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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 condition.

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.

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).

Under what conditions on $X$ can we conclude that $Skew>0\Leftrightarrow (median<mean)$?

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 condition.

For a random variable $X$ with mean zero and unit variance, the skew is defined as $S(X):=E X^3$ (for general $X$, the skew is defined as the skew of the z-scored version of $X$). 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).

Under what conditions on $X$ can we conclude that $Skew>0\Leftrightarrow (median<mean)$?

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).

Under what conditions on $X$ can we conclude that $Skew>0\Leftrightarrow (median<mean)$?