I wanted to know is there any way to calculate skewness and kurtosis of a set as a combination of skewness and kurtosis of its subsets ?


For example if my data is,

X = {1,2,5,7,8,9,0,10}
A = {1,2,5}
B = {7,8,9}
C = {0,10}

Is there any way to express the skewness and kurtosis of X as a combination of skewness and kurtosis of A,B and C ?

  • 2
    $\begingroup$ Short answer: strictly, no since the skewness of the combined sample is not only a function of the skewness of the subsamples (similarly for kurtosis). Longer answer: yes, if you add in other information. If you have all of the lower order moments, its possible. $\endgroup$
    – Glen_b
    Commented Feb 18, 2015 at 7:08

1 Answer 1


Two disjoint sets with $n_1$ and $n_2$ observations in them. The full set has $$n=n_1+n_2$$

  • The first central moment (mean):


  • The second central moment (variance) can be computed using central to non-central moment relationship: $$\sigma^2=\frac{\sum_{i=1}^nx_i^2}{n}-\mu^2=\dots$$

Use the same approach as for the mean to compute full set's non-central moment from that of its subsets: $$\sigma^2=\frac{(\sigma^2_1+\mu^2_1)n_1+(\sigma^2_2+\mu^2_2)n_2}{n_1+n_2}-\mu^2$$

  • For skewness use third non-central and central moment relationship to obtain 3rd central moment as follows: $$\gamma=\frac{1}{n}\left(\sum_ix_i^3-3\mu\sum_ix_i^2\right)+2\mu^3$$

  • The same idea works for kurtosis

This Mathematica page can be helpful

  • 2
    $\begingroup$ Are you using $\mu^3$ to mean two different things in that last equation? Normally it would represent the cube of the mean. Did you mean to write $\mu_3$ on the left hand side? I'm also concerned about the use of $\mu,\sigma$ etc for sample quantities without clearly differentiating this from the usual convention (that Greek letters are for parameters, not sample statistics), since it seems to encourage a common cause of student confusion. $\endgroup$
    – Glen_b
    Commented Feb 18, 2015 at 7:00
  • $\begingroup$ Thanks to both of you. I think the link which I am currently trying to understand uses the same approach of finding lower order moments to achieve the same. Can you comment on approach documented here ? Is it the same or requires some modifications ? $\endgroup$
    – Tusharshar
    Commented Feb 18, 2015 at 7:59
  • $\begingroup$ The parallel algorithm formulas should work. You can apply them recursively to incorporate several such groups. $\endgroup$
    – Glen_b
    Commented Feb 18, 2015 at 11:21
  • $\begingroup$ Yes that algorithm is similar in terms of using lower order moments to compute higher $\endgroup$
    – Aksakal
    Commented Feb 18, 2015 at 12:15

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