Mixing
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.
1369
$endgroup$
|
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.
1369
$endgroup$
$begingroup$
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.
1369
$endgroup$
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.
1369
asked Feb 26 '15 at 16:39
Adam I.Adam I.
1369
asked Feb 26 '15 at 16:39
Adam I.Adam I.
1369
asked Feb 26 '15 at 16:39
Adam I.Adam I.
1369
asked Feb 26 '15 at 16:39
Adam I.Adam I.
1369
1369
$begingroup$
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
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2f1166660%2fis-pxdx-equal-to-dpx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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%2f1166660%2fis-pxdx-equal-to-dpx%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
$begingroup$
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