Is there an interpretation of the hyper skewness?
$begingroup$
Let $X$ be a random variable. The standardized $n$th moment of $X$ is defined as
$$frac{E[(X-mathbb{E}[X])^n]}{mbox{Var}[X]^{n/2}}. $$
Special cases are the skewness ($k=3$) and the kurtosis $k=4$. The skewness is a measure for the asymmetry of a distribution while the kurtosis measures how peaked the distribution is. In my financial engineering project, I have to work with $k=5$ which is referred to as hyper skewness. As a benchmark, it is common to consider the normal distribution which has zero skewness and kurtosis equal to three. However, the hyper skewness of the normal distribution is also equal to zero, so at first sight it does not tell anything more about the distribution.
I was wondering if there is a useful interpretation of the hyper skewness? Does someone know any literature about this feature? If there is no any available literature then I can perhaps compute the hyper skewness for a variety of distributions and try to find an interpretation. However, some known literature would spare me some time.
Thanks in advance!
probability statistics finance
asked May 6 '16 at 14:45
SironSiron
1,205514
$endgroup$
add a comment |
$begingroup$
Let $X$ be a random variable. The standardized $n$th moment of $X$ is defined as
$$frac{E[(X-mathbb{E}[X])^n]}{mbox{Var}[X]^{n/2}}. $$
Special cases are the skewness ($k=3$) and the kurtosis $k=4$. The skewness is a measure for the asymmetry of a distribution while the kurtosis measures how peaked the distribution is. In my financial engineering project, I have to work with $k=5$ which is referred to as hyper skewness. As a benchmark, it is common to consider the normal distribution which has zero skewness and kurtosis equal to three. However, the hyper skewness of the normal distribution is also equal to zero, so at first sight it does not tell anything more about the distribution.
I was wondering if there is a useful interpretation of the hyper skewness? Does someone know any literature about this feature? If there is no any available literature then I can perhaps compute the hyper skewness for a variety of distributions and try to find an interpretation. However, some known literature would spare me some time.
Thanks in advance!
probability statistics finance
asked May 6 '16 at 14:45
SironSiron
1,205514
$endgroup$
$begingroup$
What might be more natural than the fifth central moment is the fifth cumulant: $kappa_5 = mu_5 - 10mu_3 mu_2$, where $mu_k$ is the $k$th central moment. Like the fifth central moment, the fifth cumulant is fifth-degree homogeneous and translation-invariant, but it is also "cumulative", i.e. if $X_1,ldots,X_n$ are independent random variables then $kappa_5(X_1+cdots+X_n) = kappa_5(X_1) + cdots + kappa_5(X_n)$. $qquad$
$endgroup$
– Michael Hardy
May 6 '16 at 15:39
$begingroup$
Thanks, I was aware of the cumulant function. How does that contribute to an actual interpretation of the hyper skewness? Ideally, I would obtain a similar interpretation as for the skewness/kurtosis. But perhaps that is not possible.
$endgroup$
– Siron
May 6 '16 at 17:32
1
$begingroup$
Take a look at: en.wikipedia.org/wiki/Moment_%28mathematics%29#Higher_moments
$endgroup$
– Guilherme Thompson
May 6 '16 at 18:00
add a comment |
$begingroup$
Let $X$ be a random variable. The standardized $n$th moment of $X$ is defined as
$$frac{E[(X-mathbb{E}[X])^n]}{mbox{Var}[X]^{n/2}}. $$
Special cases are the skewness ($k=3$) and the kurtosis $k=4$. The skewness is a measure for the asymmetry of a distribution while the kurtosis measures how peaked the distribution is. In my financial engineering project, I have to work with $k=5$ which is referred to as hyper skewness. As a benchmark, it is common to consider the normal distribution which has zero skewness and kurtosis equal to three. However, the hyper skewness of the normal distribution is also equal to zero, so at first sight it does not tell anything more about the distribution.
I was wondering if there is a useful interpretation of the hyper skewness? Does someone know any literature about this feature? If there is no any available literature then I can perhaps compute the hyper skewness for a variety of distributions and try to find an interpretation. However, some known literature would spare me some time.
Thanks in advance!
probability statistics finance
asked May 6 '16 at 14:45
SironSiron
1,205514
$endgroup$
Let $X$ be a random variable. The standardized $n$th moment of $X$ is defined as
$$frac{E[(X-mathbb{E}[X])^n]}{mbox{Var}[X]^{n/2}}. $$
Special cases are the skewness ($k=3$) and the kurtosis $k=4$. The skewness is a measure for the asymmetry of a distribution while the kurtosis measures how peaked the distribution is. In my financial engineering project, I have to work with $k=5$ which is referred to as hyper skewness. As a benchmark, it is common to consider the normal distribution which has zero skewness and kurtosis equal to three. However, the hyper skewness of the normal distribution is also equal to zero, so at first sight it does not tell anything more about the distribution.
