We know that a distribution with zero Skewness are symmetric. A quick Google search or looking up in textbooks says that

Symmetric distributions are distributions where the left side mirrors the right side.

In other words, they are symmetric about something. But what are they symmetric about? Their mean? Median? Mode? There probably would be no confusion if this was specified. But if someone talks about (or perhaps, if you're asked in an exam about) a symmetric distribution, what should we assume as the default?

Context: This MCQ in a test I gave --

A distribution has zero skewness if it is:

(a) Symmetrical about its mode

(b) Symmetrical about its mean

(c) Symmetrical about its median

  • $\begingroup$ 1. If the left side mirrors the right side, and you choose a random value from the distribution, what is the probability that it falls in the left side vs. the right side? How does this relate to the mean / median / mode? 2. Is the Cauchy distribution symmetric? It doesn't have a mean. 3. Is a distribution shaped like a "U" on an arbitrary interval $[a,b]$ symmetric? Where is its mode (trick question)? $\endgroup$ – jbowman May 20 '19 at 3:33
  • $\begingroup$ Also note that a distribution has zero skewness (assuming it has a third moment) if it is symmetric. You should be able to see that "symmetric" is all that is required. Thus, all three statements in the context question are true. $\endgroup$ – jbowman May 20 '19 at 3:39
  • $\begingroup$ For posts on symmetry and medians see stats.stackexchange.com/search?q=symmetric+distribution+median; for symmetry and means, stats.stackexchange.com/search?q=symmetric+distribution+mean; and for symmetry and mode, stats.stackexchange.com/search?q=symmetric+distribution+mode. For definitions and characterizations of symmetry, stats.stackexchange.com/…. $\endgroup$ – whuber May 20 '19 at 12:48
  • $\begingroup$ Although it's explained in many different places, this thread lacks a signal that skewness can be measured in many different ways, e.g. using (mean $-$ median) / SD or L-moments as well as the definition discussed in two answers so far, as a dimensionless ratio based on third and second moments. $\endgroup$ – Nick Cox May 28 '19 at 6:46

We know that a distribution with zero skewness are symmetric...

Actually, that is not correct --- symmetry implies zero skewness (assuming the coefficient of skewness exists), but zero skewness does not imply symmetry. It is possible to construct non-symmetric distributions which have zero skewness.

[For symmetric distributions] ...what are they symmetric about? Their mean? Median? Mode?

A symmetric distribution will always be symmetric about its median, which will also be equal to the mean (assuming this exists). If the distribution is unimodal then the mode will also fall at this point, but if the distribution is multimodal then the mode might occur elsewhere. In the multiple-choice question you give, the correct answer is (c).


"We know that a distribution with zero Skewness are symmetric." This is not the case. A symmetric distribution has zero skewness, but a distribution can have zero skewness and be asymmetric.

The skewness of a random variable $X$ is defined as $$\gamma_1 = \mathrm{E}\left[\left(\frac{X-\mu}{\sigma}\right)^3\right],$$ where $\mu=\mathrm{E}[X]$ and $\sigma = \sqrt{\mathrm{E}[(X - \mu)^2]}$. Notice, it's the first odd central moment of the distribution, normalized to the variance (the variance is the first even central moment).

More terminology: a distribution's moments are defined by $$E[X^n] = \int x^n f(x) \mathrm{d}\,x$$ for $f$ the probability density function of the random variable $X$. A central moment is one where the mean has been shifted away, that is $$E[(X-\mu)^n] = \int (x-\mu)^n f(x) \mathrm{d}\,x.$$ A moment is odd or even depending on if $n$ is odd or even.

Now, the mean is the first odd moment of the distribution, right? Crucially, if a distribution is even as a function about a point, then that point has to be the function's mean and median. Why is that? Well, if we integrate an odd function on an interval that is symmetric about the point the function is odd across, then we get zero. If $f$ is even about some point of symmetry $x_s$, then the quantity $(x-x_s)f(x)$ will be odd about that point. That is enough to prove that $x_s$ is the mean of the distribution (algebra left for the reader).

Showing that the median of a symmetric distribution is at the point of symmetry is fairly straightforward - the definition of the median is that half of the probability is on one side of the point, half of the other. If a function is symmetric then the integral of the function on one side of the point of symmetry has to be the same as the integral on the other (assuming the integration regions are symmetric, to).

Now, showing that the point of symmetry is not necessarily the mode is best done with an example. Consider the random variable with the pdf $$f(x) = \frac{1}{2\sqrt{2\pi}} \left(e^{-(x+2)^2/2} + e^{-(x-2)^2/2}\right).$$ The distribution is symmetric about $x=0$, but the distribution has a minimum at $x=0$, not a maximum. So, you know that the point of symmetry is a minimum or maximum, because its derivative has to vanish there (why?), but it could be a local min or local max, instead of a global max.

Constructing a distribution with vanishing skewness that is asymmetric would require a little more work. Start with the standard normal distribution $$f_N(x) = e^{-(x-\mu)^2/2\sigma^2}.$$ Now, perturb it by multiplying by (1+ax^2). You'll find that to normalize the new pdf you need to divide it by $$N_{\mathrm{new}} \sqrt{2\pi}\sigma(a \sigma^2 + a\mu^2 + 1),$$ and the new mean is $$\mu_{\mathrm{new}} = \mu \frac{3 a \sigma^2 + a\mu^2 + 1}{a\sigma^2 + a\mu^2 + 1}.$$ If you compute the third central moment you'll find that you can make it vanish when \begin{align} a & = 0 \text{ or} \\ a & = \frac{3}{\mu^2 - 3\sigma^2}. \end{align}

Now, we need $a\ge0$ for $f$ to be positive semi-definite, so the existence of a real solution will depend on whether $\mu > \sqrt{3}\sigma$ or not. The $a=0$ solution is the trivial one where the distribution is symmetric about the mean, so it doesn't pass the test of showing an asymmetric distribution with vanishing skewness.

Appendix: A function is even about a point $x_s$ if it satisfies $$f([x-x_s] + x_s) = f(-[x-x_s]+x_s)$$ and it is odd about $x_s$ if $$f([x-x_s] + x_s) = -f(-[x-x_s]+x_s).$$

  • $\begingroup$ Welcome to our site. The example involving the mode doesn't work: $0$ is still the unique mode, where the density is maximal. For analyses that are fully general, covering cases where a PDF does not exist, please visit the duplicate thread. For examples of asymmetric zero-skewness distributions see stats.stackexchange.com/questions/24853. Finally, note there is more than one quantitative way to express skewness: see the Wikipedia article. $\endgroup$ – whuber May 26 '19 at 19:31
  • $\begingroup$ @whuber Example fixed, thanks. $\endgroup$ – Sean Lake May 28 '19 at 3:41
  • $\begingroup$ Thank you (+1). One little nit-pick: even for continuous symmetric variables it's not necessarily the case that their pdf will be differentiable at their center of symmetry. What you can defensibly assert is that the center of symmetry will always be a critical point. I'm raising this issue only because so many visitors to our site (including many respondents) either neglect to examine all critical points or ignore critical points that are not zeros of a derivative (especially in maximum likelihood problems). $\endgroup$ – whuber May 28 '19 at 12:36

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