Standardized Cumulants and Excess Kurtosis

Why Standardize Cumulants?

Cumulants describe features of a distribution, but their units depend on the scale of the random variable. The third cumulant has units of the variable cubed, and the fourth has units to the fourth power. Dividing by the appropriate power of the standard deviation removes those units. The resulting quantities measure asymmetry and tail weight in a way that can be compared across scales.

The third standardized cumulant is the usual skewness. The fourth standardized cumulant is the excess kurtosis, which is Pearson kurtosis minus 3. These names are useful only when the relevant moments exist, so the moment conditions are part of the definitions, not an optional technical detail.

Definition and Moment Conditions

Definition: Let \(X\) have mean \(\mu=E[X]\), finite positive variance \(\sigma^2=\operatorname{Var}(X)>0\), and finite fourth absolute moment \(E[|X|^4]<\infty\). Write \(Z=(X-\mu)/\sigma\). The cumulants through order four are denoted \(\kappa_r(X)\), for \(r=1,2,3,4\). The standardized cumulants are \(\lambda_r=\kappa_r(Z)\). In particular, \(\lambda_1=0\), \(\lambda_2=1\), \(\lambda_3\) is skewness, and \(\lambda_4\) is excess kurtosis. For \(r\geq 2\), \(\lambda_r=\kappa_r(X)/\sigma^r\) at any order for which the \(r\)th moment exists.

The condition \(E[|X|^4]<\infty\) ensures that all moments needed for these definitions exist. Finite positive variance alone is not enough to define skewness or excess kurtosis: a distribution may have a finite variance but an infinite third or fourth absolute moment. At an order where the required moment does not exist, the corresponding cumulant and standardized cumulant are not defined by these moment formulas.

For clarity, cumulants through a finite order can be defined using a formal power series, so an exponential moment-generating function need not exist near zero. If \(m_j=E[X^j]\), form the truncated moment series \(M(t)=\sum_{j=0}^4 m_jt^j/j!\) and take its formal logarithm through degree four. The coefficient of \(t^r/r!\) in that logarithm is \(\kappa_r(X)\). When the moment-generating function is finite in a neighborhood of zero, this definition agrees with the ordinary derivatives at zero of its log, the cumulant-generating function.

Theorem: If \(X\) has finite fourth absolute moment and positive variance, and \(m_r=E[(X-E[X])^r]\) denotes its \(r\)th central moment, then \(\kappa_1(X)=E[X]\), \(\kappa_2(X)=m_2=\sigma^2\), \(\kappa_3(X)=m_3\), and \(\kappa_4(X)=m_4-3m_2^2\). Consequently, \(\lambda_3=m_3/\sigma^3\) and \(\lambda_4=m_4/\sigma^4-3\).

Proof. Center the variable by putting \(Y=X-E[X]\). Its first four moments are \(0,m_2,m_3,m_4\). In the formal moment series for \(Y\), terms through degree four are \(M_Y(t)=1+m_2t^2/2+m_3t^3/6+m_4t^4/24\). Write \(u=m_2t^2/2+m_3t^3/6+m_4t^4/24\). Through degree four, \(\log(1+u)=u-u^2/2\), because \(u^3\) has degree at least six. The degree-four part of \(u^2/2\) is \(m_2^2t^4/8\). Therefore \[ \log M_Y(t)=\frac{m_2t^2}{2}+\frac{m_3t^3}{6} +\left(\frac{m_4}{24}-\frac{m_2^2}{8}\right)t^4 \] through degree four. Matching coefficients of \(t^r/r!\) gives \(\kappa_2(Y)=m_2\), \(\kappa_3(Y)=m_3\), and \(\kappa_4(Y)=m_4-3m_2^2\). Centering does not change cumulants of order two or higher: translation multiplies the moment series by the formal series for \(e^{ct}\), so its logarithm adds \(ct\), which affects only the first cumulant. The first cumulant is the mean. Finally, scaling \(Y\) by \(1/\sigma\) multiplies its \(r\)th cumulant by \(\sigma^{-r}\). This gives the stated standardized formulas. \(\square\)

Skewness and Excess Kurtosis

The theorem identifies the two standardized cumulants most often used to describe shape. Skewness is the standardized third central moment. Positive skewness indicates that deviations above the mean contribute more to the signed third moment than deviations below it; negative skewness indicates the reverse. It is zero for a distribution symmetric about its mean, provided the third moment exists.

Pearson kurtosis is \(m_4/\sigma^4\), the fourth central moment divided by the variance squared. It is always nonnegative. Excess kurtosis subtracts the normal benchmark of 3: \[ \lambda_4=\frac{m_4}{\sigma^4}-3. \] Thus a negative excess kurtosis means Pearson kurtosis is below 3, while a positive value means it is above 3. These are moment summaries; by themselves they do not fully describe a distribution or uniquely determine its shape.

The standardization rule can also be checked directly from cumulant scaling. For an affine transformation \(aX+b\), cumulants of order \(r\geq2\) are multiplied by \(a^r\), while translation affects only the first cumulant. Dividing the cumulant of \(X\) by \(\sigma^r\) therefore removes changes in location and positive scale. Under reflection (\(a<0\)), odd standardized cumulants change sign, but even ones do not.

Worked Comparisons

Worked Example: Normal Distribution

Let \(X\sim N(\mu,\sigma^2)\), where \(\sigma^2>0\). Its cumulant-generating function is \(K(t)=\mu t+\sigma^2t^2/2\). Differentiating at zero gives \(\kappa_1=\mu\), \(\kappa_2=\sigma^2\), and \(\kappa_3=\kappa_4=0\), since \(K(t)\) has no terms of degree three or higher.

