Cumulants vs. moments
$begingroup$
In high order statistics, what is the intuition for the difference between cumulants and moments? What does any of them measure and what is the intuition to use one of them over the other?
Specifically, I am following this paper, and I am trying to understand their reasoning behind using cumulants. Why did they move from $G_n$ to $C_n$ after equation [6]? What are those cross-terms that exist in the correlation case and are not present in the cumulant case?
moment-generating-functions cumulants
asked Dec 27 '18 at 7:43
havakokhavakok
520215
$endgroup$
add a comment |
$begingroup$
In high order statistics, what is the intuition for the difference between cumulants and moments? What does any of them measure and what is the intuition to use one of them over the other?
Specifically, I am following this paper, and I am trying to understand their reasoning behind using cumulants. Why did they move from $G_n$ to $C_n$ after equation [6]? What are those cross-terms that exist in the correlation case and are not present in the cumulant case?
moment-generating-functions cumulants
asked Dec 27 '18 at 7:43
havakokhavakok
520215
$endgroup$
add a comment |
$begingroup$
In high order statistics, what is the intuition for the difference between cumulants and moments? What does any of them measure and what is the intuition to use one of them over the other?
Specifically, I am following this paper, and I am trying to understand their reasoning behind using cumulants. Why did they move from $G_n$ to $C_n$ after equation [6]? What are those cross-terms that exist in the correlation case and are not present in the cumulant case?
moment-generating-functions cumulants
asked Dec 27 '18 at 7:43
havakokhavakok
520215
$endgroup$
In high order statistics, what is the intuition for the difference between cumulants and moments? What does any of them measure and what is the intuition to use one of them over the other?
Specifically, I am following this paper, and I am trying to understand their reasoning behind using cumulants. Why did they move from $G_n$ to $C_n$ after equation [6]? What are those cross-terms that exist in the correlation case and are not present in the cumulant case?
moment-generating-functions cumulants
moment-generating-functions cumulants
asked Dec 27 '18 at 7:43
havakokhavakok
520215
asked Dec 27 '18 at 7:43
havakokhavakok
520215
edited Dec 31 '18 at 13:38
havakok
asked Dec 27 '18 at 7:43
havakokhavakok
520215
asked Dec 27 '18 at 7:43
havakokhavakok
520215
asked Dec 27 '18 at 7:43
havakokhavakok
520215
520215
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The cumulant is the part of the moment that is not "caused" by lower order moments.
To get intuition, consider the case where the measurements are all the same, $X_i=x$, Then the $n$th moment is $langle X^nrangle=x^n=langle Xrangle^n$ , whereas the cumulants would all be $0$ starting from $n=2$. If we have non identical measurements each moment can be written as a sum of products lower moments and a "new" term. This new term is the cumulant.
As an example, consider the 3rd moment $<X^3>$. We write it as $$<((X-<X>)+<X>)^3>=<(X-<X>)^3+3(X-<X>)^2<X>+3(X-<X>)<X>^2+<X>^3>$$, recalling that $<X-<X>>=0$, we can write
$$<X^3>=<(X-<X>)^3>+3<(X-<X>)^2><X>+<X>^3$$. the term $(X-<X>)^3$ cannot be written in terms of $<X>;<(X-<X>)^2>$ and is the 3 order cumulant.
answered Dec 31 '18 at 13:10
user617446user617446
4443
$endgroup$
1
$begingroup$
@hardmath I added the 3rd moment case.
$endgroup$
– user617446
Dec 31 '18 at 14:52
add a comment |
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%2f3053685%2fcumulants-vs-moments%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The cumulant is the part of the moment that is not "caused" by lower order moments.
To get intuition, consider the case where the measurements are all the same, $X_i=x$, Then the $n$th moment is $langle X^nrangle=x^n=langle Xrangle^n$ , whereas the cumulants would all be $0$ starting from $n=2$. If we have non identical measurements each moment can be written as a sum of products lower moments and a "new" term. This new term is the cumulant.
As an example, consider the 3rd moment $<X^3>$. We write it as $$<((X-<X>)+<X>)^3>=<(X-<X>)^3+3(X-<X>)^2<X>+3(X-<X>)<X>^2+<X>^3>$$, recalling that $<X-<X>>=0$, we can write
$$<X^3>=<(X-<X>)^3>+3<(X-<X>)^2><X>+<X>^3$$. the term $(X-<X>)^3$ cannot be written in terms of $<X>;<(X-<X>)^2>$ and is the 3 order cumulant.
answered Dec 31 '18 at 13:10
user617446user617446
4443
$endgroup$
1
$begingroup$
@hardmath I added the 3rd moment case.
