I would like to know how I can measure the degree of symmetry of a bimodal distribution.

Is there any a criterion like for example skewness in the case of unimodal distributions?

  • $\begingroup$ Well, you can compute most skewness measures (such as third-moment-based skewness) for a distribution with more than one mode, however, zero skewness (even for a unimodal distribution) doesn't necessarily indicate symmetry. $\endgroup$ – Glen_b Apr 8 '15 at 10:32
  • $\begingroup$ Are you talking about using data, or proceeding from some specified density/cdf? $\endgroup$ – Glen_b Apr 8 '15 at 10:41
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    $\begingroup$ @AleksandrBlekh If you specify a symmetric distribution then obviously that distribution will be symmetric -- one needn't consult any measure of symmetry for that. The point at issue is that zero skewness (by one of the usual measures of it) is not of itself a guarantee of symmetry. If you add other conditions that guarantee symmetry, then of course symmetry is guaranteed because of those additional conditions. I'm not sure how this is useful. $\endgroup$ – Glen_b Apr 8 '15 at 11:03
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    $\begingroup$ @Aleksandr That plot takes as given that you have symmetry, and given that assumption, estimates a parameter that corresponds to tail heaviness (under an assumed distribution family). Given that symmetry assumption ... you have symmetry, and the points I mentioned in my previous comment apply. $\endgroup$ – Glen_b Apr 8 '15 at 11:16
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    $\begingroup$ @Aleksandr Perhaps you're getting at the situation where you assume a distribution family where a parameter indicates the degree of asymmetry, such that 0 skewness does correspond to symmetry (e.g. a beta parameterized by the difference and average of the usual parameters $\delta=\alpha-\beta$ and $\gamma=(\alpha+\beta)/2$; the first of those new parameters being zero does indicate symmetry). If you do make such an assumption, then of course the assumption would make it work that way. The existence of such families doesn't alter the fact that on its own, zero skewness doesn't imply symmetry. $\endgroup$ – Glen_b Apr 8 '15 at 11:27

By definition, a symmetric random variable $X$ is one for which there is a constant $\mu$ for which $X-\mu$ and $\mu-X$ are identically distributed. In terms of the distribution function $F$ this is equivalent to

$$\eqalign{ F(\mu+x) &= \Pr(X \le \mu+x) = \Pr(X-\mu \le x) \\ &= \Pr(\mu-X \le x) \\ &= \Pr(X-\mu \ge -x) = \Pr(X \ge \mu - x)\\ &= 1 - \Pr(X \lt \mu- x) \\ &= 1 - F(\mu - x) + \Pr(X = \mu-x) }$$

When $F$ is continuous this simplifies to

$$F(\mu + x) + F(\mu-x) = 1$$

for all $x$ and thence (via differentiation), when $F$ is absolutely continuous with distribution function $f$,

$$f(\mu + x) = f(\mu - x)$$

for all $x$. (It's not hard to see that any of these equations uniquely determine $\mu$.)

This provides a general, flexible procedure to test for symmetry, based on any method of comparing two distributions. (Many such methods exist, ranging from comparing basic properties like moments through relative entropy, KL distances, and so on.) Specifically, take any non-negative function $\delta$ where


is intended to measure the "distance" or "dissimilarity" between distributions $F$ and $G$. All we ask of $\delta$ (besides being nonnegative) is that $\delta(F,G) = 0$ if and only if $F=G$.

For any constant $\mu$ define $F_\mu(x) = F(x-\mu)$ and $\check F_\mu(x) = F(\mu-x)$. Then merely take

$$\inf_{\mu}\, (\delta(F_\mu, \check{F}_\mu))$$

as the measure of asymmetry. This measures how close you can make $X-\mu$ and $\mu-X$ appear to be. It will be nonnegative and equal to zero only when $F$ is symmetric. Choose $\delta$ to emphasize those aspects of "closeness" important in your application, such as asymptotic tail behavior or balancing an odd moment.

As a simple example, intended to be applied to bimodal absolutely continuous distributions, let

$$\delta(F,G) = \int (f(x) - g(x))^2 dx.$$

This is the $L^2$ norm of their density functions, depending on the total area between their graphs. The illustration shows the density $f$ for a bimodal distribution at the left, followed by three graphs depicting the region between $f_\mu$ and $\check{f}_\mu$ for values of $\mu$ around the optimum $\mu=1$, where that region is the smallest in the $L^2$ sense:


This shows how the whole process lends itself to exploratory (visual) evaluation: simply make such a plot to superimpose $f_\mu$ and $\check{f}_\mu$ (or the CDFs $F_\mu$ and $\check{F}_\mu$) and vary $\mu$ until the graphs look as "alike" as possible. The visual deviations at this optimal point will not only indicate asymmetry, but they will also show the form of the asymmetry.

  • $\begingroup$ Thanks a lot, the answer is the method I was looking for. $\endgroup$ – alexi Apr 14 '15 at 16:50

While I don't think that there is a single measure of symmetry for a bimodal distribution in general, for a special case of mixture of two normal distributions, perhaps, it is possible to use one or several bi-modality measures and statistical tests. Some mixture modeling software (especially, some R packages) might have some of those measures assessment and tests implemented, so that it might be possible to determine the level of symmetry of a bimodal distribution analytically.


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