This question seems to be related to: Does a "Normal Distribution" need to have mean=median=mode?
Can a non-normal distribution have the same mean and median
distributionsprobability
Related Solutions
Denote $\mu$ the mean ($\neq$ average), $m$ the median, $\sigma$ the standard deviation and $M$ the mode. Finally, let $X$ be the sample, a realization of a continuous unimodal distribution $F$ for which the first two moments exist.
It's well known that
$$|\mu-m|\leq\sigma\label{d}\tag{1}$$
This is a frequent textbook exercise:
\begin{eqnarray} |\mu-m| &=& |E(X-m)| \\ &\leq& E|X-m| \\ &\leq& E|X-\mu| \\ &=& E\sqrt{(X-\mu)^2} \\ &\leq& \sqrt{E(X-\mu)^2} \\ &=& \sigma \end{eqnarray} The first equality derives from the definition of the mean, the third comes about because the median is the unique minimiser (among all $c$'s) of $E|X-c|$ and the fourth from Jensen's inequality (i.e. the definition of a convex function). Actually, this inequality can be made tighter. In fact, for any $F$, satisfying the conditions above, it can be shown [3] that
$$|m-\mu|\leq \sqrt{0.6}\sigma\label{f}\tag{2}$$
Even though it is in general not true (Abadir, 2005) that any unimodal distribution must satisfy either one of $$M\leq m\leq\mu\textit{ or }M\geq m\geq \mu$$ it can still be shown that the inequality
$$|\mu-M|\leq\sqrt{3}\sigma\label{e}\tag{3}$$
holds for any unimodal, square integrable distribution (regardless of skew). This is proven formally in Johnson and Rogers (1951) though the proof depends on many auxiliary lemmas that are hard to fit here. Go see the original paper.
A sufficient condition for a distribution $F$ to satisfy $\mu\leq m\leq M$ is given in [2]. If $F$:
$$F(m−x)+F(m+x)\geq 1 \text{ for all }x\label{g}\tag{4}$$
then $\mu\leq m\leq M$. Furthermore, if $\mu\neq m$, then the inequality is strict. The Pearson Type I to XII distributions are one example of family of distributions satisfying $(4)$ [4] (for example, the Weibull is one common distribution for which $(4)$ does not hold, see [5]).
Now assuming that $(4)$ holds strictly and w.l.o.g. that $\sigma=1$, we have that $$3(m-\mu)\in(0,3\sqrt{0.6}] \mbox{ and } M-\mu\in(m-\mu,\sqrt{3}]$$
and since the second of these two ranges is not empty, it's certainly possible to find distributions for which the assertion is true (e.g. when $0<m-\mu<\frac{\sqrt{3}}{3}<\sigma=1$) for some range of values of the distribution's parameters but it is not true for all distributions and not even for all distributions satisfying $(4)$.
- [0]: The Moment Problem for Unimodal Distributions. N. L. Johnson and C. A. Rogers. The Annals of Mathematical Statistics, Vol. 22, No. 3 (Sep., 1951), pp. 433-439
- [1]: The Mean-Median-Mode Inequality: Counterexamples Karim M. Abadir Econometric Theory, Vol. 21, No. 2 (Apr., 2005), pp. 477-482
- [2]: W. R. van Zwet, Mean, median, mode II, Statist. Neerlandica, 33 (1979), pp. 1--5.
- [3]: The Mean, Median, and Mode of Unimodal Distributions:A Characterization. S. Basu and A. DasGupta (1997). Theory Probab. Appl., 41(2), 210–223.
- [4]: Some Remarks On The Mean, Median, Mode And Skewness. Michikazu Sato. Australian Journal of Statistics. Volume 39, Issue 2, pages 219–224, June 1997
- [5]: P. T. von Hippel (2005). Mean, Median, and Skew: Correcting a Textbook Rule. Journal of Statistics Education Volume 13, Number 2.
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.
Best Answer
Yes.
The simplest way for a distribution to have the same mean and median is to be symmetric. There's a huge range of symmetric distributions, from Bernoulli (two points) to uniform to $t$-distributions, to multimodal distributions.
But symmetry isn't needed. You can take any distribution and make small changes to it to get the mean and median to be equal. If the mean is less than the median, you can increase the mean by adding a small probability of a very large value. If the mean is greater than the median, you can decrease the mean by adding a small probability of a large negative value.