Radius of convergence of cumulant generating function

Question

Recall that for a random variable $X$ with a moment generating function $M_X(t)$ the cumulant generating function is defined as \begin{align} K_X(t)=\log M_X(t) \end{align}

The Taylor expansion of $K_X(t)$ is given \begin{align} K_X(t)= k_1 t+ k_2 \frac{t^2}{2}+ k_3 \frac{t^3}{3!}+.. \end{align} where $k_i$ are the cumulants.

My question: Can we determine the radius of convergence of $K_X(t)$ if we know the radius of convergence of $M_X(t)$?

For example, in these slides, it is claimed both moments generating functions and cumulant generating functions have the same radius of convergence.

Answer 1

Of course, this is not correct. As a simplest example, let $X$ be a random variable which takes only values $\{0,\ldots, n\}$, then the moment generating function is a polynomial of $e^t$, of degree $n$, therefore its radius of convergence is infinite. Any polinomial with positive coefficients which add to $1$ can occur. But $\log M(t)$ has finite radius of convergence since a polynomial $P$ of degree $n\geq 1$ has some zeros in the complex plane. So $P(e^t)$ also has zeros, unless $P$ is a monomial.

In general, the radius of convergence for $K(t)$ is the distance from the origin to the closest zero of $M(t)$.