Infinitely divisible distribution

Remarks

Equivalently, for every $n \ge 1$ there is a probability measure $\mu_n$ whose $n$-fold convolution power is $\mu$; or, in terms of characteristic functions, $\varphi = \varphi_n^n$ for some characteristic function $\varphi_n$. The same definition applies to laws on $\R^d$. Example: the Poisson law with mean $\lambda$ is the law of a sum of $n$ independent Poisson variables with mean $\lambda / n$, since $e^{\lambda(e^{it} - 1)} = \left(e^{(\lambda/n)(e^{it} - 1)}\right)^n$. Example: the normal law with mean $m$ and variance $\sigma^2$ is the law of a sum of $n$ independent normal variables with mean $m/n$ and variance $\sigma^2/n$. Non-example: a Bernoulli law with parameter $p \in (0, 1)$. Suppose $Y_1 + Y_2$ has this law, with $Y_1, Y_2$ i.i.d. If $s < s'$ were two points of the support of $Y_1$, then $2s < s + s' < 2s'$ would be three values in $\set{0, 1}$. So $Y_1$ is constant, and then $Y_1 + Y_2$ is constant, a contradiction. The characteristic function of an infinitely divisible law has no zeros.