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The Bernoulli distritbution has excess kurtosis = $-2$. In addition for a symmetric beta distribution with $\alpha = \beta$ the excess kurtossis is $-6/(2\alpha+3)$ whose minimum value is again $-2$.

But, how can it be shown whether this is the minimum possible value for all conceivable probability distributions?

Davius
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1 Answers1

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Use Jensen's inequality: Let $Z = (X-\mu)/\sigma$. Then (non-excess) kurtosis is $\kappa = E(Z^4) \ge \{E(Z^2)\}^2 =1$. Subtract 3 to get excess kurtosis and you have the result. Of course this assumes finite fourth moment.

BigBendRegion
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