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How could and coskewness and cokurtosis be visualized in an easily comprehensible manner?

Mean, variances, skewness, kurtosis can easily be illustrated in density plots:

enter image description here

(Source: own *TeX-stuff)

The first cross-moment (co-variance) can be visualized by looking at joint densities:

covariance

(Source: https://www.quora.com/Is-it-possible-to-visualize-covariance-in-a-bivariate-normal-distribution-in-a-straightforward-way-similar-to-variance-in-a-univariate-normal-distribution)

How could and coskewness and cokurtosis be visualized in an easily comprehensible manner?

Whereby with coskewness (between two random variables) I mean $$\frac{\operatorname{E} \left[(X - \operatorname{E}[X])^2(Y - \operatorname{E}[Y])\right]}{\sigma_X^2 \sigma_Y}.$$ And with cokurtosis I mean both, the asymmetric $$E= {\operatorname{E}{\big[(X - \operatorname{E}[X])^3(Y - \operatorname{E}[Y])\big]} \over \sigma_X^3 \sigma_Y}$$ as well as the symetric cokurtosis $${\operatorname{E}{\big[(X - \operatorname{E}[X])^2(Y - \operatorname{E}[Y])^2\big]} \over \sigma_X^2 \sigma_Y^2}.$$

The purpose of this question is to be able to better explain these concepts to beginners

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    $\begingroup$ (1) Any good answer would first have to explain how kurtosis does not measure peakedness. I believe that point has been made in many threads on this site (but usually in comments). (2) Because "skewness" and "kurtosis" each have multiple inequivalent definitions and the terms "coskewness" and "cokurtosis" are sufficiently unusual, I think it's essential that you include clear definitions of the latter within your post. $\endgroup$
    – whuber
    Commented Apr 22, 2016 at 16:32
  • $\begingroup$ I edited my question. Is that sufficient now? $\endgroup$
    – sheß
    Commented Apr 22, 2016 at 16:46
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    $\begingroup$ Thank you. Unfortunately your edits raise important questions concerning what you mean by skewness and kurtosis in the first place, because your formulae are completely different than those when applied to one-dimensional variables: the skewness definitely is not $E[X^3]$ nor is the kurtosis $E[X^4]$! In fact, if you had a name for $E[XY]$--let's call it "Burt" for convenience--then your coskewness is the Burt of $X^2$ and $Y$ and your cokurtosis is the Burt of $X^2$ and $Y^2$, and so they introduce no new concepts or interpretations! $\endgroup$
    – whuber
    Commented Apr 22, 2016 at 16:49
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    $\begingroup$ I am not convinced that the graphs explain anything to anyone not already aware of the concepts. Graphs 1, 3 and 4 are of the same form and don't distinguish the different concepts. $\endgroup$
    – Nick Cox
    Commented Apr 22, 2016 at 17:57
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    $\begingroup$ @nick, thank you. I agree that three are not the best examples to illustrate these concepts with the help of density plots, just the ones I had at hand. if I come across better ones I'll replace them. Or do you disagree generally that density plots are of use here? $\endgroup$
    – sheß
    Commented Apr 24, 2016 at 20:27

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My case maybe a little different, recently I am running into a similar problem. I am trying to visualize a positive triple coskewness of 3 zero mean random variables, written as $\langle X_1X_2X_3\rangle$.

From the case of positive covariance, we have $\langle X_1X_2\rangle$. The computation of covariance itself does not distinguish between the cases of both $X_1$, $X_2$ are positive, or both of them are negative. As a result, if we plot the $X_1$ against $X_2$ in two dimensional space, we tend to see a line with slope = 1, as the distribution has high occurrence when both are positive or both are negative.

Now going back to the triple coskewness, there are four such degeneracies: $\langle X_1X_2X_3\rangle$, $\langle X_1-X_2-X_3\rangle$, $\langle -X_1-X_2X_3\rangle$, $\langle -X_1X_2-X_3\rangle$ It turns out, if we plot them in a three-dimensional space, with the value of $X_1$, $X_2$ and $X_3$ as the XYZ-axis, we should get a tetrahedron profile for this distribution.

Here is what it looks like when I tried to plot such distribution (with artificially added noise to simulate imperfect coskewness): enter image description here

Please note that this is merely my guess, I don't have much supporting as to whether the above is true. I am also looking for references on it, great if you happen to have anything to share.

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    $\begingroup$ Answers need to be definitive. $\endgroup$ Commented Feb 23, 2018 at 6:06

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