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I am working on a problem that I think it would be useful to express kurtosis in terms of skew (if possible) to establish some restrictions on the distributions I am handling. I was wondering if there is a way to do that?

EDIT1:

Sorry for being unclear with my question, the problem is that I want to figure out the condition that the skew of a sample need s to be in order to have sample kurtosis greater than 3

EDIT2:

Base on the answers I've received. It appears that I need additional constraints such as a defined distribution.

To further refine my problem, then, ff the sample in question is Johnson-SU distribution, is there a way to figure out the condition that the skew of this sample needs to be in order to have sample kurtosis greater than 3?

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    $\begingroup$ It's hard to figure out what you're asking for. Might it be related to en.wikipedia.org/wiki/Pearson_distribution? Perhaps you could clarify your question by explaining more about the context and background of your problem. $\endgroup$ – whuber Jun 14 '17 at 17:56
  • $\begingroup$ @whuber sorry for being unclear. I've revised my problem to add some clarity to my problem. see edit $\endgroup$ – syang Jun 14 '17 at 19:15
  • $\begingroup$ unless you have other constraints, no it's impossible. $\endgroup$ – Aksakal Jun 14 '17 at 19:20
  • $\begingroup$ @Aksakal based on the answers I've gotten i've included a distribution constraint to reformulate my problem. see edit2 $\endgroup$ – syang Jun 14 '17 at 19:50
  • $\begingroup$ Across all distributions there's no bound on skewness for $\beta_2>3$, but there would be if you had $\beta_2<3$ (from the usual bound that relates the two across all distributions; specifically $\gamma_1^2+1\leq \beta_2$) $\endgroup$ – Glen_b Jun 15 '17 at 5:30
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In general, no. However, if you are able to assume a family of distributions, then the third and fourth moments can be expressed as a function of their parameters (or as constants). For instance, a normal distribution will always have 0 skewness and 3 kurtosis. A pearson distribution has a 1-1 parameter correspondance with skewness and kurtosis, so each respectively could be set to any value irrespective of the other. A gamma distribution, however, is a two parameter family like the normal, byt with scale ($\theta$) and shape ($k$) parameters. The kurtosis and skewness are $k/6-3$ and $2/\sqrt{k}$, so doing some algebra you could express one in terms of the other and vice versa. So you have to refine the problem slightly to get yes or no answers as they apply.

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  • $\begingroup$ I have revised the question to include the distribution I am handling. See EDIT2 $\endgroup$ – syang Jun 14 '17 at 19:49
  • $\begingroup$ Thank you for the edits, syang. But that last edit does not change the validity or usefulness of this answer by AdamO, which lays out the issues and explains how you can go about finding relationships between skewness and kurtosis generally. Thus it remains a good answer even to the modified question. $\endgroup$ – whuber Jun 14 '17 at 20:18

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