroup$
@HenningMakholm: But $mathbb P(cdot |A)$ can eat in $mathcal F$, not only $mathcal F_A$... that a bit my problem...
$endgroup$
– NewMath
Jan 9 at 20:01
$begingroup$
Why is that a problem? Just ignore what it does to inputs outside of $mathcal F_A$.
$endgroup$
– Henning Makholm
Jan 9 at 22:57
$begingroup$
What you need to have a probability measure on $mathcal F_A$ is to explain how to find the probability of an arbitrary element of $mathcal F_A$. Writing "$P(;{cdot}mid A)$" supplies such an explanation clearly and succinctly. That there are also other things it might explain should not be a problem.
$endgroup$
– Henning Makholm
Jan 9 at 22:59