Mixing
A problem about notation of expectation value
When I looking expectation value in wikipedia
I find that it has several expression for expectation value
(1).$E[X]=int_Omega X(omega)dP(omega)$
(2).$E[X]=int_Bbb R xf(x)dx$
and a random variable is a function that transform a probability space to another space is built up on real number
i.e ($Omega,mathcal F,P)to(Bbb R,mathcal B,P_X$)
so,it seems (1) is a expression based on the probability space $(Omega,mathcal F,P)$
(2) is for probability space $ (Bbb R,mathcal B,P_X$)
and I know $f=frac{dP_X}{du} $ is a Radon–Nikodym derivative where $u$ is Lebesgue measure
Now I am trying to rewrite the expression from (2) to (1)
(3) $E[X]=int_Bbb R xf(x)dx=int_Bbb R xf(x)u(dx)=int_Bbb R xfdu=int_Bbb R xfrac{dP_X}{du}du=int_Bbb RxdP_x$
Is that (3) correct?
and how do I rewrite the right hand side of (3) to (1)?
probability-theory measure-theory
asked Nov 21 '18 at 8:37
Vergil Chan
334
add a comment |
When I looking expectation value in wikipedia
I find that it has several expression for expectation value
(1).$E[X]=int_Omega X(omega)dP(omega)$
(2).$E[X]=int_Bbb R xf(x)dx$
and a random variable is a function that transform a probability space to another space is built up on real number
i.e ($Omega,mathcal F,P)to(Bbb R,mathcal B,P_X$)
so,it seems (1) is a expression based on the probability space $(Omega,mathcal F,P)$
(2) is for probability space $ (Bbb R,mathcal B,P_X$)
and I know $f=frac{dP_X}{du} $ is a Radon–Nikodym derivative where $u$ is Lebesgue measure
Now I am trying to rewrite the expression from (2) to (1)
(3) $E[X]=int_Bbb R xf(x)dx=int_Bbb R xf(x)u(dx)=int_Bbb R xfdu=int_Bbb R xfrac{dP_X}{du}du=int_Bbb RxdP_x$
Is that (3) correct?
and how do I rewrite the right hand side of (3) to (1)?
probability-theory measure-theory
asked Nov 21 '18 at 8:37
Vergil Chan
334
add a comment |
When I looking expectation value in wikipedia
I find that it has several expression for expectation value
(1).$E[X]=int_Omega X(omega)dP(omega)$
(2).$E[X]=int_Bbb R xf(x)dx$
and a random variable is a function that transform a probability space to another space is built up on real number
i.e ($Omega,mathcal F,P)to(Bbb R,mathcal B,P_X$)
so,it seems (1) is a expression based on the probability space $(Omega,mathcal F,P)$
(2) is for probability space $ (Bbb R,mathcal B,P_X$)
and I know $f=frac{dP_X}{du} $ is a Radon–Nikodym derivative where $u$ is Lebesgue measure
Now I am trying to rewrite the expression from (2) to (1)
(3) $E[X]=int_Bbb R xf(x)dx=int_Bbb R xf(x)u(dx)=int_Bbb R xfdu=int_Bbb R xfrac{dP_X}{du}du=int_Bbb RxdP_x$
Is that (3) correct?
and how do I rewrite the right hand side of (3) to (1)?
probability-theory measure-theory
asked Nov 21 '18 at 8:37
Vergil Chan
334
When I looking expectation value in wikipedia
I find that it has several expression for expectation value
(1).$E[X]=int_Omega X(omega)dP(omega)$
(2).$E[X]=int_Bbb R xf(x)dx$
and a random variable is a function that transform a probability space to another space is built up on real number
i.e ($Omega,mathcal F,P)to(Bbb R,mathcal B,P_X$)
so,it seems (1) is a expression based on the probability space $(Omega,mathcal F,P)$
(2) is for probability space $ (Bbb R,mathcal B,P_X$)
and I know $f=frac{dP_X}{du} $ is a Radon–Nikodym derivative where $u$ is Lebesgue measure
Now I am trying to rewrite the expression from (2) to (1)
(3) $E[X]=int_Bbb R xf(x)dx=int_Bbb R xf(x)u(dx)=int_Bbb R xfdu=int_Bbb R xfrac{dP_X}{du}du=int_Bbb RxdP_x$
Is that (3) correct?
and how do I rewrite the right hand side of (3) to (1)?
probability-theory measure-theory
probability-theory measure-theory
asked Nov 21 '18 at 8:37
Vergil Chan
334
asked Nov 21 '18 at 8:37
Vergil Chan
334
asked Nov 21 '18 at 8:37
Vergil Chan
334
asked Nov 21 '18 at 8:37
Vergil Chan
334
asked Nov 21 '18 at 8:37
Vergil Chan
334
334
add a comment |
add a comment |
1 Answer
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$P_X$ is defined by $P_X(B)=P(X^{-1}(B))$.Ffrom this we can show that $int xdP_X=int XdP$ (by simple function approximation). It should be noted that 1) is more general than 2) in the sense it does not require existence of $f$.
answered Nov 21 '18 at 8:51
Kavi Rama Murthy
51.1k31854
add a comment |
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1 Answer
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1 Answer
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active
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$P_X$ is defined by $P_X(B)=P(X^{-1}(B))$.Ffrom this we can show that $int xdP_X=int XdP$ (by simple function approximation). It should be noted that 1) is more general than 2) in the sense it does not require existence of $f$.
answered Nov 21 '18 at 8:51
Kavi Rama Murthy
51.1k31854
add a comment |
$P_X$ is defined by $P_X(B)=P(X^{-1}(B))$.Ffrom this we can show that $int xdP_X=int XdP$ (by simple function approximation). It should be noted that 1) is more general than 2) in the sense it does not require existence of $f$.
answered Nov 21 '18 at 8:51
Kavi Rama Murthy
51.1k31854
add a comment |
$P_X$ is defined by $P_X(B)=P(X^{-1}(B))$.Ffrom this we can show that $int xdP_X=int XdP$ (by simple function approximation). It should be noted that 1) is more general than 2) in the sense it does not require existence of $f$.
answered Nov 21 '18 at 8:51
Kavi Rama Murthy
51.1k31854
$P_X$ is defined by $P_X(B)=P(X^{-1}(B))$.Ffrom this we can show that $int xdP_X=int XdP$ (by simple function approximation). It should be noted that 1) is more general than 2) in the sense it does not require existence of $f$.
answered Nov 21 '18 at 8:51
Kavi Rama Murthy
51.1k31854
answered Nov 21 '18 at 8:51
Kavi Rama Murthy
51.1k31854
answered Nov 21 '18 at 8:51
Kavi Rama Murthy
51.1k31854
answered Nov 21 '18 at 8:51
Kavi Rama Murthy
51.1k31854
51.1k31854
add a comment |
add a comment |
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