I'm trying to quantify the skewness of the distribution of random integer variable, generated in the interval from 1 to 15, with a function that I wrote in C++.

Here are the generated values:

Tested for 5000 elements, with results:

level 1: 2561  level 6: 70   level 11: 4
level 2: 1225  level 7: 44   level 12: 1
level 3: 607   level 8: 17   level 13: 0
level 4: 312   level 9: 9    level 14: 0
level 5: 147   level 10: 3   level 15: 0

From what I observe I can qualify that the distribution has positive skewness, as most of the generated elements (97%) are within the interval 1 to 5.

To quantify the skewness, I'm trying to calculate the Pearson's moment coefficient of skewness using this relation:

enter image description here

where X - random variable, mu - mean, sigma - standard deviation and E - expectation operator.

I understand that I have to subtract the mean from each random variable, divide it by the standard deviation and raise to the 3-rd power, however, I'm having difficulties understanding what is the meaning of the E operator?

Does it mean that I need to simply divide by the total number of values or something else?


Is there an easier way to quantify skewness?

P.S. apologies for the lengthy post I just wanted to show research effort.

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    $\begingroup$ If you're trying to calculate it for a sample, you need to use a calculation for sample skewness. $\endgroup$ – Glen_b Jan 10 '16 at 13:02
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    $\begingroup$ There are several possible estimators. The one used by Excel's SKEW function, for instance, is documented at support.office.com/en-us/article/…. The general situation is briefly discussed in our thread at stats.stackexchange.com/questions/157895. $\endgroup$ – whuber Jan 10 '16 at 14:44
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    $\begingroup$ In usual statistical terminology, you have just one variable from each simulation, with several values or observations. That doesn't affect your question. $\endgroup$ – Nick Cox Jan 10 '16 at 14:53
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    $\begingroup$ Wikipedia mentions three sample versions in its article on skewness (which it calls $b_1, G_1$ and $\frac{m_3}{m_2^{3/2}}$). In large samples it makes no real difference which you use. $\endgroup$ – Glen_b Jan 10 '16 at 22:29
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    $\begingroup$ There's no need to include the diamond when @-notifying me (similarly whuber). It's not part of my username -- it simply indicates we're diamond-moderators. The ♦ distinguishes elected moderators (who gain a few additional abilities along with it) from the ordinary users with moderator privileges (formally, those above 10K reputation, though users at lower reputation contribute in several ways to the moderation of the site). $\endgroup$ – Glen_b Jan 10 '16 at 23:13

You should understand the difference between the parameters and properties of a distribution and the estimators for these parameters and properties. For instance

  • The true mean, $\mu = E[X]$ is the the expected value of a stochastic variable $X$ and can not be calculated exactly.
  • The sample mean, $m = \sum{x_i/n}$ with $x_i$ your observations of $X$ is the usual estimator for $\mu$.

Chapters in text books and whole scientific articles discuss the quality of estimators. For variance

  • The true variance is $\sigma^2 = E[(X-\mu)^2]$
  • The sample variance actually is $\frac{1}{n} \sum{(x_i - m)^2}$, but it has a tendency to be smaller than $\sigma^2$, therefore it is said to be biassed. This is related to the fact that $m$ itself is estimated from the same sample.
  • The usual estimator, $s^2 = \frac{1}{n-1} \sum{(x_i - m)^2}$, does not have this disadvantage.

For skewness

  • The true skewness of the stochastic variable is $\gamma_1 = E[(\frac{X-\mu}{\sigma})^3]$

  • The sample skewness is $\frac{1}{n} \sum(\frac{x_i-m}{s})^3$, but again, it is biassed.

  • The usual estimator is $\frac{n}{(n-1)(n-2)} \sum(\frac{x_i-m}{s})^3$, in which $s$ is of course the square root of the estimator for the variance.

Have fun implementing this. For further discussion, you might consult

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