When does positive skew imply median<mean?

Question

For a random variable $X$, the skew is defined as $S(X):={\frac {E\overline{X}^3}{(E\overline{X}^2)^{3/2}}}$, where $\overline{X}=X-EX$. It is often claimed that positive (resp. negative) skew implies that the mean of $X$ is larger (resp. smaller) than the median of $X$.

However this implication can fail. For example, on a mixture of two Gaussians with appropriate chosen parameters (this can be checked directly). But it can even fail on unimodal distributions, for example the Weibull distribution (cf. Groeneveld, Skewness for the Weibull Family, 1986).

What additional conditions can we place on $X$ that will ensure $sign(Skew(X))=sign(mean(X)-median(X))$? Clearly there are trivial answers such as stipulating that $X$ belongs to a certain family of distributions, but I am wondering if there is any more general or nice condition.

Answer 1

The statement that the mean is greater than the median is simply a statement of $$F_{X}(\mathbb{E}[X])>0.5$$ for a random variable $X$ with distribution function $F_{X}(x)$.

This has nothing (directly) to do with skewness.

To impose a direct relationship between this statement and skewness (being positive) would just result in some inequality like

$$\mathbb{E}\bigg[\bigg(\frac{X-\mu}{\sigma}\bigg)^{3}\bigg]>0\quad\Rightarrow\quad \mathbb{E}[X^3]>\mathbb{E}[X](3σ^2+\mathbb{E}[X]^2)$$

where $\mathbb{E}[X]>F^{-1}_X(0.5)$ and $σ^2$ is taken to be finite.

Any distribution satisfying this inequality is what you're after.

Answer 2

This is a hard problem. Here are two special cases which seem difficult but tractable.

1. Let $$Y=\frac1n \sum_{i=1}^n X_i$$ where the $X_i$ all come from the same lognormal $LN(\mu,\sigma)$ distribution. Is it true that $$Mean[Y] > Median[Y]\ ?$$ This is an open problem for $n\ge 5$, which I answered fully for $n=2$ and with a sketch for $n=3$ and $n=4$ here.

2. Consider an exponential family of distributions defined by pdfs $$f(x|\theta)=h(x)g(\theta)e^{\eta(\theta)T(x)}$$ What are some nice conditions on $h,g,\eta,T$ which lead to $$mean>median \text{ for all }\theta\ ?$$ As formal criteria of niceness, we could require that the conditions apply in at least some cases with all four functions non-constant, and be stated without any integrals using products of the functions; asking for nice conditions informally would work too.

Answer 3

Here are some results I found in the literature. I'll assume without loss of generality that $EX=0$ and $EX^2=1$ (in particular, $Skew(X)=EX^3$). I'll also assume that $X$ has a density function $p$.

1. If $p(x)-p(-x)$ changes sign exactly once in $x$, then $$Skew>0 \iff Mean>Median$$ [Mac]. This includes all distributions from the Pearson family as well as lognormal and inverse Gaussian.

2. If $X_1, X_2, \dots$ are independent copies of $X$ and $S_n$ is the sum of the first $n$ copies, then $$\lim_{n\to\infty} Median(S_n) = \frac{-1}{6}Skew(X)$$ This has certain technical assumptions which are not too stringent and can be founded in the referenced paper. (Due to [Hall], with an earlier version proved by [Hald]). Since $$Skew(S_n)>0 \iff Skew(X)>0$$ we thus expect $$Skew(S_n)>0 \iff Median(S_n)<0$$ for sufficiently large n.

The paper actually proves a stronger statement: namely that the median goes as $-{\frac 1 6}skew+{\frac C n}$, to leading order in $n$. Here $C$ is a certain explicit expression that involves the first five moments of $X$. This allows us to estimate a finite $n$ value for which the skew-median relation will hold. Namely, it holds for $S_n$ provided that $-{\frac 1 6}|skew|+{\frac {sign(skew)C} n}<0$.

[Hall] P. Hall, On the Limiting Behaviour of the Mode and Median of a Sum of Independent Random Variables, Annals of Probability, 1980,

[Hald] J Haldane,The mode and median of nearly normal distribution with given cumulants, Biometrika, 1942.

[Mac] H MacGillivray, The mean, median, and mode inequality and skewness for a class of densities, Australian Journal of Statistics, 1982.