A problem about notation of expectation value
up vote
2
down vote
favorite
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
add a comment |
up vote
2
down vote
favorite
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
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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
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 yesterday
Vergil Chan
264
264
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
$P_X$ 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 $f$.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$P_X$ 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 $f$.
add a comment |
up vote
2
down vote
$P_X$ 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 $f$.
add a comment |
up vote
2
down vote
up vote
2
down vote
$P_X$ 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 $f$.
$P_X$ 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 yesterday
Kavi Rama Murthy
41.1k31751
41.1k31751
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007438%2fa-problem-about-notation-of-expectation-value%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown