Is p(x)dx equal to dp(x)?
$begingroup$
I'm confused with the definition of the expectation operator.
Assume a random variable $X$ having a probability distribution $p(x)$. Then the expected value of $X$ can be computed as $int xp(x)dx$.
It is noted in 1 that, given a probability space $(Omega, Sigma,
P)$ as defined in 2, the general definition of the expected value is
$int_{Omega} X dP = int_{Omega} X(w)P(dw) $
where $P$ is the probability measure returning an events probability in $Sigma$. Is this probability measure the same as the distribution $p(x)$. Additionaly. what does $X(w)$ and $P(dw)$ mean?
Also, the Eq. 1 in http://leon.bottou.org/publications/pdf/online-1998.pdf states that
$E[f(x)] = int f(x)p(x) = int f(x)dp(x)$, is this correct or simply there is a notation error?
probability probability-theory
asked Feb 26 '15 at 16:39
Adam I.Adam I.
1419
$endgroup$
|
show 2 more comments
$begingroup$
I'm confused with the definition of the expectation operator.
Assume a random variable $X$ having a probability distribution $p(x)$. Then the expected value of $X$ can be computed as $int xp(x)dx$.
It is noted in 1 that, given a probability space $(Omega, Sigma,
P)$ as defined in 2, the general definition of the expected value is
$int_{Omega} X dP = int_{Omega} X(w)P(dw) $
where $P$ is the probability measure returning an events probability in $Sigma$. Is this probability measure the same as the distribution $p(x)$. Additionaly. what does $X(w)$ and $P(dw)$ mean?
Also, the Eq. 1 in http://leon.bottou.org/publications/pdf/online-1998.pdf states that
$E[f(x)] = int f(x)p(x) = int f(x)dp(x)$, is this correct or simply there is a notation error?
probability probability-theory
asked Feb 26 '15 at 16:39
Adam I.Adam I.
1419
$endgroup$
$begingroup$
Yes, that's true.
$endgroup$
– user207710
Feb 26 '15 at 16:40
1
$begingroup$
Thanks, but if p(x)dx = dp(x), then dp(x)/dx = p(x) which is not true, no?
$endgroup$
– Adam I.
Feb 26 '15 at 16:43
$begingroup$
Apologies, that should be $p'(x)dx$.
$endgroup$
– user207710
Feb 26 '15 at 16:44
4
$begingroup$
If the measure $P$ is absolutely continuous with respect to the Lebesgue measure then you can write $P(A) = int_A f(x) dx$ for some $f$. But not all measures can be written this way. Take $P=delta_0$, for example. The notation $int X(omega) P(domega)$ is often written as $int X(omega) d P(omega)$. Think of the $P(domega)$ as the probability of an 'infinitesimal' slice :-).
$endgroup$
– copper.hat
Feb 26 '15 at 16:46
1
$begingroup$
This is a older notation. Nowadays, you write $int_Omega X(omega), dP(omega)$, not $int_Omega X(omega), P(domega)$,
$endgroup$
– ncmathsadist
Feb 26 '15 at 17:09
|
show 2 more comments
$begingroup$
I'm confused with the definition of the expectation operator.
Assume a random variable $X$ having a probability distribution $p(x)$. Then the expected value of $X$ can be computed as $int xp(x)dx$.
It is noted in 1 that, given a probability space $(Omega, Sigma,
P)$ as defined in 2, the general definition of the expected value is
$int_{Omega} X dP = int_{Omega} X(w)P(dw) $
where $P$ is the probability measure returning an events probability in $Sigma$. Is this probability measure the same as the distribution $p(x)$. Additionaly. what does $X(w)$ and $P(dw)$ mean?
Also, the Eq. 1 in http://leon.bottou.org/publications/pdf/online-1998.pdf states that
$E[f(x)] = int f(x)p(x) = int f(x)dp(x)$, is this correct or simply there is a notation error?
probability probability-theory
asked Feb 26 '15 at 16:39
Adam I.Adam I.
1419
$endgroup$
I'm confused with the definition of the expectation operator.
Assume a random variable $X$ having a probability distribution $p(x)$. Then the expected value of $X$ can be computed as $int xp(x)dx$.
It is noted in 1 that, given a probability space $(Omega, Sigma,
P)$ as defined in 2, the general definition of the expected value is
$int_{Omega} X dP = int_{Omega} X(w)P(dw) $
where $P$ is the probability measure returning an events probability in $Sigma$. Is this probability measure the same as the distribution $p(x)$. Additionaly. what does $X(w)$ and $P(dw)$ mean?
Also, the Eq. 1 in http://leon.bottou.org/publications/pdf/online-1998.pdf states that
$E[f(x)] = int f(x)p(x) = int f(x)dp(x)$, is this correct or simply there is a notation error?
probability probability-theory
probability probability-theory
asked Feb 26 '15 at 16:39
Adam I.Adam I.
1419
asked Feb 26 '15 at 16:39
Adam I.Adam I.
1419
asked Feb 26 '15 at 16:39
Adam I.Adam I.
1419
asked Feb 26 '15 at 16:39
Adam I.Adam I.
1419
asked Feb 26 '15 at 16:39
Adam I.Adam I.
1419
1419
$begingroup$
Yes, that's true.
$endgroup$
– user207710
Feb 26 '15 at 16:40
1
$begingroup$
Thanks, but if p(x)dx = dp(x), then dp(x)/dx = p(x) which is not true, no?
$endgroup$
– Adam I.
