Common Errors in Cumulant Algebra

Why Cumulant Algebra Needs Checks

Cumulants are useful partly because several calculations that are complicated for moments become simple: centering leaves cumulants of order two and above unchanged, scaling has a predictable effect at each order, and cumulants add for independent sums. Each statement has conditions and a specific meaning. A missed rate-versus-scale convention, a dropped power of a scale factor, or an unverified independence assumption can produce an answer that looks plausible but is wrong.

This tutorial focuses on how to diagnose those errors. Write $\kappa_r(X)$ for the cumulant of order $r$. When the moment-generating function $M_X(t)=E[e^{tX}]$ is finite on an open interval containing zero, define the cumulant-generating function $K_X(t)=\log M_X(t)$ and set $\kappa_r(X)=K_X^{(r)}(0)$. In particular, $\kappa_1(X)=E[X]$ and $\kappa_2(X)=\operatorname{Var}(X)$.

Definition: If $X$ has moments through order $r$, its cumulants through order $r$ can also be defined from the coefficients through degree $r$ in the formal power series $$ \log\left(1+\sum_{j=1}^{r}\frac{E[X^j]}{j!}t^j\right). $$ This definition requires no convergent moment-generating function. When the moment-generating function is finite on a neighborhood of zero, the formal definition agrees with derivatives of $K_X$. A calculation using $K_X$ as an actual function requires that neighborhood condition.

The distinction in this definition matters: having a cumulant defined from finitely many moments is not the same as having a cumulant-generating function that can be differentiated around zero. The latter is the setting for the generating-function arguments used below.

Parameterization, Centering, and Scaling

A parameter may describe a rate or a scale. These conventions are reciprocals, so inserting the same numerical parameter into formulas for both conventions can make every cumulant wrong. Centering and scaling cause a different class of errors: translating a variable affects only its first cumulant, whereas multiplying it by $a$ multiplies its $r$th cumulant by $a^r$.

Theorem: Suppose $K_X(t)$ is finite on an open interval containing zero, and let $a,b$ be real constants. Then $$ \kappa_1(aX+b)=a\kappa_1(X)+b,\qquad \kappa_r(aX+b)=a^r\kappa_r(X)\quad (r\geq 2). $$

Proof. The moment-generating function of $aX+b$ is $e^{bt}M_X(at)$ wherever the right side is finite. Its cumulant-generating function is therefore $bt+K_X(at)$. The first derivative at zero is $b+aK_X'(0)$. For $r\geq2$, the $r$th derivative of the linear term $bt$ is zero, and repeated differentiation of $K_X(at)$ gives $a^rK_X^{(r)}(at)$. Evaluating at zero proves both identities. This also covers $a=0$: the transformed variable is constant, with first cumulant $b$ and higher cumulants zero. $\square$

Worked Example: Rate Is Not Scale

Let $X$ have an exponential distribution with rate $2$, so its density is $2e^{-2x}$ for $x>0$. Calculate its first three cumulants and identify the error from treating $2$ as a scale.

Direct integration, or the exponential moment-generating function, gives $M_X(t)=2/(2-t)$ for $t<2$. Thus $K_X(t)=-\log(1-t/2)$. Differentiating, $K_X'(t)=1/(2-t)$, $K_X''(t)=1/(2-t)^2$, and $K_X'''(t)=2/(2-t)^3$.

At zero, the cumulants are $\kappa_1(X)=1/2$, $\kappa_2(X)=1/4$, and $\kappa_3(X)=1/4$. These agree with the exponential rate-$\lambda$ formula $\kappa_r=(r-1)!/\lambda^r$, here with $\lambda=2$.

If $2$ were instead a scale parameter, the corresponding exponential distribution would have rate $1/2$. Its cumulants would be $(r-1)!2^r$, giving $2,4,16$ for orders one through three. Those are not the cumulants of the stated rate-$2$ variable. The check is also visible in the mean: a rate-$2$ exponential variable has mean $1/2$, not $2$.

Worked Example: Center First, Then Check the Order

Let $X$ equal $0$ or $2$, each with probability $1/2$. Compare its cumulants through order four with those of its centered version $Z=X-1$.

