# Example of Continuous Random Variable Distribution, with Median=Mean but not Symmetric?

So I recently learned that for a random variable that has Median $=$ Mean, symmetry of the density function around the mean is not implied. I found this to be surprising as the common continuous distributions that I am aware of (Uniform, Normal, etc) that have equal median and mean are symmetric.

I had initially thought of the Gamma distribution as a possibility, however unless I am mistaken, the Mean can approach the Median under certain specifications, but will always be slightly larger.

I am curious if anyone can think of continuous random variable distributions that have an equal median and mean, but are not symmetric about the mean?

Try a density like this

$$f(x)= \begin{cases} \frac1{8} &\quad\text{if } 0 \le x \lt 4 \\ \frac14 &\quad\text{if } 4 \le x \lt 5 \\ \frac1{20} &\quad\text{if } 5 \le x \lt 10 \\ 0 &\quad\text{otherwise } \\ \end{cases}$$

which has a mean and median of $4$ and looks like

• Good suggestion. For some reason, I was thinking that the PDF would need to be continuous, but I realize now that need not be the case. Commented Jul 2, 2017 at 1:37

Looking at it from the point of view of moment problem:

There exist two distinct probability measures $$\mu_1$$, $$\mu_2$$ on $$[0, \infty)$$ such that their moment sequences coincide. Then consider the probability measure on $$\mathbb{R}$$ it's $$\frac{1}{2}(\mu_1 + \mu_2')$$ ( that is, on the positive axis $$\frac{1}{2} \mu_1$$, on the negative axis $$\frac{1}{2} \cdot \mu_2'$$, the reflexion on $$\mu_2$$. Then we see that the origin is the "median" for all of the absolute moments, yet the measure $$\mu$$ is not symmetric.