# Does mean = median imply that a unimodal distribution is symmetric?

For a unimodal distribution, if mean = median then is it sufficient to say that distribution is symmetric?

Wikipedia says in relationship between mean and median:

"If the distribution is symmetric then the mean is equal to the median and the distribution will have zero skewness. If, in addition, the distribution is unimodal, then the mean = median = mode. This is the case of a coin toss or the series 1,2,3,4,... Note, however, that the converse is not true in general, i.e. zero skewness does not imply that the mean is equal to the median."

However, it is not very straight forward (to me) to glean the information I need. Any help please.

Here is a small counterexample that is not symmetric: -3, -2, 0, 0, 1, 4 is unimodal with mode = median = mean = 0.

Edit: An even smaller example is -2, -1, 0, 0, 3.

If you want to imagine a random variable rather than a sample, take the support as {-2, -1, 0, 3} with probability mass function 0.2 on all of them except for 0 where it is 0.4.

• I believe -2, -1, 0, 0, 3 is the "smallest" discrete counterexample: we need two equal values to create a mode, adding a third distinct value would prevent mean equalling median, and a fourth value can only restore the mean equalling the median by restoring the symmetry. I also suspect these numbers are the "smallest" (close to 0) integers possible, since 3 is the smallest whole number we can write as the sum of two distinct whole numbers. Distinct is vital, as trying -1, -1, 0, 0, 2 would no longer be unimodal! Naturally, we can scale and translate to get a new data set with this property. Nov 24, 2014 at 23:04

This began as a comment but grew too long; I decided to make it into more of an answer.

Alexis' fine answer deals with the immediate question (in short: i. that logically $${A\implies B}$$ doesn't mean $$B\implies A$$; and ii. the reverse statement is actually false in general), and Silverfish gives counterexamples.

I'd like to deal with some additional issues and point out some extensive answers already here which are related to some extent.

1. The statement on the Wikipedia page that you quote is not strictly true either. Consider, for example, the Cauchy distribution, which is certainly symmetric about its median, but which doesn't have a mean. The statement needs a qualifier such as 'provided the mean and skewness are finite'. Even if we reduce it to the weaker statement in the first half of the first sentence, it still needs "provided the mean is finite".

2. Your question partly conflates symmetry with zero skewness (I assume you intend third moment skewness, but a similar discussion could be written for other skewness measures). Having 0 skewness doesn't imply symmetry. The later part of your quote and the section from Wikipedia quoted by Alexis mention this, though the explanation given in the second quote could use some tweaking.

This answer shows that the relationship between third moment skewness and the direction of the relationship between mean and median is weak (third moment skewness and second-Pearson skewness needn't correspond).

Item 1. on this answer gives a discrete counterexample, similar to but different from the one given by Silverfish.

Edit: I finally dug up the unimodal example I was actually looking for earlier.

In this answer I mention the following family:

$$\frac{1}{24}\exp(-x^{1/4}) [1 -\alpha \sin(x^{1/4})]$$

Taking two specific members (say the blue and green densities in the specific example at that linked answer, which have $$\alpha=0$$, and $$\alpha=\frac{_1}{^2}$$ respectively), and flipping one about the x-axis and taking a 50-50 mixture of the two, we would get a unimodal asymmetric density with all odd moments zero: (grey lines show the blue density flipped about the x-axis to make the asymmetry plain)

Whuber gives another example here with zero skewness that's continuous, unimodal and asymmetric. I've reproduced his diagram: which shows the example and the same flipped about the mean (to clearly show the asymmetry) but you should go read the original, which contains a lot of useful information.

[Whuber's answer here gives another asymmetric continuous family of distributions with all the same moments. Doing the same "choose two, flip one and take a 50-50 mixture" trick has the same outcome of asymmetric with all odd moments zero, but I think it doesn't give unimodal results here (though perhaps there are some examples).]

The answer here discusses the relationship between mean, median and mode.

This answer discusses hypothesis tests of symmetry.

No.

If, in addition, the distribution is unimodal, then the mean = median = mode.

In the same way that "If the baby animal is a chicken, then its origin is an egg" does not imply that "If the origin is an egg, then the baby animal is a chicken."

