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I have a confusion about the main statistical indicators for the shape of a distribution. Let me consider first the Fisher index defined as $$\rho=\frac{\frac{1}{n}\sum_{i=1}^n(x_i-\bar x)^3}{\sigma^3},$$ where $\sigma$ is the standard deviation of the statistical distribution and $\bar x$ is the total average. Is it true that

  1. If $\rho>0$ then the distribution is positively asymmetric (it is not true the inverse implication)

  2. If the distribution is symmetric then $\rho=0$ (it is not true the reverse implication). There exists a counterexample about the inverse implication?

  3. If $\rho<0$ then the distribution is negatively asymmetric (it is not true the inverse implication)

Am I correct?

What about the index for the curtosis of a distribution? $$\beta=\frac{\frac{1}{n}\sum_{i=1}^n(x_i-\bar x)^4}{\sigma^4}?$$

  1. If $\beta>3$ then the distribution is leptokurtic (it is not true the inverse implication)

  2. If $\beta=3$ then the distribution is mesokurtic (it is not true the inverse implication)

  3. If $\beta<3$ then the distribution is platikurtic (it is not true the inverse implication)

Could someone tell me if I am correct and clarify my ideas?

kjetil b halvorsen
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user268193
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    These look tautological to me: $\rho$ *defines* whatever "positively/negatively asymmetric" might mean and $\kappa$ *defines* whatever "lepto/platykurtic" might mean. As argued extensively elsewhere on this site, such moment-based statistics indicate exceptionally little about a distribution's shape (because they inherently depend on the frequencies of extreme values). – whuber Sep 28 '21 at 13:39
  • Thank you for the answer. I am trying to understand well these things. So $\rho>0$ implies positively asymmetric (by definition), while $\rho<0$ implies negatively asymmetric (by definition). While $\rho=0$ doesn't imply anything. Is this correct? – user268193 Sep 28 '21 at 13:44
  • The same happens for the Kurtosis index? This means that $\beta=3$ doesn't imply anything? If you could suggest some link in this page in which these thinngs are discussed I would be very greatful. I am new in the statistics world and I really want to learn. – user268193 Sep 28 '21 at 13:46
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    What I am saying is that these statements are not implications; they are *definitions.* They imply little about the shapes of the distributions (except when they achieve extreme values, such as $\kappa=1$). It's easy to find a lot of material about skewness and kurtosis here on CV: [search kurtosis shape](https://stats.stackexchange.com/search?q=kurtosis+shape) and [search skewness shape](https://stats.stackexchange.com/search?q=skewness+shape). – whuber Sep 28 '21 at 13:51
  • This means that a distribution is said negatively asymptotics if $\rho<0$. Therefore if $\rho<0$ then the distribution is negatively asymptotics. On the other hand if it is a definition means that if a distribution is negatively asymmetric tehn $\rho<0$. Therefore a distribution is negatively asymmetric if and only if $\rho<0$. The same should be true for positive asymetry. It is not clear to me how is it possible that $\rho=0$ doesn't imply symmetry. – user268193 Sep 28 '21 at 14:09
  • This is because a distribution can be symmetric, negative asymmetric, positive asymmetric but can be also none of the previous options. So can be asymmetric (not positively and not negatively)? – user268193 Sep 28 '21 at 14:10
  • That's right. See https://stats.stackexchange.com/questions/24853 for examples of the latter (asymmetric but with $\rho=0$). – whuber Sep 28 '21 at 15:02
  • Thank you very much – user268193 Sep 28 '21 at 15:30
  • Sorry! I have another question about the Skweness. I can define the positive/negative asimmetry also trough the Pearson index $a_P=\frac{\bar x-Moda}{\sigma}$? In the sense that I can say that a distribution is positevely (negatively) asymmetric when $a_p>0(<0)$? – user268193 Sep 28 '21 at 15:50

1 Answers1

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While use of skewness and kurtosis as measures of shape can be dubious, there are precise visual and mathematical interpretations for both that provide correct insights and information. These regard tail behavior, rather than shape.

Skewness:

Let $Z = (X - \mu)/\sigma$. Let $Z_3 = Z^3$. Consider the pdf of $Z_3$, call it $p_3(.)$. The skewness of $X$ is the mean of the distribution of $Z_3$, hence it is the "point of balance" of $p_3$.

Now, place a fulcrum at $0.0$ under the graph of $p_3$. If the skewness of $X$ is $>0$, then the graph "falls to the right" of the fulcrum. Thus, the right tail of $X$ is more extreme, in terms of mass and/or extremity, than the left tail, where extremity refers to amplification by the third power. A similar definition holds for skewness $<0$. If skewness is $=0$ it does not imply symmetry, but instead equal leverage of the tails, again as amplified through the third power.

Kurtosis:

Let $Z_4 = Z^4$. Consider the pdf of $Z_4$, call it $p_4(.)$. The kurtosis of $X$ is the mean of the distribution of $Z_4$, hence it is the "point of balance" of $p_4$.

If $X$ has a normal distribution, then this point of balance is exactly 3.0.

So place a fulcrum at 3.0 under the graph of $p_4$. If the kurtosis is $>3$, then the graph falls to the right, indicating that $X$ has greater tail leverage than the normal distribution. Here, "tail" refers to extremity of the $X$ values as amplified through the fourth power.

Note that kurtosis $>3$ tells you nothing about the shape of the peak, or even about the probability within a standard deviation of the mean: The graph falls the the right, not to the left.

If the kurtosis is $<3$, then the graph "falls to the left," and distribution of $X$ has less tail leverage than the normal distribution. This of course tells you nothing about the shape of the "peak." (It is certainly not "flat" - the Beta(.5,1) distribution provides a simple counterexample).

If kurtosis $=3$, then the tail leverage of $X$ (as amplified by the fourth power) is the same as that of the normal distribution, but it certainly does not imply that the distribution is normal, or even bell-shaped.

BigBendRegion
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