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It is a classical and well known problem that a random variable $X$ is not uniquely determined by its moments $\mathbb{E}(X_n)$. The moment problem is the problem of determining the probability density of a random variable in terms of its moments, as well as the uniqueness of such a density given by the moments. I have read this from the Algebraic Combinatorics and Computer Science of H. Crapo and D. Senato:

"In the first half of this century, the method of moments was replaced by Paul Levy by a more pliable method relying upon the characteristic function $\mathbb{E}(\mathrm{e}^{\mathrm{i}tX })$, which is used to this day to derive the limit theorems of probability in their sharpest form. There is however one drawback to the characteristic function: it has no obvious probabilistic significance. My teacher William Feller was aware that the religious invocation of characteristic functions is extraneous to probabilistic reasoning. He managed to avoid characteristic functions in his treatise on probability. To be sure, characteristic functions made an occasional appearance in the second volume, when he just could not do without them. However, it was his intention to write a third volume, dealing with Brownian motion and diffusion processes, in which characteristic functions would be relegated to the dustbin of history. Unfortunately, he died before he could accomplish this task."

Other important objects that can better represent many properties of random variables than moments are cumulants.

For $n\geq 1$, we consider a vector of real-valued random variables $X_{[n]}= (X_1,\ldots, X_n)$ such that $\mathbb{E}(|X_j|^{n}) <\infty, \forall\ j = 1,\ldots,n$. For every subset $b = \{j_1,\ldots, j_k\} \subset [n]=\{1,\ldots,n\}$ , one writes $X_b=(X_{j_1},\ldots, X_{j_k})$ and $X^{b}=X_{j_1}\times \ldots\times X_{j_k}$. From the $k$-dimensional vector $X_b$, one cane define the multivariate characteristicfunction of this vector as $$\phi_{X_b}(z_1,\ldots,z_k)=\mathbb{E}\Bigg[\exp\Big(\mathrm{i}\sum_{l=1}^{k}z_l X_{j_l}\Big)\Bigg].$$ The joint cumulant of the components of the vector $X_b$ is defined as \begin{align*} k(X_b)=(-\mathrm{i})^{k} \frac{\partial^{k} }{\partial z_1\ldots\partial z_k}\log\phi_{X_b}(z_1,\ldots,z_k)|_{z_1=\ldots=z_k=0}. \end{align*}

I know that cumulants are polynomials in moments, invariants by the translation and additive if the the components of the vector $X_b$ are partly independents. They have also common properties with characteristic and generating function in the sense they characterize uniquely the distribution of a random variable as also they can characterize the independence of random variables.

My question: why is a random variable is better described by its cumulants, which are combinatoric nature, than by its generating function or its characteristic function, which are analytical nature.

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    $\begingroup$ "Better described"? Why do you think so? $\endgroup$ Commented May 12, 2017 at 19:27
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    $\begingroup$ "Better" is an operation that takes in three inputs: a thing X, a thing Y it's being compared to, and a purpose P the comparison is being made for. $\endgroup$ Commented May 12, 2017 at 19:35

3 Answers 3

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The quote actually belongs to C.G.Rota.

Because cumulant sequences are closed under addition while moment sequences are not. That makes cumulant a more tractable algebraic structure altogether. Although umbral calculus provides a algebraic framework for both cumulants and factorial moments. [Di Nardo et.al]

Secondly, cumulants are more naturally regarded as tensors of components of the random variables. But you can still argue that moments can also be regarded as tensors since they have one-to-one transformation correspondence. [Morton&Lim]

Thirdly, cumulants are Mobius inversions from partition lattice ordered by inclusion of the sample space. [Speed] Thus adoption of cumulants actually simplifies many combinatorial proofs. [Rota]

But cumulants come from characteristic functions, which are Fourier transforms of random variables. Moments are coming from Laplace transforms of random variables, so you can argue that Fourier transforms are generally "better than" Laplace because they always exist...But I can also argue that Laplace is better because they are more tractable...(to me).

Reference

[Di Nardo et.al]Di Nardo, Elvira, and Domenico Senato. "An umbral setting for cumulants and factorial moments." European Journal of Combinatorics 27.3 (2006): 394-413.

[Morton&Lim] Morton, Jason, and Lek-Heng Lim. "Principal cumulant component analysis." preprint (2009). https://www.stat.uchicago.edu/~lekheng/work/pcca.pdf

[Speed] Speed, T. P. "Cumulants and partition lattices." Australian & New Zealand Journal of Statistics 25.2 (1983): 378-388.

[Rota] Rota, Gian-Carlo, and Jianhong Shen. "On the combinatorics of cumulants." Journal of Combinatorial Theory, Series A 91.1-2 (2000): 283-304.

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  • $\begingroup$ @Heny, thanks, notably for these references. But, my interrest is to compare the cumulants to characteristic function or generating function. For example why the characteristic function has no obvious probabilistic significance. $\endgroup$
    – Iliyo
    Commented May 13, 2017 at 7:17
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I think it is a bit misleading to contrast cumulants as combinatorial quantities with the characteristic function as analytic object. The characteristic function is an analytical device containing information about the moments. In the same way there is an analytic object (namely the logarithm of the characteristic function) which contains information about the cumulants. So I would say that there are moments and there are cumulants, dealing with them has often a combinatorial flavor, and there are analytic reformulations of moments and of cumulants which allow the use of more analytic tools (and, in particular, allow to deal with situations where no moments/cumulants exist). In some cases moments are better suited for the problem at hand, in some cases cumulants are.

In the classical case, the closeness between the analytic avatars of moments and of cumulants (the first is the characteristic function, the second is the logarithm of the characteristic function) might be a reason that one usually does not talk so much about the analytic version of cumulants. In free probability theory the difference between the Cauchy transform (the analytic function for moments) and the $R$-transform (the analytic function for free cumulants) is much bigger and the parallelism between the combinatorial and the analytical side of moments/cumulants is more visible.

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  • $\begingroup$ If one drops the positvity property to the variables which are operators in the non-commutative case, by the classial umbral calculus, it is preferable to work with cumulants than the characteristic function. The later is optimised by definition for anatycal tasks rather than combinatorial ones. $\endgroup$
    – Iliyo
    Commented Oct 29, 2017 at 23:52
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I would argue for the opposite, at least for stable distributions the random variable is better described by its characteristic function, which has a simple closed-form expression, than by its cumulants (which do not exist). In this case the characteristic function is also more helpful than the probability density, which lacks a closed-form expression.

The OP also asks for the "probabilistic significance" of the characteristic function. In the context of stochastic processes of Lévy type, the characteristic function has the socalled Lévy–Khintchine representation, which describes the time dependence of the random variable in terms of a linear drift, a Brownian motion and a superposition of independent Poisson processes with different jump sizes.

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    $\begingroup$ +1, A nice complement $\endgroup$
    – Henry.L
    Commented May 13, 2017 at 16:00
  • $\begingroup$ The cumulants have nice algebraic properties, especially the translation invariance. Notice that these are truly algebraic relations, which has very little to do with the positivity attribute of random variables (though in probability theory or statistics, positivity is extremely important). But, they are endowed with the algebraic spirit of a random variable. $\endgroup$
    – Iliyo
    Commented May 15, 2017 at 8:56

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