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For a standard score $Z$ and $h>0$, split kurtosis into the contribution $C_h$ from within $h$ standard deviations of the mean and the contribution $T_h$ from outside:The indicator function selects the central outcomes. The bound follows directly from $Z^4\le h^4$ there. Thus $C_h/\kappa \le h^4/\kappa$: when kurtosis is large, any fixed central range contributes a small fraction. This is a precise reason to emphasize extreme standard scores rather than peak height.^[ref]

For the rare-outlier three-point distribution with $q=1/100$, $C_1=0$ and $T_1=100$. At the other endpoint $q=1$, all probability mass lies at $Z=\pm 1$, so $C_1=\kappa =1$: central dominance is possible when kurtosis is small. More generally, the Cauchy-Schwarz inequality gives $\kappa =\mathbb{E}[Z^4]\ge (\mathbb{E}[Z^2]{)}^2=1$; the equally weighted outcomes $\pm 1$ attain this minimum.

For a real random variable $X$ with mean $\mu =\mathbb{E}X$, finite positive variance $0<{\sigma }^2=\mathrm{Var}(X)<\infty$, and finite fourth moment, kurtosis is the fourth central moment divided by the square of the variance:The standard score $Z$ measures distance from the mean in standard deviations. Its fourth power gives a value ten standard deviations away a weight $10^4=10000$ inside the expected value. Thus a small probability mass far from the mean can strongly affect kurtosis. It is useful for describing the contribution of extreme standard scores, but does not by itself describe the height or shape of a probability density function near its peak.^[ref]

The normal distribution has kurtosis $3$. A random scale can increase kurtosis even when the conditional distribution is normal. A constant random variable has zero variance, so this ratio is undefined. If the variance is finite and positive but the fourth central moment is infinite, the ratio can be written $+\infty$ in an extended sense; it is not a finite kurtosis. If the variance itself is infinite or the mean is undefined, the displayed normalization is unavailable.