# What conditions must satisfiy a positively skewed density function to ensure that median is greater than mean

We are collecting environmental Air Quality data. When we validate data, we always plot ECDF and compute basic statistics and percentiles.

Our experimental distributions are far away from normality. They are positive definite (concentrations) and positively skewed. Some distribution (NO2) almost never show median greater than mean (one occurrence in 30 years). Then I was wondering about a link between skewness, median and mean such as:

$$\mu_3^* > 0 \Rightarrow x_{50} < \mu_1$$

Unfortunately, I just read the following on Wikipedia (Skewness):

"The skewness is not strictly connected with the relationship between the mean and median: a distribution with negative skew can have the mean greater than or less than the median, and likewise for positive skew."

And then I would ask the following question: What additional conditions must satisfy a positively skewed density function to ensure that median is greater than mean?

Edit: I have found a very clear paper detailing this affirmation in T. von Hippel, Mean, Median and Skew: Correcting a Textbook rule .

Rewording the question:

What additional properties must have $f(x)$, knowing:

$$f(x) > 0 \, \forall x \in \mathbb{R},\\ \int\limits_0^{\infty}{f(x)\mathrm{d}x} = 1,\\ \mu_3^* = \int\limits_0^{\infty}{(x-\mu_1)^3f(x)\mathrm{d}x} > 0$$

to validate the following constraint: $$x_{50} = \int\limits_0^{x_{50}}{f(x)\mathrm{d}x} < \int\limits_0^{\mu_1}{f(x)\mathrm{d}x} = \mu_1$$

Or equivalently: $F(x_{50}) < F(\mu_1)$ or $x_{50} < \mu_1$.

• The relative 'strength' of the skew ness doesn't affect this comparison. There will always be counterexamples, by choosing a suitable distribution. – Calvin Lin May 12 '14 at 13:20
• @Calvin Lin: Is there a better criterion than skewness then? – jlandercy May 12 '14 at 13:38