Cumulant Scaling in Independent Sums

Why Cumulant Scaling Matters

The central limit theorem describes the leading normal approximation to a standardized sum. Edgeworth methods refine that approximation by organizing nonnormal features according to their size as the sample size $n$ grows. Cumulants make this organization transparent: for iid observations, the standardized third cumulant is of order $n^{-1/2}$, the fourth is of order $n^{-1}$, and each higher-order cumulant has a correspondingly smaller scale, provided it exists.

This hierarchy identifies which features can affect a first- or second-order correction. It does not, by itself, establish that an Edgeworth approximation is valid or accurate; those questions require additional conditions. Here the goal is to derive the scaling and its consequences for the order of the correction terms.

Cumulants and Standardization

When the moment-generating function exists in a neighborhood of zero, the cumulant-generating function is its logarithm. For a real random variable $X$, write $K_X(t)=\log E[e^{tX}]$. The $r$th cumulant is the $r$th derivative of $K_X$ at zero. More generally, cumulants through order $r$ can be defined algebraically from moments through order $r$, even when the moment-generating function does not exist near zero.

Definition: If $E[|X|^r]<\infty$, the cumulants $\kappa_1(X),\ldots,\kappa_r(X)$ are defined by the coefficients through degree $r$ in the formal power series $$ \log\left(\sum_{j=0}^{r} E[X^j]\frac{t^j}{j!}\right) =\sum_{j=1}^{r}\kappa_j(X)\frac{t^j}{j!} \quad\text{through degree }r. $$ In particular, $\kappa_1(X)=E[X]$, $\kappa_2(X)=\operatorname{Var}(X)$, and, when the relevant moments exist, $\kappa_3(X)=E[(X-E[X])^3]$ and $\kappa_4(X)=E[(X-E[X])^4]-3\operatorname{Var}(X)^2$.

The phrase “through degree $r$” means that coefficients above degree $r$ are ignored. The definition uses only moments up to order $r$, so it is available whenever the absolute $r$th moment is finite. If the moment-generating function exists near zero, the formal definition agrees with the derivative definition.

For a variable with mean $\mu$ and finite positive variance $\sigma^2$, its standardized version is $Z=(X-\mu)/\sigma$. Its first two cumulants are $0$ and $1$. For $r\geq 3$, its standardized cumulant is often written $\lambda_r=\kappa_r(X)/\sigma^r$, provided the $r$th moment exists.

Additivity and Scaling

Two basic cumulant rules explain the sample-size dependence. If $X$ and $Y$ are independent, the cumulants of their sum add. If $a$ and $b$ are constants, multiplying by $a$ multiplies the $r$th cumulant by $a^r$, while adding $b$ changes only the first cumulant. These rules follow directly from the formal moment series.

Theorem: Let $X_1,\ldots,X_n$ be independent random variables with finite absolute $r$th moments. For each $1\leq j\leq r$, $$ \kappa_j\left(\sum_{i=1}^n X_i\right)=\sum_{i=1}^n\kappa_j(X_i). $$ For constants $a,b$, $\kappa_1(aX+b)=a\kappa_1(X)+b$, and for $j\geq2$, $\kappa_j(aX+b)=a^j\kappa_j(X)$.

Proof. Form the moment series $M_X(t)=\sum_{j=0}^r E[X^j]t^j/j!$, interpreted only through degree $r$. Independence gives $E[(X+Y)^k]=\sum_{j=0}^k {k\choose j}E[X^j]E[Y^{k-j}]$ for $k\leq r$, so $M_{X+Y}(t)=M_X(t)M_Y(t)$ through degree $r$. Taking formal logarithms gives $\log M_{X+Y}=\log M_X+\log M_Y$ through that degree, proving additivity. For scaling and translation, the moment series of $aX+b$ is $e^{bt}M_X(at)$ through degree $r$. Its formal logarithm is $bt+\log M_X(at)$. Comparing coefficients gives the stated rules. Repeated application gives the result for any finite independent sum. $\square$