$endgroup$
– user617446
Dec 31 '18 at 14:52
add a comment |
$begingroup$
The cumulant is the part of the moment that is not "caused" by lower order moments.
To get intuition, consider the case where the measurements are all the same, $X_i=x$, Then the $n$th moment is $langle X^nrangle=x^n=langle Xrangle^n$ , whereas the cumulants would all be $0$ starting from $n=2$. If we have non identical measurements each moment can be written as a sum of products lower moments and a "new" term. This new term is the cumulant.
As an example, consider the 3rd moment $<X^3>$. We write it as $$<((X-<X>)+<X>)^3>=<(X-<X>)^3+3(X-<X>)^2<X>+3(X-<X>)<X>^2+<X>^3>$$, recalling that $<X-<X>>=0$, we can write
$$<X^3>=<(X-<X>)^3>+3<(X-<X>)^2><X>+<X>^3$$. the term $(X-<X>)^3$ cannot be written in terms of $<X>;<(X-<X>)^2>$ and is the 3 order cumulant.
answered Dec 31 '18 at 13:10
user617446user617446
4443
$endgroup$
1
$begingroup$
@hardmath I added the 3rd moment case.
$endgroup$
– user617446
Dec 31 '18 at 14:52
add a comment |
$begingroup$
The cumulant is the part of the moment that is not "caused" by lower order moments.
To get intuition, consider the case where the measurements are all the same, $X_i=x$, Then the $n$th moment is $langle X^nrangle=x^n=langle Xrangle^n$ , whereas the cumulants would all be $0$ starting from $n=2$. If we have non identical measurements each moment can be written as a sum of products lower moments and a "new" term. This new term is the cumulant.
As an example, consider the 3rd moment $<X^3>$. We write it as $$<((X-<X>)+<X>)^3>=<(X-<X>)^3+3(X-<X>)^2<X>+3(X-<X>)<X>^2+<X>^3>$$, recalling that $<X-<X>>=0$, we can write
$$<X^3>=<(X-<X>)^3>+3<(X-<X>)^2><X>+<X>^3$$. the term $(X-<X>)^3$ cannot be written in terms of $<X>;<(X-<X>)^2>$ and is the 3 order cumulant.
answered Dec 31 '18 at 13:10
user617446user617446
4443
$endgroup$
The cumulant is the part of the moment that is not "caused" by lower order moments.
To get intuition, consider the case where the measurements are all the same, $X_i=x$, Then the $n$th moment is $langle X^nrangle=x^n=langle Xrangle^n$ , whereas the cumulants would all be $0$ starting from $n=2$. If we have non identical measurements each moment can be written as a sum of products lower moments and a "new" term. This new term is the cumulant.
As an example, consider the 3rd moment $<X^3>$. We write it as $$<((X-<X>)+<X>)^3>=<(X-<X>)^3+3(X-<X>)^2<X>+3(X-<X>)<X>^2+<X>^3>$$, recalling that $<X-<X>>=0$, we can write
$$<X^3>=<(X-<X>)^3>+3<(X-<X>)^2><X>+<X>^3$$. the term $(X-<X>)^3$ cannot be written in terms of $<X>;<(X-<X>)^2>$ and is the 3 order cumulant.
answered Dec 31 '18 at 13:10
user617446user617446
4443
edited Dec 31 '18 at 14:51
answered Dec 31 '18 at 13:10
user617446user617446
4443
answered Dec 31 '18 at 13:10
user617446user617446
4443
answered Dec 31 '18 at 13:10
user617446user617446
4443
4443
1
$begingroup$
@hardmath I added the 3rd moment case.
$endgroup$
– user617446
Dec 31 '18 at 14:52
add a comment |
1
$begingroup$
@hardmath I added the 3rd moment case.
$endgroup$
– user617446
Dec 31 '18 at 14:52
1
1
$begingroup$
@hardmath I added the 3rd moment case.
$endgroup$
– user617446
Dec 31 '18 at 14:52
$begingroup$
@hardmath I added the 3rd moment case.
$endgroup$
– user617446
Dec 31 '18 at 14:52
add a comment |
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%2f3053685%2fcumulants-vs-moments%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
