= Rare-outlier three-point distribution
{title2=$\mathbb P(X=0)=1-q,\quad\mathbb P(X=\pm1)=q/2$}
For $0<q\leq1$, put <probability mass> $1-q$ at zero and $q/2$ at each of $-1$ and $1$. This <symmetric probability distribution> has <mean> zero. Its second and <fourth moments> are both $q$, because the nonzero outcomes have square and fourth power equal to one:
$$
\operatorname{Var}(X)=q,\qquad \mathbb E[X^4]=q,\qquad \kappa(X)=\frac{q}{q^2}=\frac1q.
$$
For $q=1/4$ the <kurtosis> is $4$; for $q=1/100$ it is $100$. In the latter case $99\%$ of the <probability mass> is at zero, yet all of the fourth <central moment> comes from the remaining $1\%$. The <standard score> $Z=X/\sqrt q$ takes values $0$ and $\pm q^{-1/2}$, with <variance> one. Each nonzero outcome contributes
$$
\frac q2\left(q^{-1/2}\right)^4=\frac1{2q}
$$
to <kurtosis>. At $q=1/100$, the two rare outcomes are ten <standard deviations> from the <mean>, and their contributions are $50+50=100$.
\Image[/kurtosis-rare-outlier-mass-and-contributions.png]
{title=<Probability mass> and fourth-<moment> contributions for a <rare-outlier three-point distribution>}
{description=The left panels keep the original <support of a probability distribution> $\{-1,0,1\}$; the right panels use <standard scores>. A rare outcome can have small <probability mass> but a large contribution to <kurtosis>. Each row is a different specified <probability distribution>, not a <statistical sample>.}
Every member has finite <support of a probability distribution>, so all its <moments> are finite. Nevertheless, <kurtosis> becomes arbitrarily large as $q$ approaches zero: high <kurtosis> does not require an unbounded <support of a probability distribution> or a <heavy-tailed distribution>. At $q=0$ the <variance> vanishes and <kurtosis> is undefined. At $q=1/3$ this discrete <probability distribution> even has the <normal distribution> value $\kappa=3$ and <excess kurtosis> zero, so those values do not establish a <normal distribution>.
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