The mean is $E[X]=1$. Since $Z$ equals $-1$ or $1$ with equal probabilities, $E[Z]=0$, $E[Z^2]=1$, $E[Z^3]=0$, and $E[Z^4]=1$. The cumulant formulas through order four are $\kappa_1=E[X]$, $\kappa_2=E[(X-E[X])^2]$, $\kappa_3=E[(X-E[X])^3]$, and $\kappa_4=E[(X-E[X])^4]-3\operatorname{Var}(X)^2$.

For $Z$, these give $\kappa_1(Z)=0$, $\kappa_2(Z)=1$, $\kappa_3(Z)=0$, and $\kappa_4(Z)=1-3=-2$. The affine transformation rule gives the same cumulants of orders two through four for $X=Z+1$, while $\kappa_1(X)=1$.

A common centering error is to assume that translating by one leaves the first cumulant unchanged too. It does not: the first cumulant is the mean and changes by the translation amount. A different error is to identify the fourth cumulant with the fourth central moment. Here that would give $1$, whereas the fourth cumulant is $-2$.

Worked Example: Scaling Raises the Factor to the Cumulant Order

Let $X$ be exponential with rate $1$, and set $Y=3X$. Find the first three cumulants of $Y$.

For rate $1$, $K_X(t)=-\log(1-t)$, so differentiation at zero gives $\kappa_1(X)=1$, $\kappa_2(X)=1$, and $\kappa_3(X)=2$. By the scaling rule, $\kappa_1(Y)=3$, $\kappa_2(Y)=3^2(1)=9$, and $\kappa_3(Y)=3^3(2)=54$.

The result can be checked from $K_Y(t)=K_X(3t)=-\log(1-3t)$, whose first three derivatives at zero are $3$, $9$, and $54$. Multiplying every cumulant merely by $3$ would give an incorrect variance and third cumulant.

Existence: Moments Are Not the Same as a CGF

A cumulant calculation must use a definition that is available for the variable in question. If moments through order $r$ exist, the cumulant through order $r$ is defined by the corresponding moment polynomial. But a derivation that differentiates $K_X(t)$ as a moment-generating function requires $M_X(t)$ to be finite on an open interval containing zero. Finite moments alone do not guarantee this.

Worked Example: A Lognormal Variable Has Moments but No CGF Near Zero

Let $Z\sim N(0,1)$ and $X=e^Z$. For every positive integer $r$, the normal moment-generating function gives $E[X^r]=E[e^{rZ}]=e^{r^2/2}$. Thus every positive integer moment of $X$ is finite, and its cumulants can be defined from moments at every finite order.

However, for $t>0$, $$ E[e^{tX}] =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \exp\left(te^z-\frac{z^2}{2}\right)\,dz. $$ As $z$ tends to infinity, $te^z-z^2/2$ tends to infinity. In particular, the integrand is eventually greater than $1$, so its integral over that infinite tail diverges. Hence $M_X(t)=\infty$ for every $t>0$, and there is no open interval around zero on which the moment-generating function is finite.

The correction is not to say that the lognormal has no moment-defined cumulants. It does have cumulants from its finite moments. Rather, one cannot justify a cumulant calculation for this distribution by assuming a CGF exists near zero. For a variable with a Cauchy distribution, the problem is stronger: even its first absolute moment is infinite, so the ordinary moment-defined cumulants are not available.

Additivity Requires Independence

The cumulant-generating function of a sum is not generally the sum of the individual cumulant-generating functions. If $X$ and $Y$ are independent and both moment-generating functions are finite near zero, then $M_{X+Y}(t)=M_X(t)M_Y(t)$; taking logarithms yields $K_{X+Y}(t)=K_X(t)+K_Y(t)$. Differentiating gives additivity of every order. Without independence, the joint CGF contains mixed terms that must not be discarded.

