# Can a non-symmetrical distribution have the same areas under the PDF in the two sides around the mean?

I was thinking about symmetrical vs. non-symmetrical distributions and I found myself stuck in a thought that I had never thought before. We know that symmetrical distributions like Normal, Laplace, etc. have the same area under the curve around the mean (which in the symmetrical case, is equal to the median).

However, can a non-symmetrical distribution (e.g, GEV https://w.wiki/AfKx) also have this property, if the area under the tail is so that it equals the area of the main lobe?

Can the parameters of such a distribution be found analytically?

• It absolutely can. Examples are actually quite easy to construct, especially in the simple case of a discrete random variable. Try putting probabilities on the numbers from 1 to 5 and see whether you can come up with an example. And while you are at it, please add the self-study tag to the question and read its wiki. Thank you! Commented Jul 14 at 15:08
• Take a look at math.stackexchange.com/questions/4706494/… for one way to construct distributions that are asymmetrical and have the same mean and median.
– JimB
Commented Jul 15 at 3:18
• Every such distribution function $F$ can be obtained by selecting a median $m,$ choosing a distribution $G$ supported on $[0,\infty)$ and a different distribution $H$ supported on $(-\infty, 0],$ and setting $F(x)=(G(x-m)+H(x-m))/2$ for all $x.$ There are myriad examples of such $G$ and $H.$
– whuber
Commented Jul 15 at 12:56
• 0,0,0,1,1,1,4 has mean and median 1 and is asymmetric. This is a trivial example, but that's the point too. More worrying are the textbooks that imply or even state that mean and median always differ in an asymmetric distribution. Anyone who objects to this example has contrived can fall back on binomial examples as in the answer from @Thomas Lumley. Commented Jul 16 at 11:21

Your title question is identical to asking "Can an asymmetric distribution have mean equal to median" to which the answer is "yes" and many examples are to be found on site already. It's a simple matter to construct as many as you like

In the case of the GEV, the Gumbel and Frechet cannot, but the reversed Weibull can; you need the Weibull shape parameter to be about $$k=3.4395431137$$ which is a function of the value of $$\xi$$ in the GEV. I'm referring to the Wikipedia parameterizations there. You can just take the expressions for the mean and median of the GEV for the $$\xi<0$$ case, or indeed for the Weibull itself, in either case from the sidebar of their wikipedia page and equate the expressions for the mean and median to get a nonlinear equation to solve (which is easy enough in software, or by standard root-finding approaches 'by hand' if you have the functions available -- or even by a little trial and error if you have a minute or two).

This example isn't especially satisfying because the Weibull is hard to visually distinguish from symmetry in this region but it is never symmetric.

A more satisfying example: As noted before I wanted a continuous example with continuous density that was not multimodal (albeit its mode is the whole interval from 0 to m). I constructed this by taking two non-increasing densities on non-negative values, each with mean 1 and and with their heights at $$0$$ matched up. I then flipped one of them about $$x=0$$, and took a 50-50 mixture of the two producing the desired result that mean and median are both $$0$$.

The triangular was simple enough to set up, the other is itself a mix of uniform and a shifted exponential, designed so the heights meet up at their join point ($$m$$). This was made with just enough free parameters that I could both set the mean to 1 and match heights with the other density. This sort of approach with other choices of the leftmost and rightmost density (and judicious choice of which one to shift-and-combine with a uniform) should work for a wide variety of examples.

The function of the uniform piece is simply to add a degree of freedom to be able to get continuity of the combined density when they wouldn't otherwise line up at the median. If this is not required, you can use this trick with pretty much any two distributions with the same mean but different shapes by taking a 50-50 mixture after flipping one around 0.

As noted on the diagram, the triangular part is on $$(-3,0]$$ and has height $$\frac13$$ (making its area $$\frac12$$). The uniform is on $$(0,m]$$ where $$m=(3-\sqrt{3})/2$$ and has a mixing proportion of $$\frac12-\frac{\sqrt{3}}{6}$$, giving it a height of $$\frac13$$, and the shifted exponential has mixing proportion $$\frac{\sqrt{3}}{6}$$, rate parameter $$\lambda =\frac{2\sqrt{3}}{3}$$ and shift parameter $$m$$, also giving it height $$\frac13$$. It's a simple matter to confirm that the means and medians are then such that the entire distribution has mean $$0$$ and median $$0$$.

This distribution is relatively easy to simulate values from which allowed a useful final reasonableness check on the working, both for the density function actually meeting up and having the desired shape and of the values for mean and median.

(This seems to be simple enough that finding the constraints to make mean and median both $$0$$ could be set as a student exercise in probability, but there's enough in it that laxness of notation or careless work would lead to difficulties.)

If an example with a mode at a single point was required, a mixture of this distribution with a symmetric unimodal distribution with mean, median and mode all at zero should work (e.g. a normal, or a Laplace, or a shifted and scaled symmetric beta for example). However, I'm content with this one; I think it's both clear and compelling.

[I have a second one in mind I might make, though, using a different approach.]

• +1 An intuitive way to think about Glen_b's opening sentence is to consider the shape of one side of an asymmetric distribution and (a) imagine shrinking its area under the curve towards zero (perhaps while retaining an infinitely log tail, perhaps not), and (b) imagine increasing its area under the curve towards one (same 'perhaps' here). Now recognize that you could (c) do likewise for the other side of the distribution, and (d) you could shrink or grow each side until its area was exactly one half. Commented Jul 14 at 15:54
• @Glen_b Thanks for your response. Yes, other way of writing my question would be in terms of the mean equaling the median. Based on yours and others reply, then we can say that the following definition in the investopedia website is wrong, right? "Asymmetrical distribution is a situation in which the values of variables occur at irregular frequencies and the mean, median, and mode occur at different points. " investopedia.com/terms/a/asymmetrical-distribution.asp (since the mode and the median of an assymetric distribution can indeed be equal) Commented Jul 14 at 21:14
• Yes that's certainly wrong. Asymmetry can occur when mean median and mode are all equal and with the moment skewness being zero ... I have examples. Indeed we could pile on other single-number measures of asymmetry as well, though construction becomes harder. To describe asymmetry in a way that avoids this requires a function. Investopedia has more things wrong. Avoid relying on it heavily Commented Jul 14 at 22:49
• Douglas, I have added a relatively nice example - continuous, very clearly asymmetric, but still with mean and median at 0. Commented Jul 15 at 5:22
• @Douglas a symmetric distribution has identical mean, median, and (if unimodal) mode. An asymmetric distribution can have all of them different, and loosely speaking probably has all of them different. But it's not impossible for them to be equal :) Commented Jul 15 at 17:05

A nice example is that any Binomial distribution with integer mean has median equal to its mean (this is hard to prove in general, but easy in any specific example).