Rare-outlier three-point distribution ($\mathbb{P}(X=0)=1-q,\mathbb{P}(X=\pm 1)=q/2$)

For $0<q\le 1$, 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: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 contributesto kurtosis. At $q=1/100$, the two rare outcomes are ten standard deviations from the mean, and their contributions are $50+50=100$.

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.