From the same Wikipedia article:

In cases where one tail is long but the other tail is fat, skewness does not obey a simple rule. For example, a zero value indicates that the tails on both sides of the mean balance out, which is the case both for a symmetric distribution, and for asymmetric distributions where the asymmetries even out, such as one tail being long but thin, and the other being short but fat.

• Note that in the unimodal symmetric case, you can have (easy to construct examples) mean = median = point with minimal (0) density! Sep 15, 2015 at 13:37
• @kjetilbhalvorsen, no such examples come to my mind. Could you hint at how to approach this? Jan 25, 2022 at 11:52
• @RichardHardy: I do not understand now what I was thinking, so I will delete the comment! Feb 13, 2022 at 23:38

Interesting and easy to understand examples come from the binomial distribution.

Here are binomial probabilities for 0(1)5 successes in 5 trials when the probability of success is 0.2. It's immediate that the mean is 0.2 $\times$ 5 $=$ 1, which inspection of probabilities confirms as also the median and the (single) mode, but the distribution is clearly not symmetric. There are naturally many other examples of skewed binomials with mean a positive integer.

            1        2
+-------------------+
1 |       0   .32768  |
2 |       1    .4096  |
3 |       2    .2048  |
4 |       3    .0512  |
5 |       4    .0064  |
6 |       5   .00032  |
+-------------------+


Stata code for this display was mata : (0..5)' , binomialp(5, (0..5), 0.2)' and presumably it is as simple or simpler in any statistical software worth mentioning.

As a matter of psychology rather than logic, this example can't be convincingly dismissed as pathological (as in other problems one might discount distributions for which certain moments do not even exist) or as a bizarre or trivial example contrived for the purpose (as for example the invented data described by @Silverfish or 0, 0, 1, 1, 1, 3).

This answer follows the same idea as Glen B, but with some slightly different story and visual examples

The median and the mean are both measures that can be seen as splitting a distribution into two parts that have equal weights on both sides.

For the mean and the median, these weights on both sides are different measures. They consider different absolute partial moments. They are both integrals of the absolute difference $$|x-m|$$ but with different powers. ### Symmetric distributions

For symmetric distributions, these two sides are automatically the same when the split is made in the plane of symmetry.

It works the same for mean as for the median which are integrals of a left side and a right side that become equal if the two sides have the same shape.

So the point of the plane of symmetry is equal to the mean and it is equal to the median. And the median and mean will be equal (but they do not need this symmetry to be equal)

### Asymmetric distributions

For asymmetric distribution, we do not need to have automatically that the dividing plane for the median (giving equal weights of probability on both sides) is also the dividing plane for the mean (giving equal weights of average distance on both sides), and vice-versa.

It is also very typical for asymmetric distributions to have unequal mean and variance.

The only counter-example among common asymmetric distributions that comes to my mind is the binomial distribution where $$np$$ is an integer and $$p \neq 0.5$$ (a worked-out example is in Nick Cox's answer with $$n=5$$ and $$p=0.2$$) such that median and mean are the same while the distribution is asymmetric.

For continuous distributions, I do not know a common distribution that is both asymmetric and has equal mean and median.

However, it is not difficult to construct a counter-example. The only thing that is needed is to transform a distribution and scale the distances on the left and right sides appropriately such that they have both equal mass and also the equal average distance.

Below is a counterexample where we have a hypothetical distribution that is composed of an equal fifty-fifty mixture of two distributions, on the right side a $$\chi^2$$-distribution and on the left side a Weibull distribution. By selecting parameters of these distributions such that the means are equal, we get that this left and right side have the same weights. The distribution is obviously asymmetric but both sides have the same absolute 0-th and 1-th partial moment, namely both sides have 50% of the probability mass and the average absolute distance from the center is 5.

In this sense, the question looks a bit similar like, "Is it possible to have distributions with different shapes but with the same mean?".

### Unimodal

Edit: I missed the 'unimodal' specification. To get this we can do the same trick and use a mixture distribution. But this time we need to have both sides with the same mode as well. To find this example I took three distributions with each a mean equal to 1 (exponential distribution, half-logistic distribution scaled by $$log(4)$$, half-normal distribution scaled by $$\sqrt{2/\pi}$$) and add two of them together in order to get the same peak height. 