Theorem: If $X,Y$ are independent and their moment-generating functions are finite on neighborhoods of zero, then $$ \kappa_r(X+Y)=\kappa_r(X)+\kappa_r(Y),\qquad r\geq1. $$ For dependent variables with a joint CGF finite near $(0,0)$, the variance identity is instead $$ \operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)+2\operatorname{Cov}(X,Y). $$

Proof. For independent $X,Y$, $E[e^{t(X+Y)}]=E[e^{tX}]E[e^{tY}]$. Taking logarithms gives the stated identity for the CGFs on a sufficiently small neighborhood of zero, and differentiating $r$ times at zero proves cumulant additivity. For the variance identity, expand $\operatorname{Var}(X+Y)=E[((X-E[X])+(Y-E[Y]))^2]$. The square has two individual squared terms and the cross term $2E[(X-E[X])(Y-E[Y])]=2\operatorname{Cov}(X,Y)$, proving the formula. Independence makes this covariance zero, but dependence need not. $\square$

Worked Example: The Same Variable Added to Itself

Let $X$ be exponential with rate $1$, and set $Y=X$. Then $Y$ and $X$ have identical marginal distributions, but they are not independent. Since $X+Y=2X$, the scaling rule gives $\kappa_1(X+Y)=2$, $\kappa_2(X+Y)=4$, and $\kappa_3(X+Y)=8\kappa_3(X)=16$, using $\kappa_3(X)=2$.

If one incorrectly applied the independent-sum rule, one would obtain $2$, $2$, and $4$, respectively. The variance error can be checked directly: $\operatorname{Var}(2X)=4$, while the two marginal variances sum to $1+1=2$. The missing term is $2\operatorname{Cov}(X,X)=2\operatorname{Var}(X)=2$.

For the third cumulant, the joint CGF makes the omitted terms explicit. It is $K_{X,Y}(s,t)=K_X(s+t)$. The mixed third-order cumulants $\kappa(X,X,Y)$ and $\kappa(X,Y,Y)$, defined by the corresponding mixed derivatives of this joint CGF, both equal $K_X'''(0)=2$. Differentiating $K_{X+Y}(t)=K_{X,Y}(t,t)$ three times gives $\kappa_3(X+Y)=\kappa_3(X)+\kappa_3(Y)+3\kappa(X,X,Y)+3\kappa(X,Y,Y)=2+2+6+6=16$. The independent-sum formula is valid when the variables are independent; matching marginal distributions is not enough.

Common Mistakes and a Reliable Check

Some incorrect calculations come from importing rules for moments into cumulant algebra. For example, raw second moments are not additive even for independent variables: if $X,Y$ are independent with finite second moments, then $E[(X+Y)^2]=E[X^2]+E[Y^2]+2E[X]E[Y]$. Cumulant additivity for independent variables concerns cumulants, not every moment. In particular, the second cumulant is a variance, whereas the raw second moment also reflects the mean.

Before accepting a cumulant identity, check the parameterization, the order of the cumulant, and which definition is being used. Then verify any independence assumption and confirm that the required moments or CGFs exist. A compact error-checking sequence is:

Fix the convention.
For distributions described by a rate or scale, write which parameter is being used before substituting into a formula.

Track the order.
Under multiplication by $a$, check for $a^r$ at order $r$; under translation, check that only the first cumulant changes.

Check existence and assumptions.
Determine whether the calculation uses finite moments or a CGF near zero, and do not use independent-sum additivity without independence.

These checks are useful because a false identity can still produce a tidy expression. Testing it on a constant, a centered variable, or a dependent pair such as $X+X$ often reveals the missing condition immediately.

Key takeaway: Cumulant identities are precise, conditional rules—not automatic analogies with moments. Keep rate and scale distinct, use the order-specific affine rule, match existence assumptions to the definition, and include dependence terms unless independence has been established.

Check Your Understanding

For each question, identify the relevant assumption or algebraic rule before calculating.

An exponential distribution is specified with rate $4$. What is its variance, and what mistaken value could result from treating $4$ as its scale?
If $Y=X-5$, which cumulant orders change, and by how much does the first cumulant change?
If $Y=-2X$, express $\kappa_3(Y)$ in terms of $\kappa_3(X)$. Why is multiplying by $-2$ insufficient?
A random variable has finite moments of every integer order, but its moment-generating function is infinite for every positive argument. Can its cumulants be defined from moments? Can a CGF-neighborhood argument be used?
Two variables have the same distribution. Is that enough to conclude that the cumulants of their sum equal twice the cumulants of either variable? State the missing condition.