The $n$th **cumulant** $\kappa\_n$ of a probability distribution for $n\ge2$ is functional that is a polynomial in the first $n$ **moments** of the distribution, that has the properties of $(1)$ homogeneity, $(2)$ translation invariance, and $(3)$ additivity.


If $X\_1,\ldots,X\_m$ are independent random variables and $c$ is a constant (so $c$ is *not random*) then


* $\kappa\_n(cX\_1) = c^n \kappa\_n(X\_1),$
* $\kappa\_n(c+X\_1) = \kappa\_n(X\_1),$
* $\kappa\_n(X\_1+\cdots+X\_m) = \kappa\_n(X\_1)+\cdots + \kappa\_n(X\_m).$


For example, the fourth cumulant of the distribution of a random variable $X$ is the fourth central moment minus three times the square of the second central moment:
\begin{align}
\kappa\_4(X) & = \operatorname E\big( (X-\operatorname E(X))^4\big) - 3\big(\operatorname{var}(X)\big)^2 \\[4pt]
& = \operatorname E\big( (X-\operatorname E(X))^4\big) - 3\big(\operatorname E\big((X-\operatorname E(X)\big)^2\big)^2.
\end{align}


(These properties plus one trivial property characterize the cumulants. That one trivial property is this: In the aforementioned polynomial, the coefficient of the $n$th central moment is $1.$)


In about 1970 David Brillinger proved the **[law of total cumulance](https://en.wikipedia.org/wiki/Law_of_total_cumulance).**


The **Poisson distribution** with expected value $\lambda$ assigns to each nonnegative integer $n$ the probability $\dfrac{\lambda^n e^{-\lambda}} {n!}.$


Suppose $X\_1,X\_2,X\_3,\ldots$ are independent identically distributed random variables, with finite moments of all orders, and $N$ is a random variable with a Poisson distribution with expected value $\lambda.$


Let $\displaystyle Y= \sum\_{n=0}^n X\_n.$


Then $Y$ has a **compound Poisson distribution** and the distribution of any of $X\_1,X\_2,X\_3,\ldots$ is the distribution that gets compounded.


A corollary of the law of total cumulance is this:



> 
> The sequence of cumulants of the compound Poisson distribution is the same as the sequence of moments of the distribution that gets compounded. (So in particular, for example, the cumulants of even order of a compound Poisson distribution are nonnegative.)
> 
> 
> 


**My question is** whether this corollary appeared in the literature before I added it to the linked Wikipedia article in 2005?



Was this proposition on cumulants of compound Poisson distributions known before I put it into a Wikipedia article?
This is a simple fact, below is a short proof. It is certainly very-well known for quite some time, a sample reference is formula (6.6) in


* Cacoullos T. (1989) *Generating Functions. Characteristic Functions*. In: *Exercises in Probability*. Problem Books in Mathematics. Springer, New York, NY. [DOI:10.1007/978-1-4612-4526-1\_6](https://doi.org/10.1007/978-1-4612-4526-1_6)




---


We have $\log \mathbb E e^{t N} = e^t - 1$. Thus, if we denote the moment generating function of $X$ by $M(t) = \mathbb E e^{tX}$, then the cumulant generating function of $Y$ is
$$\begin{aligned} \log \mathbb E e^{tY} & = \log \mathbb E \biggl( \prod\_{i = 1}^N e^{t X\_i} \biggr) \\ & = \log \mathbb E \biggl( \mathbb E \biggr( \prod\_{i = 1}^N e^{t X\_i} \bigg| N \biggr) \biggr) \\ & = \log \mathbb E (M(t))^N = M(t) - 1 . \end{aligned}$$


