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I am trying to get some ideas on how to test for an implicit relationship, if any, between variance and skewness. That is, given a very large data set (e.g 90 years of monthly returns), is there a way to generally test if skewness is likely to increase with increasing variance or vice versa?

I will be very grateful for any ideas or suggestions regarding this topic.

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    $\begingroup$ In general there can be no such relations. Consider the following two examples. Example 1: Let $X = \pm n$ with probability 1/2 each. What is the variance? (It will vary with $n$.) What is the skewness? (It will be constant.) Example 2: Let $X = -n$ with probability $p = 1 / 3n^2$, $X = 2n$ with probability $p/2$ and $X = 0$ otherwise. What is the variance? (It will be constant.) What is the skewness? (It will vary with $n$.) Conclusion? $\endgroup$
    – cardinal
    Commented Oct 9, 2012 at 23:49
  • $\begingroup$ @cardinal That also depends on how skewness is defined. $\endgroup$
    – user10525
    Commented Oct 19, 2012 at 13:19
  • $\begingroup$ I`m performing a downside risk capm research for the Brazilian market. (50 IBOVESPA COMPANIES, daily returns, 7 years period) After providing proofs that returns distribution is not normally distributed (komolgorov-smirnov test), I found a statistical and significant relationship between variance a skewness of returns. Further control test are required, but it could mean that the relationship between returns and skewness ( I tested it too) may be partly explained by variance. Any comments would be appreciated. GD $\endgroup$
    – user37331
    Commented Jan 15, 2014 at 7:26

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In general it cannot. Take normal distribution as an example. Variance can take any positive value but skewness is still zero.

And there are other similar questions:

Does skewness predict variance? Or does variance predict kurtosis (since variance is second and kurtosis is fourth moment)? Do answers depend on type of distribution?

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    $\begingroup$ While variance and skewness are unrelated in the normal distribution by definition, that does not mean this is always the case... $\endgroup$
    – analystic
    Commented Oct 10, 2012 at 6:21

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