Can positive values with sd > mean have skewness = 0?

Question

I'm trying to create an example of a distribution with all positive values, standard deviation > mean, and skewness =0 (third moment). I cannot. Is that possible? Can you prove it mathematically? Thanks.

Answer 1

This is impossible. Suppose $X$ is the distribution, and let $a,b,c$ be its first three moments, $a=E[X], b=E[X^2], c=E[X^3]$.

We derive three relationships for these variables, and show that they can not be satisfied simulateously.

First, by the Cauchy-Schwarz inequality: \begin{align} E[\sqrt{X^\phantom{1}}\!\! \sqrt{X^3}] &\le \sqrt{E[X]}\sqrt{E[X^3]}\\ (E[X^2])^2 &\le E[X]E[X^3]\\ b^2 &\le ac \\ \end{align} where $\sqrt{X}$ and $\sqrt{X^3}$ are positive by the positivity of $X$.

Second, since skewness is 0, \begin{align} \phantom{skewness = skewness = = = }0 &= E[(X-a)^3]\\ &= E[X^3]-3aE[X^2]+3a^2E[X]-a^3\\ &= c-3ab+3a^3-a^3\\ &= c-3ab+2a^3 \end{align}

Third, since the standard deviation is greater than the mean, \begin{align} \text{variance} &> \text{mean}^2 \phantom{+mean} \\ b - a^2 &> a^2\\ b &> 2a^2\\ \end{align} and, since $a^2$ is positive, also $b>a^2$.

Together these yield $$0\ge b^2-ac=(b-2a^2)(b-a^2)>0$$ which is impossible.

Answer 2

There is in fact a generic lower bound on skewness of strictly positive data:

$$ g_1 > \frac\sigma\mu - \frac\mu\sigma. $$

This shows nicely why you are running into difficulties: the coefficient of variation $\sigma/\mu$ of your desired distribution exceeds $1$, so the positive term in this lower bound is bigger than the negative term. See also this answer.

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In response to @cdalitz's comment below, please see this PubPeer post for an elementary (if lengthy) proof.