# Looking for an appealing counterexample in probability

There is a commonly-encountered-but-wrong rule of thumb that says something like

If a probability distribution is positively skewed, its mean is greater than its median.

(You sometimes also see it phrased in terms of mean and mode). I'm looking for a nice counterexample to this rule. What do I mean by "nice"? Hard to say exactly, of course, but something like:

1. It must be continuously differentiable.
2. It must have only one maximum and at most two points of inflection.
3. Its graph must be "obviously" positively skewed and the mean must be "obviously" less than the median.

I'm aware of the Weibull distribution, which can satisfy (1) and (2) - but for me the graph appears too close to being unskewed and the mean appears too close to the median for it to be an intuitively obvious counterexample. So I'm looking for something where the properties are more easily visible.

Alternatively: if such a distribution did exist it seems to me that it would be quite well-known. So it seems plausible that the rule of thumb might be almost true, in the sense that there might exist a true statement of the form

If a probability distribution is positively skewed, its mean can be at most __ less than its median.

I'm also very interested to see answers of this form.

• From Wikipedia: the mean and the median never differ by more than one standard deviation (en.wikipedia.org/wiki/…). It may be possible to do better than this with information on skewness, I don't know. May 8 '11 at 23:18

## 1 Answer

You are looking for random variables $X$ of median $m(X)$ such that $$E((X-E(X))^3)>0\quad\mbox{and}\quad E(X)< m(X).$$ Assume there exists two nonnegative random variables $Y$ and $Z$ and an independent Bernoulli sign $B=\pm1$ with $E(B)=0$ such that $X=Y$ on $[B=1]$ and $X=-Z$ on $[B=-1]$.

Then $m(X)=0$ and $E(X)=\frac12(E(Y)-E(Z))$. Furthermore, assuming $E(Y)< E(Z)$, one can prove that $$E(Y^3)>E(Z^3)\quad\implies\quad E((X-E(X))^3)>0.$$ To prove this implication, expand $(X-E(X))^3$ using $Y$ and $Z$ to see that $E((X-E(X))^3)>0$ if and only if $$2(E(Y^3)-E(Z^3))+2(E(Z)^3-E(Y)^3)+3(E(Z)-E(Y))(\mbox{var}(Y)+\mbox{var}(Z))>0.$$ If $E(Z)> E(Y)$, the second and the third terms are positive, hence if the first term is positive as well, the whole LHS is positive. Thus the implication holds.

Finally, every couple $(Y,Z)$ such that $$E(Y)< E(Z),\qquad E(Y^3)>E(Z^3)$$ is a solution. Note that every $X$ such that $m(X)=0$ can be written as above.

First example If $Y$ is distributed like $y$ times a reduced exponential and $Z$ is distributed like $z\sqrt{\pi/2}$ times the absolute value of a reduced normal, then $E(Y)=y$, $E(Y^3)=6y^3$, $E(Z)=z$ and $E(Z^3)=2z^3$, hence every couple of parameters $(y,z)$ such that $y< z< \sqrt[3]{3}y$ yields a solution.

Second example If $Y$ is distributed like $y$ times a reduced exponential and $Z$ is distributed like $z$ times the sum of two independent reduced exponentials, then $E(Y)=y$, $E(Y^3)=6y^3$, $E(Z)=2z$ and $E(Z^3)=24z^3$, hence every couple of parameters $(y,z)$ such that $\sqrt[3]{4}z< y< 2z$ yields a solution.