Theorem: Let $X_1,\ldots,X_n$ be iid with mean $\mu$, variance $0<\sigma^2<\infty$, and finite absolute $r$th moment. For $2\leq j\leq r$, define $$ W_n=\frac{\sum_{i=1}^n X_i-n\mu}{\sigma\sqrt n}. $$ Then $\kappa_1(W_n)=0$, $\kappa_2(W_n)=1$, and for $3\leq j\leq r$, $$ \kappa_j(W_n)=n^{1-j/2}\frac{\kappa_j(X_1)}{\sigma^j}. $$

Proof. By additivity and identical distribution, the $j$th cumulant of $\sum_{i=1}^n X_i$ is $n\kappa_j(X_1)$. Centering removes the first cumulant and leaves all higher cumulants unchanged. Scaling by $1/(\sigma\sqrt n)$ multiplies the $j$th cumulant by $(\sigma\sqrt n)^{-j}$. Thus for $j\geq2$, $$ \kappa_j(W_n)=n\kappa_j(X_1)(\sigma\sqrt n)^{-j} =n^{1-j/2}\frac{\kappa_j(X_1)}{\sigma^j}. $$ The first cumulant is zero by centering. For $j=2$, the formula gives $n\sigma^2/(\sigma^2 n)=1$. $\square$

Worked Example: Bernoulli Observations

Let $X_i$ be iid Bernoulli with success probability $p=1/4$. For a Bernoulli variable, $\mu=p$, $\sigma^2=p(1-p)$, $\kappa_3=p(1-p)(1-2p)$, and $\kappa_4=p(1-p)\{1-6p(1-p)\}$.

Here $\mu=1/4$ and $\sigma^2=(1/4)(3/4)=3/16$, so $\sigma=\sqrt{3}/4$. The third cumulant is $$ \kappa_3(X_i)=\frac14\frac34\left(1-\frac12\right)=\frac{3}{32}. $$ The standardized third cumulant is therefore $$ \lambda_3=\frac{3/32}{(\sqrt{3}/4)^3} =\frac{3/32}{3\sqrt{3}/64} =\frac{2}{\sqrt{3}}. $$ For the fourth cumulant, $$ \kappa_4(X_i)=\frac{3}{16}\left(1-\frac{18}{16}\right) =\frac{3}{16}\left(-\frac18\right) =-\frac{3}{128}. $$ Since $\sigma^4=(3/16)^2=9/256$, its standardized fourth cumulant is $$ \lambda_4=\frac{-3/128}{9/256}=-\frac23. $$ Consequently, $W_n$ has third cumulant $2/(\sqrt{3}\sqrt n)$ and fourth cumulant $-2/(3n)$. For $n=100$, these are $1/(5\sqrt3)\approx0.11547$ and $-1/150\approx-0.006667$, respectively.

Worked Example: Exponential Observations

Suppose $X_i$ are iid exponential variables with rate $2$. Their cumulant-generating function is $K_X(t)=-\log(1-t/2)$, for $t<2$. Expanding gives $K_X(t)=\sum_{j=1}^{\infty}t^j/(j2^j)$, hence $\kappa_j(X)=(j-1)!/2^j$. In particular, $\mu=1/2$, $\sigma^2=1/4$, $\kappa_3=1/4$, and $\kappa_4=3/8$.

The standardized cumulants are $$ \lambda_3=\frac{1/4}{(1/2)^3}=2, \qquad \lambda_4=\frac{3/8}{(1/2)^4}=6. $$ Thus the standardized sum has third cumulant $2/\sqrt n$ and fourth cumulant $6/n$. At $n=25$, these are $2/5=0.4$ and $6/25=0.24$. The fourth-cumulant contribution is smaller in its dependence on $n$, though its numerical size at a particular sample size also depends on the distribution-specific value of $\lambda_4$.

Worked Example: Normal Observations

If $X_i\sim N(\mu,\sigma^2)$, then $K_{X_i}(t)=\mu t+\sigma^2t^2/2$. There are no terms of degree three or higher, so every cumulant $\kappa_j(X_i)$ for $j\geq3$ is zero. The scaling theorem gives $\kappa_j(W_n)=0$ for all $j\geq3$, for every $n$. In this case the standardized sum is exactly $N(0,1)$, not merely approximately normal.

Which Cumulants Enter Each Edgeworth Order?