The standardized cumulants are therefore \(\lambda_1=0\), \(\lambda_2=1\), \(\lambda_3=0/\sigma^3=0\), and \(\lambda_4=0/\sigma^4=0\). In moment terminology, its fourth central moment is \(3\sigma^4\), so Pearson kurtosis is \(3\sigma^4/\sigma^4=3\) and excess kurtosis is \(3-3=0\).

Worked Example: Bernoulli Distribution

Let \(X\sim\operatorname{Bernoulli}(p)\), with \(0<p<1\), and write \(q=1-p\). The strict inequalities ensure positive variance. Its cumulant-generating function is \(K(t)=\log(q+pe^t)\). Define \(s(t)=pe^t/(q+pe^t)\). Then \(K'(t)=s(t)\) and \(s'(t)=s(t)(1-s(t))\). At zero, \(s(0)=p\), so \[ \kappa_1=p,\qquad \kappa_2=pq,\qquad \kappa_3=pq(1-2p),\qquad \kappa_4=pq(1-6pq). \] For the last two identities, differentiate \(K''=s(1-s)\): \(K'''=s(1-s)(1-2s)\), and \(K''''=s(1-s)(1-6s(1-s))\); then set \(s=p\).

Since \(\sigma^2=pq\), the standardized cumulants are \[ \lambda_3=\frac{pq(1-2p)}{(pq)^{3/2}} =\frac{1-2p}{\sqrt{pq}},\qquad \lambda_4=\frac{pq(1-6pq)}{(pq)^2} =\frac{1-6pq}{pq}. \] For the concrete choice \(p=1/4\), \(q=3/4\), and \(pq=3/16\). Thus \(\lambda_3=(1/2)/(\sqrt{3}/4)=2/\sqrt{3}\). Also \(1-6pq=1-18/16=-1/8\), so \(\lambda_4=(-1/8)/(3/16)=-2/3\). The skewness is positive, while the excess kurtosis is negative.

Worked Example: Poisson Distribution

Let \(X\sim\operatorname{Poisson}(\lambda)\), with \(\lambda>0\). Its cumulant-generating function is \(K(t)=\lambda(e^t-1)\). Every derivative of positive order at zero equals \(\lambda\), so in particular \(\kappa_1=\kappa_2=\kappa_3=\kappa_4=\lambda\). The variance is \(\sigma^2=\lambda\), and therefore \[ \lambda_3=\frac{\lambda}{\lambda^{3/2}}=\frac{1}{\sqrt{\lambda}}, \qquad \lambda_4=\frac{\lambda}{\lambda^2}=\frac{1}{\lambda}. \]

For \(\lambda=4\), the variance is \(4\), the standard deviation is \(2\), the skewness is \(1/2\), and the excess kurtosis is \(1/4\). Both standardized cumulants are positive for every \(\lambda>0\); as \(\lambda\) increases, their values decrease toward zero.

Reading the Comparisons Carefully

The three examples show how the standardized quantities reflect distributional structure. The normal distribution has zero skewness and zero excess kurtosis. A Bernoulli distribution is bounded and can be strongly asymmetric when \(p\) is far from \(1/2\); at \(p=1/2\) its skewness is zero, and its excess kurtosis is \(-2\). A Poisson distribution has positive skewness and positive excess kurtosis, with both decreasing as its rate grows.

A zero skewness does not prove symmetry, and zero excess kurtosis does not prove normality. These are single-number summaries of moments, not complete characterizations of distributions. Likewise, the sign of excess kurtosis should not be used as a complete description of tail probabilities: it reports a fourth-moment comparison with the normal benchmark, not the behavior of every tail event.

Common Mistake: Confusing Kurtosis with Excess Kurtosis

A frequent error is to call \(m_4/\sigma^4\) “the fourth cumulant” or to label it the fourth standardized cumulant. That ratio is Pearson kurtosis. The fourth cumulant is \(\kappa_4=m_4-3\sigma^4\), so the standardized fourth cumulant is \(m_4/\sigma^4-3\), the excess kurtosis. For a normal variable the distinction is immediate: Pearson kurtosis is 3, whereas the fourth cumulant and excess kurtosis are both 0.

Another error is to define all standardized cumulants merely from finite variance. The third and fourth standardized cumulants require the corresponding higher moments. A finite positive variance permits standardization, but it does not guarantee that skewness or excess kurtosis exists.

Key takeaway: With a finite fourth absolute moment and positive variance, skewness is \(\kappa_3/\sigma^3=m_3/\sigma^3\), and excess kurtosis is \(\kappa_4/\sigma^4=m_4/\sigma^4-3\). Check both the moment conditions and whether a reported fourth-order quantity is Pearson kurtosis or excess kurtosis.

Check Your Understanding

Use the definitions and calculations above to answer the following.

Why does finite positive variance alone not ensure that skewness exists?
For a distribution with variance \(\sigma^2\) and fourth central moment \(m_4\), write Pearson kurtosis and excess kurtosis in terms of \(m_4\) and \(\sigma\).
For \(X\sim\operatorname{Bernoulli}(1/2)\), calculate its skewness and excess kurtosis from the formulas in the tutorial.
For \(X\sim\operatorname{Poisson}(9)\), find its skewness and excess kurtosis.
Under the transformation \(Y=-X+b\), what happens to skewness and excess kurtosis, assuming the required moments exist?