I was wondering if there is a useful interpretation of the hyper skewness? Does someone know any literature about this feature? If there is no any available literature then I can perhaps compute the hyper skewness for a variety of distributions and try to find an interpretation. However, some known literature would spare me some time.
Thanks in advance!
probability statistics finance
probability statistics finance
asked May 6 '16 at 14:45
SironSiron
1,205514
asked May 6 '16 at 14:45
SironSiron
1,205514
edited May 6 '16 at 14:56
Siron
asked May 6 '16 at 14:45
SironSiron
1,205514
asked May 6 '16 at 14:45
SironSiron
1,205514
asked May 6 '16 at 14:45
SironSiron
1,205514
1,205514
$begingroup$
What might be more natural than the fifth central moment is the fifth cumulant: $kappa_5 = mu_5 - 10mu_3 mu_2$, where $mu_k$ is the $k$th central moment. Like the fifth central moment, the fifth cumulant is fifth-degree homogeneous and translation-invariant, but it is also "cumulative", i.e. if $X_1,ldots,X_n$ are independent random variables then $kappa_5(X_1+cdots+X_n) = kappa_5(X_1) + cdots + kappa_5(X_n)$. $qquad$
$endgroup$
– Michael Hardy
May 6 '16 at 15:39
$begingroup$
Thanks, I was aware of the cumulant function. How does that contribute to an actual interpretation of the hyper skewness? Ideally, I would obtain a similar interpretation as for the skewness/kurtosis. But perhaps that is not possible.
$endgroup$
– Siron
May 6 '16 at 17:32
1
$begingroup$
Take a look at: en.wikipedia.org/wiki/Moment_%28mathematics%29#Higher_moments
$endgroup$
– Guilherme Thompson
May 6 '16 at 18:00
add a comment |
$begingroup$
What might be more natural than the fifth central moment is the fifth cumulant: $kappa_5 = mu_5 - 10mu_3 mu_2$, where $mu_k$ is the $k$th central moment. Like the fifth central moment, the fifth cumulant is fifth-degree homogeneous and translation-invariant, but it is also "cumulative", i.e. if $X_1,ldots,X_n$ are independent random variables then $kappa_5(X_1+cdots+X_n) = kappa_5(X_1) + cdots + kappa_5(X_n)$. $qquad$
$endgroup$
– Michael Hardy
May 6 '16 at 15:39
$begingroup$
Thanks, I was aware of the cumulant function. How does that contribute to an actual interpretation of the hyper skewness? Ideally, I would obtain a similar interpretation as for the skewness/kurtosis. But perhaps that is not possible.
$endgroup$
– Siron
May 6 '16 at 17:32
1
$begingroup$
Take a look at: en.wikipedia.org/wiki/Moment_%28mathematics%29#Higher_moments
$endgroup$
– Guilherme Thompson
May 6 '16 at 18:00
$begingroup$
What might be more natural than the fifth central moment is the fifth cumulant: $kappa_5 = mu_5 - 10mu_3 mu_2$, where $mu_k$ is the $k$th central moment. Like the fifth central moment, the fifth cumulant is fifth-degree homogeneous and translation-invariant, but it is also "cumulative", i.e. if $X_1,ldots,X_n$ are independent random variables then $kappa_5(X_1+cdots+X_n) = kappa_5(X_1) + cdots + kappa_5(X_n)$. $qquad$
$endgroup$
– Michael Hardy
May 6 '16 at 15:39
$begingroup$
What might be more natural than the fifth central moment is the fifth cumulant: $kappa_5 = mu_5 - 10mu_3 mu_2$, where $mu_k$ is the $k$th central moment. Like the fifth central moment, the fifth cumulant is fifth-degree homogeneous and translation-invariant, but it is also "cumulative", i.e. if $X_1,ldots,X_n$ are independent random variables then $kappa_5(X_1+cdots+X_n) = kappa_5(X_1) + cdots + kappa_5(X_n)$. $qquad$
$endgroup$
– Michael Hardy
May 6 '16 at 15:39
$begingroup$
Thanks, I was aware of the cumulant function. How does that contribute to an actual interpretation of the hyper skewness? Ideally, I would obtain a similar interpretation as for the skewness/kurtosis. But perhaps that is not possible.
$endgroup$
– Siron
May 6 '16 at 17:32
$begingroup$
Thanks, I was aware of the cumulant function. How does that contribute to an actual interpretation of the hyper skewness? Ideally, I would obtain a similar interpretation as for the skewness/kurtosis. But perhaps that is not possible.
