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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?

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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

mean equals median

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  • $\begingroup$ 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. $\endgroup$
    – user345
    Commented Jul 2, 2017 at 1:37
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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.

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