Write $\lambda_j=\kappa_j(X_1)/\sigma^j$ for $j\geq3$. The scaling theorem says that the cumulants of $W_n$ satisfy $\kappa_j(W_n)=\lambda_j n^{1-j/2}$. In particular, the third cumulant is $\lambda_3n^{-1/2}$, the fourth is $\lambda_4n^{-1}$, and the fifth is $\lambda_5n^{-3/2}$.

This scaling can be translated into an order-by-order expansion of the characteristic function. The following is an algebraic expansion near zero, or equivalently a formal expansion organized by powers of $n^{-1/2}$. It identifies the terms that occur; it does not assert a uniform approximation or provide the regularity conditions for a valid Edgeworth expansion.

Result: Through order $n^{-1}$, the characteristic-function expansion has the formal form $$ \phi_{W_n}(t) =e^{-t^2/2}\left[ 1+\frac{\lambda_3(it)^3}{6\sqrt n} +\frac{1}{n}\left\{ \frac{\lambda_4(it)^4}{24} +\frac{\lambda_3^2(it)^6}{72} \right\} \right] +\text{terms of higher formal order}. $$ Thus the first-order term is driven by skewness $\lambda_3$. At second order, the contributing terms are the fourth cumulant $\lambda_4$ and the square of the third cumulant $\lambda_3^2$.

To derive the bookkeeping, use the cumulants in the logarithm of the characteristic function. Its quadratic term is $-t^2/2$; the next terms are $\lambda_3(it)^3/(6\sqrt n)$ and $\lambda_4(it)^4/(24n)$. Exponentiating these terms gives, through order $n^{-1}$, $$ \exp\left\{\frac{\lambda_3(it)^3}{6\sqrt n} +\frac{\lambda_4(it)^4}{24n}+\cdots\right\} =1+\frac{\lambda_3(it)^3}{6\sqrt n} +\frac{\lambda_4(it)^4}{24n} +\frac12\left\{\frac{\lambda_3(it)^3}{6\sqrt n}\right\}^2 +\text{higher formal orders}. $$ The squared term is $\lambda_3^2(it)^6/(72n)$, which proves the stated order classification. The fifth cumulant is of order $n^{-3/2}$, and products involving the third and fourth cumulants are also of order $n^{-3/2}$; neither contributes through order $n^{-1}$.

Moment Conditions and a Common Pitfall

Cumulant scaling is conditional on existence. For each $r$, the formula for the $r$th cumulant requires moments through order $r$; a convenient sufficient condition is $E[|X_1|^r]<\infty$. Finite variance alone justifies the first two cumulants, but it says nothing about whether the third or fourth cumulant exists.

For example, a Pareto variable with tail $P(X>x)=x^{-\alpha}$ for $x\geq1$, where $2<\alpha<3$, has finite second moment but infinite third absolute moment. Its variance exists, so it can be centered and standardized, but its third cumulant is not finite. It is therefore incorrect to assert that every finite-variance iid sum has a third cumulant equal to a constant times $n^{-1/2}$. The correct statement is: whenever the $r$th cumulant exists, its standardized-sum scaling is $n^{1-r/2}$ times the standardized single-observation cumulant.

A second pitfall is to treat the scaling hierarchy as a proof that a particular Edgeworth approximation is valid. The hierarchy only tracks formal orders. Turning the expansion into an approximation to a density or distribution function requires additional assumptions and justification for the relevant analytic steps.

Key takeaway: For iid observations with finite positive variance, and whenever the required cumulants exist, the $r$th cumulant of the standardized sum is $n^{1-r/2}\kappa_r(X_1)/\sigma^r$. This makes skewness a first-order feature and makes kurtosis and squared skewness second-order features.

Check Your Understanding

Use the scaling rules and moment conditions to answer the following.

If the standardized third cumulant of one observation is $\lambda_3$, what is the third cumulant of the standardized sum of $n$ iid observations?
Why does the second-order characteristic-function expansion contain a term involving $\lambda_3^2$, even though it is a third-cumulant effect?
If the fourth cumulant of one observation is zero but its third cumulant is nonzero, which second-order contribution can remain?
What moment condition is sufficient to define and scale the fourth cumulant?
Why is finite variance alone insufficient to claim an $n^{-1/2}$ scaling for the third cumulant?