Factorial Cumulants for Count Distributions

Why Use Factorial Cumulants?

For a count-valued random variable, the probability-generating function is naturally centered at $s=1$: its derivatives there give factorial moments. Taking its logarithm instead produces quantities that describe the distribution’s cumulant structure in the same way that the ordinary cumulant-generating function does, but in coordinates adapted to counts. This is especially revealing for a Poisson law: its log probability-generating function is linear in $s-1$, so all factorial cumulants beyond the first vanish.

The distinction between ordinary and factorial cumulants matters in applications. Ordinary cumulants are derivatives of the log moment-generating function at zero, while factorial cumulants are derivatives of the log probability-generating function at one. Both summarize the same distribution when the required moments exist, but they organize information differently. The conversions between them involve Stirling numbers.

Definition and Existence

Let $X$ be a nonnegative integer-valued random variable, and let its probability-generating function be $G(s)=E[s^X]$ where this expectation is finite. Its log probability-generating function, when defined, is $H(u)=\log G(1+u)$. The variable $u$ shifts the argument so that derivatives at $u=0$ are derivatives at $s=1$.

Definition: If $G(s)$ is finite in a neighborhood of $s=1$ and $G(1)=1$, the $r$th factorial cumulant of $X$, for integer $r\geq 1$, is $$ \kappa_{[r]}(X)=H^{(r)}(0) =\left.\frac{d^r}{ds^r}\log G(s)\right|_{s=1}. $$ Here brackets distinguish factorial cumulants from ordinary cumulants $\kappa_r(X)$.

A sufficient regularity condition for all the derivatives and power-series manipulations below is that $E[e^{\delta X}]<\infty$ for some $\delta>0$. Indeed, then $G(s)$ is finite for real $s$ in an interval containing 1, and its power series is analytic in a neighborhood of 1. Since $G(1)=1$, continuity ensures that $G$ is nonzero in a sufficiently small neighborhood of 1, so its logarithm has derivatives there. The factorial cumulants through order $r$ require only the corresponding derivatives to exist; the stronger condition is a convenient common assumption for the full sequence.

For comparison, the $r$th factorial moment is $E[(X)_r]$, where $(X)_r=X(X-1)\cdots(X-r+1)$. Differentiating $G$ gives $G^{(r)}(1)=E[(X)_r]$ when the expectation is finite. Factorial cumulants instead differentiate $\log G$; starting at order two, they are generally not factorial moments.

Conversion to Ordinary Cumulants

Assume the moment-generating function of $X$ is finite in a neighborhood of zero, and write $K(t)=\log E[e^{tX}]$. Its derivatives at zero are the ordinary cumulants: $\kappa_r(X)=K^{(r)}(0)$. Since $G(e^t)=E[e^{tX}]$, the two log-generating functions are connected by $K(t)=H(e^t-1)$. This composition gives the Stirling-number conversion.

Theorem: Suppose the moment-generating function of $X$ is finite in a neighborhood of zero. For every integer $r\geq1$, $$ \kappa_r(X)=\sum_{j=1}^{r} S(r,j)\,\kappa_{[j]}(X), \qquad \kappa_{[r]}(X)=\sum_{j=1}^{r}s(r,j)\,\kappa_j(X). $$ Here $S(r,j)$ are Stirling numbers of the second kind, defined by $x^r=\sum_{j=0}^r S(r,j)(x)_j$, and $s(r,j)$ are the signed Stirling numbers of the first kind, defined by $(x)_r=\sum_{j=0}^r s(r,j)x^j$. Terms with $j=0$ do not contribute for $r\geq1$.

Proof. Analyticity under the stated moment-generating-function condition permits Taylor expansions near zero. Expand $H$ at zero and substitute $e^t-1$:

K(t)=H(e^t-1) =\sum_{j=1}^{\infty}\kappa_{[j]}(X)\frac{(e^t-1)^j}{j!}.

The exponential generating function for the Stirling numbers of the second kind is $(e^t-1)^j/j!=\sum_{r=j}^{\infty}S(r,j)t^r/r!$. Consequently, the coefficient of $t^r/r!$ in the displayed expansion of $K(t)$ is $\sum_{j=1}^r S(r,j)\kappa_{[j]}(X)$. That coefficient is $K^{(r)}(0)=\kappa_r(X)$, proving the first identity.

For the inverse identity, substitute $t=\log(1+u)$ into $K(t)=H(e^t-1)$, obtaining $H(u)=K(\log(1+u))$. Expand $K$, then use the signed first-kind Stirling-number identity $(\log(1+u))^j/j!=\sum_{r=j}^{\infty}s(r,j)u^r/r!$. The coefficient of $u^r/r!$ is $\sum_{j=1}^r s(r,j)\kappa_j(X)$, and it is also $H^{(r)}(0)=\kappa_{[r]}(X)$. This proves the inverse identity. $\square$

The first few conversions make the structure concrete:

\begin{aligned} \kappa_1&=\kappa_{[1]},\\ \kappa_2&=\kappa_{[1]}+\kappa_{[2]},\\ \kappa_3&=\kappa_{[1]}+3\kappa_{[2]}+\kappa_{[3]},\\ \kappa_4&=\kappa_{[1]}+7\kappa_{[2]}+6\kappa_{[3]}+\kappa_{[4]}. \end{aligned}

Because $\kappa_1=E[X]$ and $\kappa_2=\operatorname{Var}(X)$, the first two factorial cumulants satisfy $\kappa_{[1]}=E[X]$ and $\kappa_{[2]}=\operatorname{Var}(X)-E[X]$. Thus the second factorial cumulant records the difference between variance and mean for a count distribution.

