What's the definition of a symmetric distribution? Someone told me that a random variable $X$ came from a symmetric distribution if and only if $X$ and $-X$ has the same distribution. But I think this definition is partly true. Because I can present a counterexample $X\sim N(\mu,\sigma^{2})$ and $\mu\neq0$. Obviously, it has a symmetric distribution, but $X$ and $-X$ have different distribution! Am I right? Do you guys ever think about this question? What's the exact definition of symmetric distribution?

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    $\begingroup$ When you say, a "distribution is symmetric", you have to specify with respect to what point is symmetric. In the case of the normal distribution you present, the symmetry is given around $\mu$. In this case $X-\mu$ and $-(X-\mu)$ have the same distribution. In terms of the density this can be expressed as: $f$ is symmetric about $\mu$ if $f(\mu-x)=f(\mu+x)$. BTW, it is good manners to accept answers when you are satisfied with one of them. $\endgroup$ – user10525 May 23 '12 at 10:43
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    $\begingroup$ Yes, we guys have thought about this question. Symmetric generally means symmetric about $0$, and, to forestall further counterexamples, the claim about distributions being symmetric is not something that is true about the cumulative probability distribution function. Your "counterexample" has symmetry about the point $\mu \neq 0$, not about the point $0$. $\endgroup$ – Dilip Sarwate May 23 '12 at 10:44
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    $\begingroup$ @Dilip When a definition depends on one way of describing something, but that definition can be shown to be an intrinsic property of that something, then it makes no sense to apply the definition to a different form of description. In this case, symmetry is a property of a distribution, but that does not imply that all descriptions of that distribution (including the PDF and CDF) must be "symmetric" in the same ways. By applying the symmetry of the PDF to the CDF, your comment confuses the question rather than clarifying it. $\endgroup$ – whuber May 23 '12 at 15:58
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    $\begingroup$ shijing, @Procrastinator has observed that you have asked many questions without accepting any answers. That suggests you may be unfamiliar with how this site works. To clear up any misunderstanding, would you please read the relevant part of our FAQ all the way through? It will take only a couple of minutes and following its guidance will enhance the value of our site to you. $\endgroup$ – whuber May 23 '12 at 16:06
  • $\begingroup$ @whuber The CDF is one of the few descriptions in which the word distribution actually occurs in the name, and I was trying to clarify that the symmetry property did not hold for the CDF. $\endgroup$ – Dilip Sarwate May 23 '12 at 16:17

Briefly: $X$ is symmetric when $X$ and $2a-X$ have the same distribution for some real number $a$. But arriving at this in a fully justified manner requires some digression and generalizations, because it raises many implicit questions: why this definition of "symmetric"? Can there be other kinds of symmetries? What is the relationship between a distribution and its symmetries, and conversely, what is the relationship between a "symmetry" and those distributions that might have that symmetry?

The symmetries in question are reflections of the real line. All are of the form

$$x \to 2a-x$$

for some constant $a$.

So, suppose $X$ has this symmetry for at least one $a$. Then the symmetry implies

$$\Pr[X \ge a] = \Pr[2a-X \ge a] = \Pr[X \le a]$$

showing that $a$ is a median of $X$. Similarly, if $X$ has an expectation, then it immediately follows that $a = E[X]$. Thus we usually can pin down $a$ easily. Even if not, $a$ (and therefore the symmetry itself) is still uniquely determined (if it exists at all).

To see this, let $b$ be any center of symmetry. Then applying both symmetries we see that $X$ is invariant under the translation $x \to x + 2(b-a)$. If $b-a \ne 0$, the distribution of $X$ must have a period of $b-a$, which is impossible because the total probability of a periodic distribution is either $0$ or infinite. Thus $b-a=0$, showing that $a$ is unique.

More generally, when $G$ is a group acting faithfully on the real line (and by extension on all its Borel subsets), we could say that a distribution $X$ is "symmetric" (with respect to $G$) when

$$\Pr[X \in E] = \Pr[X \in E^g]$$

for all measurable sets $E$ and elements $g \in G$, where $E^g$ denotes the image of $E$ under the action of $g$.

