Bound for cumulants of bounded random variables

Question

Let $X$ be a random variable taking values in $[-1,1]$. The cumulant generating function is defined as $$ K(t) = \log \mathbb{E} [e^{tX}], $$ and the cumulants of $X$ are $$ \kappa_n = K^{(n)}(0). $$

I would like to know the best possible bounds on $\kappa_n$. I would be interested in any bound of the form $$ |\kappa_n| \leq C^n $$ which holds for bounded random variables $X$. Does this bound hold?

With my limited understanding of cumulants, I would guess that the biggest cumulants would come from the Bernoulli random variable $X$ which takes the values $\pm 1$ equally often. In this case one has $K(t)=\tanh(t)$, and the cumulants decay to $0$ as $n\to\infty$. I expect that there should be worse examples, but I cannot find them.

Answer 1

Using the relationship between the cumulants and the moments,

$$\kappa(n) = \sum_{k=1}^n(-1)^{k-1}(k-1)!B_{n,k}(\mu_1, \dots, \mu_{n-k+1})$$

where $\mu_1, \dots, \mu_n$ are the central moments and $B_{n,k}(x_1, \dots, x_{n-k+1})$ are the incomplete Bell polynomials. We have $|\mu_i| \leq 1$ for all $i$ for random variables bounded in $[-1,+1]$, and the Bell polynomials can be upper bounded combinatorially. We have

$$B_{n,k}(1,\dots,1) = \genfrac{\{}{\}}{0pt}{}{n}{k}$$

where $\genfrac{\{}{\}}{0pt}{}{n}{k}$ is the Stirling number of the second kind. The quantity $k! \genfrac{\{}{\}}{0pt}{}{n}{k}$ counts the number of surjections from $[n]$ to $[k]$. It can be upper bounded by $k^n$. Overall, we obtain $$ |\kappa(n)| \leq n^n.$$

Maybe this argument could be improved to show that the balanced Bernoulli distribution is maximal, as you predict. On the other hand, for that distribution, $\mu_i = 1$ if $i$ is even and $0$ if $i$ is odd, so we can compute that $|\kappa(n)| \approx B_{n,n/2}(0,1,\dots,0,1) \approx n^{n/2}$. This shows that a bound of the form $|\kappa(n)| \leq C^n$ is not possible.