Relationship between median and mean of a left skewed distribution

Question

Question

If I see graphically that my distribution is left-skewed. How to concretely (still qualitatively) conclude that the Mean is less than the Median? At least in nonpathological examples?

There is a possible duplicate, but the accepted answer or other answers there didn't solve my doubt, infact they were focused on saying, this statement isn't true in the most general setting.

Thoughts When I try to look it up, I always find one argument saying that since the data is left-skewed, the tail goes far left, and some outliers will bring down the mean more than the median. But that sounds like just a restatement of my doubt. Why, if the mean is brought down, it gets below the median?

Answer 1

Let $F$ be an absolutely continuous CDF with density $f$ and assume that $f$ has a unique mode $M$. Denote the (unique) median by $m$ and the finite mean by $\mu$.

A sufficient condition for the mean-median-mode (MMM) inequality $$\mu\leq m \leq M $$ is $$ F(m-x)+F(m+x)\geq 1\ \forall\ x, $$ and MMM is strict if $m\neq M$.

Some special cases include:

1. If there exists a $x_0>0$ such that $$f(m+x)-f(m-x) \begin{cases}\geq0 & \text{for }0\leq x < x_0 \\ \leq 0 & \text{for } x>x_0, \end{cases}$$ the MMM inequality holds. This condition is essentially reflecting the pdf about the median $m$. If the reflected curve crosses the original density from above exactly once, then the MMM inequality holds. See case 3 of this example of the technique applied to the $\chi^2$ distribution.

2. If $f$ is differentiable and $$f'(x_1) < |f'(x_2)|\text{ for all } x_1 \text{ and } x_2 \text{ such that }f(x_1)=f(x_2)>0\text{ and } x_1<M<x_2,$$ the MMM inequality holds. Intuitively, this condition says if we chop the pdf horizontally at any height, the steepness at the cutting boundary on one side is always larger than the other side.

3. If $$f\left[F^{-1}(t)\right] \leq f\left[F^{-1}(1-t)\right]\text{ for all }0<t<\frac12,$$ then MMM inequality holds. This says that if we vertically chop off equal tail probabilities $t=F(x_1)=1-F(x_2)$ from each tail of the pdf at $x_1$ and $x_2$ for $x_1<m<x_2$, the height of one cutting boundary is always higher than the other boundary height, i.e., $f(x_1)\leq f(x_2)$.