As an example, let $G$ still be a group of order $2$, but now let its action be to take the reciprocal of a real number (and let it fix $0$). The standard lognormal distribution is symmetric with respect to this group. This example can be understood as an instance of a reflection symmetry where a nonlinear re-expression of the coordinates has taken place. This suggests focusing on transformations that respect the "structure" of the real line. The structure essential to probability must be related to Borel sets and Lebesgue measure, both of which can be defined in terms of (Euclidean) distance between two points.

A distance-preserving map is, by definition, an isometry. It is well known (and easy, albeit a little involved, to demonstrate) that all isometries of the real line are generated by reflections. Whence, when it is understand that "symmetric" means symmetric with respect to some group of isometries, the group must be generated by at most one reflection and we have seen that reflection is uniquely determined by any symmetric distribution with respect to it. In this sense, the preceding analysis is exhaustive and justifies the usual terminology of "symmetric" distributions.

Incidentally, a host of multivariate examples of distributions invariant under groups of isometries is afforded by considering "spherical" distributions. These are invariant under all rotations (relative to some fixed center). These generalize the one-dimensional case: the "rotations" of the real line are just the reflections.

Finally, it is worth pointing out that a standard construction--averaging over the group--gives a way to produce loads of symmetric distributions. In the case of the real line, let $G$ be generated by the reflection about a point $a$, so that it consists of the identity element $e$ and this reflection, $g$. Let $X$ be any distribution. Define the distribution $Y$ by setting

$${\Pr}_Y[E] = \frac{1}{|G|}\sum_{g \in G} {\Pr}_X[E^g] = ({\Pr}_X[E] + {\Pr}_X[E^g])/2$$

for all Borel sets $E$. This is manifestly symmetric and it's easy to check that it remains a distribution (all probabilities remain nonnegative and the total probability is $1$).


Illustrating the group averaging process, the PDF of a symmetrized Gamma distribution (centered at $a=2$) is shown in gold. The original Gamma is in blue and its reflection is in red.

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    $\begingroup$ (+1) I would like to add that, in the multivariate setting, the definition of symmetry is not unique. In this book there are 8 possible definitions of symmetric multivariate distributions. $\endgroup$ – user10525 May 23 '12 at 16:09
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    $\begingroup$ @Procrastinator I'm curious about what you might mean by "not unique." AFAIK, anything justifying the name "symmetry" ultimately refers to a group action on a space. It would be interesting to see what different kinds of actions statisticians have found useful. Because that book is out of print and not available on the Web, could you give a quick example of two really different kinds of symmetry considered in that book? $\endgroup$ – whuber May 23 '12 at 16:13
  • $\begingroup$ Your intuition is correct, this is related to statistical features: Central symmetry ${\bf X}-\mu\stackrel{d}{=}-({\bf X}-\mu)$; Spherical symmetry $X-\mu\stackrel{d}{=}{\bf O}({\bf X}-\mu)$ for all orthogonal matrix ${\bf O}$. I cannot recall the rest, but I will try to borrow the book on these days. In this link you can find some of them. $\endgroup$ – user10525 May 23 '12 at 16:24
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    $\begingroup$ @Procrastinator Thanks. Note that the two examples you offer are both special cases of the general definition I have supplied: the central symmetry generates a two-element group of isometries and the spherical symmetries are also a subgroup of all isometries. The "elliptical symmetry" in the link is a spherical symmetry after an affine transformation, and so exemplifies the phenomenon I pointed to with the lognormal example. The "angular symmetries" again form a group of isometries. The "half-space symmetry" [sic] is not a symmetry, but allows for discrete departures therefrom: that's new. $\endgroup$ – whuber May 23 '12 at 16:36

The answer will depend on what you mean by symmetry. In physics the notion of symmetry is fundamental and has become very general. Symmetry is any operation that leaves the system unchanged. In the case of a probability distribution this could be translated to any operation $X \to X'$ that returns the same probability $P(X) = P(X')$.

In the simple case of the first example you are referring to the reflection symmetry about the maximum. If the distribution were sinusoidal then you could have the condition $X \to X + \lambda$, where $\lambda$ is the wavelength or period. Then $P(X) = P(X + \lambda)$ and would still fit a more general definition of symmetry.


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