Worked Example: Factorial Cumulants of a Poisson Variable

Let $X\sim\operatorname{Poisson}(3)$. Its probability-generating function is $G(s)=\exp(3(s-1))$, so

H(u)=\log G(1+u)=\log\{\exp(3u)\}=3u.

Therefore $H'(0)=3$, and every derivative $H^{(r)}(0)$ for $r\geq2$ is zero. The factorial cumulants are $\kappa_{[1]}=3$ and $\kappa_{[r]}=0$ for every $r\geq2$.

The conversion theorem checks this pattern against ordinary cumulants. For any $r\geq1$, only the $j=1$ term remains, and $S(r,1)=1$, so $\kappa_r=3$. Thus vanishing higher factorial cumulants does not mean vanishing higher ordinary cumulants.

Worked Example: A Binomial Count

Let $X\sim\operatorname{Binomial}(5,0.2)$. Its probability-generating function is $G(s)=(0.8+0.2s)^5$. Writing $s=1+u$ gives

H(u)=5\log(1+0.2u).

The derivative of order $r$ of $\log(1+au)$, at zero, is $(-1)^{r-1}(r-1)!a^r$. Taking $a=0.2$, the first three factorial cumulants are

\begin{aligned} \kappa_{[1]}&=5(0.2)=1,\\ \kappa_{[2]}&=-5(0.2)^2=-0.2,\\ \kappa_{[3]}&=5\cdot2!(0.2)^3=0.08. \end{aligned}

The ordinary variance is $\kappa_2=\kappa_{[1]}+\kappa_{[2]}=1-0.2=0.8$, matching $np(1-p)=5(0.2)(0.8)=0.8$. The third ordinary cumulant is $\kappa_3=\kappa_{[1]}+3\kappa_{[2]}+\kappa_{[3]}=1-0.6+0.08=0.48$. This agrees with the binomial third cumulant $np(1-p)(1-2p)=5(0.2)(0.8)(0.6)=0.48$.

Worked Example: A Negative-Binomial Count

Let $X$ count the failures before two successes, with success probability $p=0.5$. Its probability-generating function is $G(s)=[p/(1-(1-p)s)]^2$. Set $u=s-1$; since $1-(1-p)(1+u)=p-(1-p)u$, this becomes

H(u)=-2\log\left(1-\frac{1-p}{p}u\right)=-2\log(1-u).

Differentiating at zero gives $\kappa_{[r]}=2(r-1)!$ for every $r\geq1$. In particular, $\kappa_{[1]}=2$, $\kappa_{[2]}=2$, and $\kappa_{[3]}=4$. The mean is $2$, while the variance is $\kappa_2=\kappa_{[1]}+\kappa_{[2]}=4$. The third ordinary cumulant is $\kappa_3=2+3(2)+4=12$. These values agree with the negative-binomial formulas $E[X]=2(1-p)/p=2$, $\operatorname{Var}(X)=2(1-p)/p^2=4$, and $\kappa_3=2(1-p)(2-p)/p^3=12$.

Why the Poisson Pattern Matters

A Poisson variable has factorial cumulants only at order one because its log probability-generating function is linear in $s-1$. This is a structural feature, not merely a convenient calculation. For other count distributions, nonzero higher factorial cumulants describe departures from that pattern. In particular, the second factorial cumulant compares variance with the mean: it is negative for the binomial example, zero for a Poisson variable, and positive for the negative-binomial example.

Factorial cumulants are useful when the count structure is central to the problem, because factorial powers $(X)_r$ are the quantities generated by derivatives of the probability-generating function at one. Their logarithmic counterparts retain a cumulant-like organization, and the Stirling conversions let you move back to ordinary cumulants whenever those are the more convenient summaries.

Common Pitfall

A frequent error is to treat factorial cumulants as though they were ordinary cumulants, or to infer that all cumulants vanish above order one when the higher factorial cumulants vanish. For $X\sim\operatorname{Poisson}(\lambda)$, the correct statements are $\kappa_{[1]}=\lambda$, $\kappa_{[r]}=0$ for $r\geq2$, and $\kappa_r=\lambda$ for every $r\geq1$. The conversion formula explains the difference: each ordinary cumulant receives a contribution from the first factorial cumulant because $S(r,1)=1$.

Key takeaway: Factorial cumulants are derivatives of $\log G(s)$ at $s=1$. Stirling numbers convert them to ordinary cumulants, and the Poisson law is characterized by a log probability-generating function linear in $s-1$, making all factorial cumulants above the first equal to zero.

Check Your Understanding

Use the definitions and conversion identities to answer these questions.

What function is differentiated to obtain the $r$th factorial cumulant, and at what point is it differentiated?
For a count distribution with mean $6$ and variance $9$, what is its second factorial cumulant?
If a Poisson variable has mean $\lambda$, state its factorial cumulants and its ordinary cumulants.
For the binomial example above, verify the variance using $\kappa_2=\kappa_{[1]}+\kappa_{[2]}$.
Why does a zero third factorial cumulant not, by itself, imply a zero third ordinary cumulant?