Cumulants vs. moments
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
havakok
510215
add a comment |
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
havakok
510215
add a comment |
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
havakok
510215
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
havakok
510215
asked Dec 27 '18 at 7:43
havakok
510215
edited 2 days ago
asked Dec 27 '18 at 7:43
havakok
510215
asked Dec 27 '18 at 7:43
havakok
510215
asked Dec 27 '18 at 7:43
havakok
510215
510215
add a comment |
add a comment |
1 Answer
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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 2 days ago
user617446
4443
1
@hardmath I added the 3rd moment case.
– user617446
2 days ago
add a comment |
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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 2 days ago
user617446
4443
1
@hardmath I added the 3rd moment case.
– user617446
2 days ago
add a comment |
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 2 days ago
user617446
4443
1
@hardmath I added the 3rd moment case.
– user617446
2 days ago
add a comment |
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 2 days ago
user617446
4443
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 2 days ago
user617446
4443
edited 2 days ago
answered 2 days ago
user617446
4443
answered 2 days ago
user617446
4443
answered 2 days ago
user617446
4443
4443
1
@hardmath I added the 3rd moment case.
– user617446
2 days ago
add a comment |
1
@hardmath I added the 3rd moment case.
– user617446
2 days ago
1
1
@hardmath I added the 3rd moment case.
– user617446
2 days ago
@hardmath I added the 3rd moment case.
– user617446
2 days ago
add a comment |
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