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What is skewness of a distribution?

I ask it why any particular indices seem indecisive about symmetry, and in some case also about asymmetry.

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    $\begingroup$ Interestingly enough, we don't seem to have a question on this simple question. This one comes close: What does positive skew show? However the only answer so far is misleading (claiming that positive skewness means that the mean is larger than the median), and just having downvoted it, I cannot suggest that question as a duplicate. $\endgroup$ – Stephan Kolassa May 12 '16 at 10:57
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    $\begingroup$ There is a fundamental paper discussing such concepts: jstor.org/stable/4615828?seq=1#page_scan_tab_contents WEhen(if I get time I will try to write an answer based on that! $\endgroup$ – kjetil b halvorsen May 12 '16 at 11:44
  • $\begingroup$ A skewed distribution is one that is not symmetric. $\endgroup$ – whuber May 12 '16 at 13:38
  • $\begingroup$ The main problem is that asymmetry (absent additional restrictions, such as to some class of distributions) does not admit a partial ordering. If you try to measure amount&direction of skewness by a single number then you don't capture it properly (e.g. you'll end up with non-symmetric distributions which your measure will assign zero skewness). For some discussion on the last sentence of the question, see here. ... (ctd) $\endgroup$ – Glen_b -Reinstate Monica May 17 '16 at 23:54
  • $\begingroup$ ctd ... Some relevant discussion also in parts of this answer. This answer offers some links to other discussions. There are many other useful discussions on site. $\endgroup$ – Glen_b -Reinstate Monica May 17 '16 at 23:57
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Skewness is related to the symmetry of a distribution.

Note that I do not write that "skewness measures symmetry" or some such. The specific relationship between symmetry and skewness is a bit complicated.

A symmetric distribution will have zero skewness, for usual definitions of skewness. (Yes, there are multiple ones.) For instance, in Pearson's moment skewness, the third power in the formula implies that probability masses to the left and the right of the mean cancel out.

However, the converse is not true. You can easily create distributions that are not symmetric but whose Pearson's moment skewness is zero - we just need the densities to cancel out. In fact, you can do this for unimodal distributions as well. The same applies to other skewness measures, like Pearson's mode skewness or median skewness.

However, for practical purposes zero skewness is usually treated as equivalent to symmetry, and unless you purposely create a pathological example, a zero skew distribution will usually be close enough to symmetry that you will be fine.

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  • $\begingroup$ Thaks you for your answer ! However I understood that "The specific relationship between symmetry and skewness is a bit complicated" but we have a formal precise and unambiguous definition of symmetry in probability and statistics ? Or no ? $\endgroup$ – markowitz May 12 '16 at 12:36
  • $\begingroup$ We do. See Definition 1 in the paper I linked to. $\endgroup$ – Stephan Kolassa May 12 '16 at 12:40
  • $\begingroup$ This answer is a very strong argument that stats.stackexchange.com/questions/2899 is a duplicate. $\endgroup$ – whuber May 12 '16 at 13:39
  • $\begingroup$ The paper talking about location and skewness in term of "comparable" distribution. The definition is always related to two dinstribution, and the symmetry isn't definite in absolute sense ... or at least it seem to me. Furthermore the definition is complicated. Maybe this simpler definition is possible: if F(a-x)=1-F(a+x) for all x, and one perticular value a (median?) , where F() is the CDF, then the distribution is symmetric. What do you think ? $\endgroup$ – markowitz May 13 '16 at 7:47
  • $\begingroup$ I don't see how the definition is related to two distributions: "$X$ is symmetrically distributed if there is a $\mu$ such that $X-\mu$ and $-(X-\mu)$ are identically distributed." In the case of an absolutely continuous distribution, this seems to boil down to the definition you propose (but it's a bit more general, since it also works for not absolutely continuous distributions). $\endgroup$ – Stephan Kolassa May 13 '16 at 7:57

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