Mixing
probability measure of Random Variable on probability space
A random variable is a function that transform a probability space to another space is built up on real number
i.e ($Omega,mathcal F,P)to(Bbb R,mathcal B,P_X$)
and a random variable is call measurable if $forall B in mathcal B$ $X^{-1}(B)inmathcal F$
How do we guarantee that$P(A)$ $forall Ain mathcal F $
is equal to $P_X(B)forall Bin mathcal B$
or How we show that $int _AdP=int _BdP_X$
where $A=$ $X^{-1}(B)$ $forall Bin mathcal B$
I know this maybe intuitive but I want have a strict mathematics proof to show this two things is equal
probability probability-theory measure-theory
asked Nov 22 '18 at 1:10
Vergil ChanVergil Chan
334
add a comment |
A random variable is a function that transform a probability space to another space is built up on real number
i.e ($Omega,mathcal F,P)to(Bbb R,mathcal B,P_X$)
and a random variable is call measurable if $forall B in mathcal B$ $X^{-1}(B)inmathcal F$
How do we guarantee that$P(A)$ $forall Ain mathcal F $
is equal to $P_X(B)forall Bin mathcal B$
or How we show that $int _AdP=int _BdP_X$
where $A=$ $X^{-1}(B)$ $forall Bin mathcal B$
I know this maybe intuitive but I want have a strict mathematics proof to show this two things is equal
probability probability-theory measure-theory
asked Nov 22 '18 at 1:10
Vergil ChanVergil Chan
334
add a comment |
A random variable is a function that transform a probability space to another space is built up on real number
i.e ($Omega,mathcal F,P)to(Bbb R,mathcal B,P_X$)
and a random variable is call measurable if $forall B in mathcal B$ $X^{-1}(B)inmathcal F$
How do we guarantee that$P(A)$ $forall Ain mathcal F $
is equal to $P_X(B)forall Bin mathcal B$
or How we show that $int _AdP=int _BdP_X$
where $A=$ $X^{-1}(B)$ $forall Bin mathcal B$
I know this maybe intuitive but I want have a strict mathematics proof to show this two things is equal
probability probability-theory measure-theory
asked Nov 22 '18 at 1:10
Vergil ChanVergil Chan
334
A random variable is a function that transform a probability space to another space is built up on real number
i.e ($Omega,mathcal F,P)to(Bbb R,mathcal B,P_X$)
and a random variable is call measurable if $forall B in mathcal B$ $X^{-1}(B)inmathcal F$
How do we guarantee that$P(A)$ $forall Ain mathcal F $
is equal to $P_X(B)forall Bin mathcal B$
or How we show that $int _AdP=int _BdP_X$
where $A=$ $X^{-1}(B)$ $forall Bin mathcal B$
I know this maybe intuitive but I want have a strict mathematics proof to show this two things is equal
probability probability-theory measure-theory
probability probability-theory measure-theory
asked Nov 22 '18 at 1:10
Vergil ChanVergil Chan
334
asked Nov 22 '18 at 1:10
Vergil ChanVergil Chan
334
asked Nov 22 '18 at 1:10
Vergil ChanVergil Chan
334
asked Nov 22 '18 at 1:10
Vergil ChanVergil Chan
334
asked Nov 22 '18 at 1:10
Vergil ChanVergil Chan
334
334
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
That's because the construction of the probability measure $P^{X}$.
You can see that $forall Bin mathcal{B}, hspace{0.3 cm}$ $P^{X}(B):=P(X^{-1}(B))$ is indeed a probability measure in the space $(mathbb{R},mathcal{B})$. (Check that it satisfies the axioms of probability)
Therefore,
$ forall Bin mathcal{B} hspace{0.2 cm}$ if we define $A=X^{-1}(B)in mathcal{F}$
we have that
$P^{X}(B)=P(X^{-1}(B))=P(A)$
edited Nov 22 '18 at 2:41
Graham Kemp
84.7k43378
answered Nov 22 '18 at 1:45
GuilleAlAlGuilleAlAl
212
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
That's because the construction of the probability measure $P^{X}$.
You can see that $forall Bin mathcal{B}, hspace{0.3 cm}$ $P^{X}(B):=P(X^{-1}(B))$ is indeed a probability measure in the space $(mathbb{R},mathcal{B})$. (Check that it satisfies the axioms of probability)
Therefore,
$ forall Bin mathcal{B} hspace{0.2 cm}$ if we define $A=X^{-1}(B)in mathcal{F}$
we have that
$P^{X}(B)=P(X^{-1}(B))=P(A)$
edited Nov 22 '18 at 2:41
Graham Kemp
84.7k43378
answered Nov 22 '18 at 1:45
GuilleAlAlGuilleAlAl
212
add a comment |
That's because the construction of the probability measure $P^{X}$.
You can see that $forall Bin mathcal{B}, hspace{0.3 cm}$ $P^{X}(B):=P(X^{-1}(B))$ is indeed a probability measure in the space $(mathbb{R},mathcal{B})$. (Check that it satisfies the axioms of probability)
Therefore,
$ forall Bin mathcal{B} hspace{0.2 cm}$ if we define $A=X^{-1}(B)in mathcal{F}$
we have that
$P^{X}(B)=P(X^{-1}(B))=P(A)$
edited Nov 22 '18 at 2:41
Graham Kemp
84.7k43378
answered Nov 22 '18 at 1:45
GuilleAlAlGuilleAlAl
212
add a comment |
That's because the construction of the probability measure $P^{X}$.
You can see that $forall Bin mathcal{B}, hspace{0.3 cm}$ $P^{X}(B):=P(X^{-1}(B))$ is indeed a probability measure in the space $(mathbb{R},mathcal{B})$. (Check that it satisfies the axioms of probability)
Therefore,
$ forall Bin mathcal{B} hspace{0.2 cm}$ if we define $A=X^{-1}(B)in mathcal{F}$
we have that
$P^{X}(B)=P(X^{-1}(B))=P(A)$
edited Nov 22 '18 at 2:41
Graham Kemp
84.7k43378
answered Nov 22 '18 at 1:45
GuilleAlAlGuilleAlAl
212
That's because the construction of the probability measure $P^{X}$.
You can see that $forall Bin mathcal{B}, hspace{0.3 cm}$ $P^{X}(B):=P(X^{-1}(B))$ is indeed a probability measure in the space $(mathbb{R},mathcal{B})$. (Check that it satisfies the axioms of probability)
Therefore,
$ forall Bin mathcal{B} hspace{0.2 cm}$ if we define $A=X^{-1}(B)in mathcal{F}$
we have that
$P^{X}(B)=P(X^{-1}(B))=P(A)$
edited Nov 22 '18 at 2:41
Graham Kemp
84.7k43378
answered Nov 22 '18 at 1:45
GuilleAlAlGuilleAlAl
212
edited Nov 22 '18 at 2:41
Graham Kemp
84.7k43378
edited Nov 22 '18 at 2:41
Graham Kemp
84.7k43378
edited Nov 22 '18 at 2:41
Graham Kemp
84.7k43378
84.7k43378
answered Nov 22 '18 at 1:45
GuilleAlAlGuilleAlAl
212
answered Nov 22 '18 at 1:45
GuilleAlAlGuilleAlAl
212
answered Nov 22 '18 at 1:45
GuilleAlAlGuilleAlAl
212
212
add a comment |
add a comment |
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