$endgroup$
– Siron
May 6 '16 at 17:32
1
1
$begingroup$
Take a look at: en.wikipedia.org/wiki/Moment_%28mathematics%29#Higher_moments
$endgroup$
– Guilherme Thompson
May 6 '16 at 18:00
$begingroup$
Take a look at: en.wikipedia.org/wiki/Moment_%28mathematics%29#Higher_moments
$endgroup$
– Guilherme Thompson
May 6 '16 at 18:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here's the method to construct an interpretation for anything. Think of relevant English words. Map Mathmatical ideas to English descriptions in a one-to-one fashion.
The expectation of a random variable is called $E(X)$.
English: $E(X)$ is the average value of $X$.
The variation from mean is called $(X-E(X))$.
English: $(X-E(X))$ is the difference between the random variable and its average value.
$X^n$ is power redistribution of the bariable.
English: $X^n$ compresses low values and stretches high values relative to $1$. More so, if $n$ is large. Less so if $n$ is small.
Put these translations together to get the "meaning" of $E((X-E(X))^n)$. The other part is just a constant.
$endgroup$
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%2f1774211%2fis-there-an-interpretation-of-the-hyper-skewness%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
$begingroup$
Here's the method to construct an interpretation for anything. Think of relevant English words. Map Mathmatical ideas to English descriptions in a one-to-one fashion.
The expectation of a random variable is called $E(X)$.
English: $E(X)$ is the average value of $X$.
The variation from mean is called $(X-E(X))$.
English: $(X-E(X))$ is the difference between the random variable and its average value.
$X^n$ is power redistribution of the bariable.
English: $X^n$ compresses low values and stretches high values relative to $1$. More so, if $n$ is large. Less so if $n$ is small.
Put these translations together to get the "meaning" of $E((X-E(X))^n)$. The other part is just a constant.
$endgroup$
add a comment |
$begingroup$
Here's the method to construct an interpretation for anything. Think of relevant English words. Map Mathmatical ideas to English descriptions in a one-to-one fashion.
The expectation of a random variable is called $E(X)$.
English: $E(X)$ is the average value of $X$.
The variation from mean is called $(X-E(X))$.
English: $(X-E(X))$ is the difference between the random variable and its average value.
$X^n$ is power redistribution of the bariable.
English: $X^n$ compresses low values and stretches high values relative to $1$. More so, if $n$ is large. Less so if $n$ is small.
Put these translations together to get the "meaning" of $E((X-E(X))^n)$. The other part is just a constant.
$endgroup$
add a comment |
$begingroup$
Here's the method to construct an interpretation for anything. Think of relevant English words. Map Mathmatical ideas to English descriptions in a one-to-one fashion.
The expectation of a random variable is called $E(X)$.
English: $E(X)$ is the average value of $X$.
The variation from mean is called $(X-E(X))$.
English: $(X-E(X))$ is the difference between the random variable and its average value.
$X^n$ is power redistribution of the bariable.
English: $X^n$ compresses low values and stretches high values relative to $1$. More so, if $n$ is large. Less so if $n$ is small.
Put these translations together to get the "meaning" of $E((X-E(X))^n)$. The other part is just a constant.
$endgroup$
Here's the method to construct an interpretation for anything. Think of relevant English words. Map Mathmatical ideas to English descriptions in a one-to-one fashion.
The expectation of a random variable is called $E(X)$.
English: $E(X)$ is the average value of $X$.
The variation from mean is called $(X-E(X))$.
English: $(X-E(X))$ is the difference between the random variable and its average value.
$X^n$ is power redistribution of the bariable.
English: $X^n$ compresses low values and stretches high values relative to $1$. More so, if $n$ is large. Less so if $n$ is small.
Put these translations together to get the "meaning" of $E((X-E(X))^n)$. The other part is just a constant.
answered May 6 '16 at 15:03
community wiki
Zach466920
add a comment |
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%2f1774211%2fis-there-an-interpretation-of-the-hyper-skewness%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
$begingroup$
What might be more natural than the fifth central moment is the fifth cumulant: $kappa_5 = mu_5 - 10mu_3 mu_2$, where $mu_k$ is the $k$th central moment. Like the fifth central moment, the fifth cumulant is fifth-degree homogeneous and translation-invariant, but it is also "cumulative", i.e. if $X_1,ldots,X_n$ are independent random variables then $kappa_5(X_1+cdots+X_n) = kappa_5(X_1) + cdots + kappa_5(X_n)$. $qquad$
$endgroup$
– Michael Hardy
May 6 '16 at 15:39
$begingroup$
Thanks, I was aware of the cumulant function. How does that contribute to an actual interpretation of the hyper skewness? Ideally, I would obtain a similar interpretation as for the skewness/kurtosis. But perhaps that is not possible.
$endgroup$
– Siron
May 6 '16 at 17:32
1
$begingroup$
Take a look at: en.wikipedia.org/wiki/Moment_%28mathematics%29#Higher_moments
$endgroup$
– Guilherme Thompson
May 6 '16 at 18:00