Project Euclid

A probability distribution function $F$ is said to be infinitely divisible if, for every integer $n$, there is a distribution function $F_n$ such that $F$ is the $n$-fold convolution of $F_n$. If $F$ is infinitely divisible its characteristic function can be written in the Khinchin-Levy canonical form: \begin{equation*}\tag{1}f(u) = \exp\big\{i\gamma u + \int^\infty_{-\infty}\big(e^{iux} - 1 - \frac{iux}{1 + x^2}\big)\frac{1 + x^2}{x^2} dG(x)\big\},\end{equation*} where $\gamma$ is a constant and $G$ is a bounded, non-decreasing function. Hartman and Wintner [2] proved that if $G$ is discrete, then the distribution function $F$ is pure, i.e., it is either absolutely continuous or discrete or continuous singular, and an example was given of each of these types of pure distributions which was determined by a discrete $G$. In the example of a discrete $G$ producing a continuous singular $F, G$ was given jumps of size $(1/N)^{2j}$ at $\pm 1/N^j, j = 1, 2, \cdots$, where $N$ is a positive integer $\geqq 2$. It was first proved that $F$ must be continuous; then it was proved that $f(u)$ does not converge to zero as $|u| \rightarrow \infty$, thus violating the conclusion of the Riemann-Lebesgue lemma. The main purpose of this paper is to give sufficient conditions that a continuous singular $F$ be obtained from a discrete $G$. These conditions are not too broad; for example, they do not include the example cited above. However, these conditions should be of considerable interest in that they are obtained by purely probabilistic methods, there being no use made of the Riemann-Lebesgue lemma, and thus supply more insight into the structure of continuous singular infinitely divisible distributions. In Section 2 a lemma is proved which might be of independent interest. This lemma is used to prove Theorem 1 in Section 3, which gives sufficient conditions that $F$ be continuous singular. In Section 4 a theorem is proved, giving sufficient conditions that every $m$-fold convolution of $F, F^{\ast m}$, be continuous singular.