Let $X$ be a random variable such that $E(X^2)<\infty$. Let $m$ denote a median of $X$ (that is to say any number $m$ such that $P(X\leq m)=P(X>m)=\frac 12$) and $\sigma$ its standard deviation. It's known that $|E(X)-m|\leq \sigma$ (see this for example).

Out of curiosity, I'm looking for an example of a distribution such that $E(X)-m = \sigma$ (that is to say it has maximal mean absolute deviation)

If $E(X)-m = \sigma$, it's not hard to prove (see my post for example) that we must have $$P(X-E(X)\leq -\sigma)=\frac12$$

The last equality is quite intuitive: in a discrete setting, if there's one outlier much larger than the rest of the values, $E(X)$ will be much higher than $m$, hence $X-E(X)$ will be mostly negative.

I can't come up with an example of such distribution off the top of my head.


1 Answer 1


The measure you are talking about is the so-called nonparametric skew, a measure of the skewness of a random variable's distribution that is defined as


An example of distribution with $S$ near to $1$ may be the following. Consider the discrete distribution given by a set of $N$ numbers (with $N$ odd) that can assume only the values $0$ or $1$, and where the frequencies of $0$ and $1$ are $\frac {N+1}{2}\,$ and $\frac{N-1}{2}\,$. In this distribution the mean $\mu$, the median $m$, and the standard deviation $\sigma$ are

$$ \mu=\frac{1}{2}-\frac{1}{2N}\,$$ $$m=0$$ $$\sigma=\sqrt{\frac{N+1}{4N}}$$

It is not difficult to show that, for $N \rightarrow \infty\,\,$, both $\mu$ and $\sigma$ tend to $1/2\,$, and then $S \rightarrow 1\,\,$. For example, with $N=11\,\,$ we have $S\approx 0.8704\,\,\,$, for $N=1001\,\,$ we have $S\approx 0.9985\,\,\,$, for $N=10001\,\,$ we have $S\approx 0.9998\,\,\,\,$, and so on.

Lastly, consider that if we use a very similar measure of skewness, equal to the nonparametric skew but with the variance in the denominator (let us call it $S'$) you can find more easily some distribution with $S' =1$. For instance, the Cantor distribution has mean $1/2\,$, median ranging between $1/3\,$ and $2/3\,$, and variance $1/8\,$. Thus, a Cantor distribution with median $3/8\,$ has $S'=1\,$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .