It is known that every infinitely divisible random variable is the limit in law of a sequence of compound Poisson random variables (see for instance Theorem 1.2.18 of Lévy Processes and Stochastic Calculus).

This can be shown as follows. Let $X$ be a infinitely divisible random variable with characteristic function $\Phi_X(t) = \mathbb{E}[\mathrm{e}^{\mathrm{i} t X}]$. Since $X$ is infinitely divisible, $\Phi^{1/n}$ is a valid characteristic function. Hence, for every $n \in \mathbb{N}$, we can define $X_n$ the compound Poisson random variable* with characteristic function $\exp\left(n\left(\Phi_X^{1/n}(t)-1 \right)\right)$. Moreover, we have $\Phi_{X_n}(t) \rightarrow \Phi_X(t)$ for all $t\in\mathbb{R}$, therefore $X_n \rightarrow X$ in law.

*Reminder: a compound Poisson random variable has a characteristic function of the form $\exp\left(\lambda\left(\Phi(t)-1\right)\right)$ where $\Phi$ is a characteristic function itself and $\lambda >0$.

Can we quantify the rate at which the convergence of $X_n$ to $X$ holds?

I am thinking, for instance, of something in the spirit of the Berry-Esseen theorem, bounding $\sup_x \lvert F_n(x) - F(x) \rvert $ with $F$ (resp. $F_n$) the cdf of $X$ (resp. $X_n$).

More generally, I would be interested by any result that quantifies the rate of convergence in law.

  • $\begingroup$ Hi, if you have not found anything by now (which I doubt) maybe this paper and references therein could be helpful to you: link.springer.com/article/10.1007/s10986-014-9225-0 $\endgroup$
    – Gytis
    Nov 10, 2017 at 15:54
  • $\begingroup$ $$\begin{align} & \text{“It is known that every infinitely divisible} \\ & \phantom{“} \text{random variable is the limit in law of a} \\ & \phantom{“} \text{sequence of compound Poisson random} \\ & \phantom{\text{“}} \text{variables”} \end{align}$$ A compound Poisson random variable cannot have a negative cumulant of even degree because its cumulants are equal to the moments of the distribution that is compounded. Would this mean that distributions with nonnegative even-degree cumulants can converge in law to a distribution with a negative even-degree cumulant? So there's a discontinuity? $\endgroup$ Jan 1 at 19:55
  • $\begingroup$ (My comment above presupposes the existence of an infinitely divisible law with a negative cumulant of even degree. I don't know an example off hand. But once I asked about this, I think on sci.math.research, and if I recall correctly, it may have been Herman Rubin who posted an example. But I don't remember what it was.) $\endgroup$ Jan 1 at 19:58

2 Answers 2


You might want to check this recent preprint:


and Section 4 in particular (Theorems 4.2, 4.4 and 4.5 and Proposition 4.4).


The book "Approximation Methods in Probability Theory" by Vydas Čekanavičius seems to be what you are looking for. From introduction:

"This book presents a wide range of well-known and less common methods used for estimating the accuracy of probabilistic approximations, including the Esseen type inversion formulas, the Stein method as well as the methods of convolutions and triangle function."



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.