Gauss's inequality

A first important result is Gauss's inequality.^[7] Gauss's inequality gives an upper bound on the probability that a value lies more than any given distance from its mode. This inequality depends on unimodality.

Vysochanskiï–Petunin inequality

A second is the Vysochanskiï–Petunin inequality,^[8] a refinement of the Chebyshev inequality. The Chebyshev inequality guarantees that in any probability distribution, "nearly all" the values are "close to" the mean value. The Vysochanskiï–Petunin inequality refines this to even nearer values, provided that the distribution function is continuous and unimodal. Further results were shown by Sellke and Sellke.^[9]

Mode, median and mean

Gauss also showed in 1823 that for a unimodal distribution^[10]

${\displaystyle \sigma \leq \omega \leq 2\sigma }$

and

${\displaystyle |\nu -\mu |\leq {\sqrt {\frac {3}{4}}}\omega ,}$

where the median is ν, the mean is μ and ω is the root mean square deviation from the mode.

It can be shown for a unimodal distribution that the median ν and the mean μ lie within (3/5)^1/2 ≈ 0.7746 standard deviations of each other.^[11] In symbols,

${\displaystyle {\frac {|\nu -\mu |}{\sigma }}\leq {\sqrt {\frac {3}{5}}}}$

where | . | is the absolute value.

In 2020, Bernard, Kazzi, and Vanduffel generalized the previous inequality by deriving the maximum distance between the symmetric quantile average ${\displaystyle {\frac {q_{\alpha }+q_{(1-\alpha )}}{2}}}$ and the mean,^[12]

${\displaystyle {\frac {\left|{\frac {q_{\alpha }+q_{(1-\alpha )}}{2}}-\mu \right|}{\sigma }}\leq \left\{{\begin{array}{cl}{\frac {{\sqrt[{}]{{\frac {4}{9(1-\alpha )}}-1}}{\text{ }}+{\text{ }}{\sqrt[{}]{\frac {1-\alpha }{1/3+\alpha }}}}{2}}&{\text{for }}\alpha \in \left[{\frac {5}{6}},1\right)\!,\\{\frac {{\sqrt[{}]{\frac {3\alpha }{4-3\alpha }}}{\text{ }}+{\text{ }}{\sqrt[{}]{\frac {1-\alpha }{1/3+\alpha }}}}{2}}&{\text{for }}\alpha \in \left({\frac {1}{6}},{\frac {5}{6}}\right)\!,\\{\frac {{\sqrt[{}]{\frac {3\alpha }{4-3\alpha }}}{\text{ }}+{\text{ }}{\sqrt[{}]{{\frac {4}{9\alpha }}-1}}}{2}}&{\text{for }}\alpha \in \left(0,{\frac {1}{6}}\right]\!.\end{array}}\right.}$

It is worth noting that the maximum distance is minimized at ${\displaystyle \alpha =0.5}$ (i.e., when the symmetric quantile average is equal to ${\displaystyle q_{0.5}=\nu }$), which indeed motivates the common choice of the median as a robust estimator for the mean. Moreover, when ${\displaystyle \alpha =0.5}$, the bound is equal to ${\displaystyle {\sqrt {3/5}}}$, which is the maximum distance between the median and the mean of a unimodal distribution.

A similar relation holds between the median and the mode θ: they lie within 3^1/2 ≈ 1.732 standard deviations of each other:

${\displaystyle {\frac {|\nu -\theta |}{\sigma }}\leq {\sqrt {3}}.}$

It can also be shown that the mean and the mode lie within 3^1/2 of each other:

${\displaystyle {\frac {|\mu -\theta |}{\sigma }}\leq {\sqrt {3}}.}$

Skewness and kurtosis

Rohatgi and Szekely claimed that the skewness and kurtosis of a unimodal distribution are related by the inequality:^[13]

${\displaystyle \gamma ^{2}-\kappa \leq {\frac {6}{5}}=1.2}$

where κ is the kurtosis and γ is the skewness. Klaassen, Mokveld, and van Es showed that this only applies in certain settings, such as the set of unimodal distributions where the mode and mean coincide.^[14]

They derived a weaker inequality which applies to all unimodal distributions:^[14]

${\displaystyle \gamma ^{2}-\kappa \leq {\frac {186}{125}}=1.488}$

This bound is sharp, as it is reached by the equal-weights mixture of the uniform distribution on [0,1] and the discrete distribution at {0}.