5
$\begingroup$

Inspired by this answer, I have following question: Is it enough to know just skewness and kurtosis in order to determine distribution that data comes from? Is there any theorem that implies this? Moreover, can we replace skewness and kurtosis by any other pair of moments (for example expected value and variance)?

$\endgroup$
  • 6
    $\begingroup$ If this were true, then all distributions could be continuously parameterized with four real numbers (the mean, variance, skewness, and kurtosis). How, then, could we contemplate even discrete distributions defined on six or more values, which require five or more real parameters? $\endgroup$ – whuber Apr 22 '13 at 13:03
  • $\begingroup$ They require 11 parameters: 6 locations and 5 probabilities. $\endgroup$ – StasK Apr 23 '13 at 0:28
8
$\begingroup$

No it's not enough. These are only the third and fourth standardized moments, & distributions may differ in higher-order moments. Note also that some moments may not exist.

$\endgroup$
  • $\begingroup$ So is it correct to determine type of distribution with only skewness and kurtusis as answered here ? $\endgroup$ – Miroslav Sabo Apr 22 '13 at 12:05
  • 5
    $\begingroup$ "Correct" is not the right way to look at it. It's one way (of many) to decide between distributions that are more-or-less consistent with the data and ones that clearly aren't. There is literally no way in which to decide what the distribution that generated the data was, if you don't have the entire population. $\endgroup$ – Glen_b -Reinstate Monica Apr 22 '13 at 12:17
  • 1
    $\begingroup$ It's incorrect to think you're determining the type of distribution from skew & kurtosis rather than comparing sample estimates of skew & kurtosis to theoretical values for particular distributions & using that as a guide to assess which distributions might adequately model your data generating process. $\endgroup$ – Scortchi - Reinstate Monica Apr 22 '13 at 12:19
5
$\begingroup$

The general answer is no, they are just moments. To keep it simple: Does the mean (first moment) determine the type of distribution ?

This works only if you are working on a certain family of distribution. For example you can define a Gaussian with only the mean and the variance. This is what happened in your example, they are fitting models.

On finite interval, the knowledge of all the moments define a unique distribution if and only if the Hankel matrices $H_n$ are all positive, where $H_{n i,j} =m_{i+j} $ and $m_i$ the ith moment.

On an infinite interval, there is some condition too but no equivalence. For example, the log-normal law is not defined by its moments.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.