Feb 26 '15 at 16:43
$begingroup$
Apologies, that should be $p'(x)dx$.
$endgroup$
– user207710
Feb 26 '15 at 16:44
4
$begingroup$
If the measure $P$ is absolutely continuous with respect to the Lebesgue measure then you can write $P(A) = int_A f(x) dx$ for some $f$. But not all measures can be written this way. Take $P=delta_0$, for example. The notation $int X(omega) P(domega)$ is often written as $int X(omega) d P(omega)$. Think of the $P(domega)$ as the probability of an 'infinitesimal' slice :-).
$endgroup$
– copper.hat
Feb 26 '15 at 16:46
1
$begingroup$
This is a older notation. Nowadays, you write $int_Omega X(omega), dP(omega)$, not $int_Omega X(omega), P(domega)$,
$endgroup$
– ncmathsadist
Feb 26 '15 at 17:09
|
show 2 more comments
$begingroup$
Yes, that's true.
$endgroup$
– user207710
Feb 26 '15 at 16:40
1
$begingroup$
Thanks, but if p(x)dx = dp(x), then dp(x)/dx = p(x) which is not true, no?
$endgroup$
– Adam I.
Feb 26 '15 at 16:43
$begingroup$
Apologies, that should be $p'(x)dx$.
$endgroup$
– user207710
Feb 26 '15 at 16:44
4
$begingroup$
If the measure $P$ is absolutely continuous with respect to the Lebesgue measure then you can write $P(A) = int_A f(x) dx$ for some $f$. But not all measures can be written this way. Take $P=delta_0$, for example. The notation $int X(omega) P(domega)$ is often written as $int X(omega) d P(omega)$. Think of the $P(domega)$ as the probability of an 'infinitesimal' slice :-).
$endgroup$
– copper.hat
Feb 26 '15 at 16:46
1
$begingroup$
This is a older notation. Nowadays, you write $int_Omega X(omega), dP(omega)$, not $int_Omega X(omega), P(domega)$,
$endgroup$
– ncmathsadist
Feb 26 '15 at 17:09
$begingroup$
Yes, that's true.
$endgroup$
– user207710
Feb 26 '15 at 16:40
$begingroup$
Yes, that's true.
$endgroup$
– user207710
Feb 26 '15 at 16:40
1
1
$begingroup$
Thanks, but if p(x)dx = dp(x), then dp(x)/dx = p(x) which is not true, no?
$endgroup$
– Adam I.
Feb 26 '15 at 16:43
$begingroup$
Thanks, but if p(x)dx = dp(x), then dp(x)/dx = p(x) which is not true, no?
$endgroup$
– Adam I.
Feb 26 '15 at 16:43
$begingroup$
Apologies, that should be $p'(x)dx$.
$endgroup$
– user207710
Feb 26 '15 at 16:44
$begingroup$
Apologies, that should be $p'(x)dx$.
$endgroup$
– user207710
Feb 26 '15 at 16:44
4
4
$begingroup$
If the measure $P$ is absolutely continuous with respect to the Lebesgue measure then you can write $P(A) = int_A f(x) dx$ for some $f$. But not all measures can be written this way. Take $P=delta_0$, for example. The notation $int X(omega) P(domega)$ is often written as $int X(omega) d P(omega)$. Think of the $P(domega)$ as the probability of an 'infinitesimal' slice :-).
$endgroup$
– copper.hat
Feb 26 '15 at 16:46
$begingroup$
If the measure $P$ is absolutely continuous with respect to the Lebesgue measure then you can write $P(A) = int_A f(x) dx$ for some $f$. But not all measures can be written this way. Take $P=delta_0$, for example. The notation $int X(omega) P(domega)$ is often written as $int X(omega) d P(omega)$. Think of the $P(domega)$ as the probability of an 'infinitesimal' slice :-).
$endgroup$
– copper.hat
Feb 26 '15 at 16:46
1
1
$begingroup$
This is a older notation. Nowadays, you write $int_Omega X(omega), dP(omega)$, not $int_Omega X(omega), P(domega)$,
$endgroup$
– ncmathsadist
Feb 26 '15 at 17:09
$begingroup$
This is a older notation. Nowadays, you write $int_Omega X(omega), dP(omega)$, not $int_Omega X(omega), P(domega)$,
$endgroup$
– ncmathsadist
Feb 26 '15 at 17:09
|
show 2 more comments
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$begingroup$
Yes, that's true.
$endgroup$
– user207710
Feb 26 '15 at 16:40
1
$begingroup$
Thanks, but if p(x)dx = dp(x), then dp(x)/dx = p(x) which is not true, no?
$endgroup$
– Adam I.
Feb 26 '15 at 16:43
$begingroup$
Apologies, that should be $p'(x)dx$.
$endgroup$
– user207710
Feb 26 '15 at 16:44
4
$begingroup$
If the measure $P$ is absolutely continuous with respect to the Lebesgue measure then you can write $P(A) = int_A f(x) dx$ for some $f$. But not all measures can be written this way. Take $P=delta_0$, for example. The notation $int X(omega) P(domega)$ is often written as $int X(omega) d P(omega)$. Think of the $P(domega)$ as the probability of an 'infinitesimal' slice :-).
$endgroup$
– copper.hat
Feb 26 '15 at 16:46
1
$begingroup$
This is a older notation. Nowadays, you write $int_Omega X(omega), dP(omega)$, not $int_Omega X(omega), P(domega)$,
$endgroup$
– ncmathsadist
Feb 26 '15 at 17:09