To define the property I am interested, let me introduce a notion of convolution measure on a measurable monoid $\mathcal{S}$. However, I am predominantly interested in the case where $\mathcal{S} \subseteq \mathbb{R}$ and the binary operation is just the addition ($+$).
Definition. (Convolution Measure) Let $\mathcal{S}$ be a measurable monoid, in the sense that $\mathcal{S}$ is equipped with
- a $\sigma$-algebra $\Sigma$, and
- a binary operation $ (a, b) \mapsto a \oplus b$ which is measurable.
If $\mu$ and $\nu$ are $\sigma$-finite measures on $\mathcal{S}$, then the convolution $\mu \ast \nu$ is defined by the relation
$$ (\mu \ast \nu)(A) = \int_{\mathcal{S}\times\mathcal{S}} \mathbf{1}_{A}(x \oplus y) \, \mu(\mathrm{d}x)\nu(\mathrm{d}y) \quad \forall A \in \Sigma. $$
Note that, when $\mu$ and $\nu$ are probability measures, and if $X \sim \mu$ and $Y\sim \nu$ are independent, then $\mu\ast\nu$ is the law of $X \oplus Y$.
Now let me introduce the property I am interested in:
Definition. We say that a family $\mathcal{M}$ of probability measures on a measurable monoid $\mathcal{S}$ has property P if the following conditions hold:
- $\mathcal{M}$ is closed under convolution. That is, if $\mu, \nu \in \mathcal{M}$, then their convolution $\mu \ast \nu$ also lies in $\mathcal{M}$.
- $\mathcal{M}$ is closed under deconvolution. That is, if $\mu \in \mathcal{M}$ can be written as $\mu = \nu_1 \ast \nu_2$ for probability measures $\nu_1$ and $\nu_2$, then $\nu_1, \nu_2 \in \mathcal{M}$.
Item 1 is satisfied by many naturally-occurring families of distributions, but Item 2 seems much harder to be satisfied. So my question is:
$\color{blue}{\textbf{Question.}}\,$ Are there any "naturally-occurring, non-trivial" examples of a family of probability measures on $\mathbb{R}$ having property P?
Let me not articulate the precise meaning of naturality and nontriviality and rather leave it to your own discretion.
Below are two examples I found:
Example 1. (Poisson distributions) The family $\mathcal{M}_{\text{Poisson}}$ of all Poisson distributions on $\mathbb{N}_0$, including the Dirac delta $\delta_0$, has property P.
First, it is well-known that the sum of two independent Poisson random variables again has Poisson distribution. Now, let $Z$ has a Poisson distribution with rate $\lambda$, and assume $Z = X + Y$ for some independent $X$ and $Y$ taking values in $\mathbb{N}_0$. If we write $p_X(z) = \mathbf{E}[z^X] = \sum_{n=0}^{\infty} z^n \mathbf{P}(X = n)$ and likewise for $p_Y(z)$, then
\begin{align*} p_X(z)p_Y(z) = \mathbf{E}[z^Z] = \sum_{n=0}^{\infty} \frac{z^n \lambda^n}{n!} e^{-\lambda} = e^{(z-1)\lambda}. \end{align*}
Since $p_Y(z) \geq p_Y(1) = 1$ for all $z \geq 1$, the above identity shows that $p_X(|z|) \leq e^{(|z|-1)\lambda}$ whenever $|z| \geq 1$. So, $p_X(z)$ defines an entire function on $\mathbb{C}$ of order at most $1$. Moreover, neither of $p_X(z)$ nor $p_Y(z)$ has a zero on $\mathbb{C}$. From this, we conclude that $p_X(z)$ is an exponential function satisfying $p_X(1) = 1$. This then implies that $p_X(z) = e^{\eta(z-1)}$ for some $\eta \geq 0$ and hence $X$ has a Poisson distribution with rate $\eta$. The same argument applies to $p_Y(z)$, and so, the desired conclusion follows.
Example 2 (normal distributions). The family $\mathcal{M}_{\text{normal}}$ of all normal distributions on $\mathbb{R}$, including degenerate distributions, has property P.
Indeed, it is well-known that the sum of two independent normal RVs is again normal. Then item 2 follows from Lévy-Cramér theorem.
The above two examples may give an impression that a family of infinitely divisible distributions might be a good guess to start with. However, we have some non-examples:
Non-Example 1 (gamma distributions). Let $\mathcal{M}_{\text{Gamma}}$ be the family of all Gamma distributions (with unit shape parameter) on $[0, \infty)$, $$ \text{Gamma}(\mathrm{d}x; \lambda) = \frac{x^{\lambda-1}e^{-x}}{\Gamma(\lambda)} \, \mathrm{d}x, \qquad \lambda > 0, $$ together with $\text{Gamma}(\mathrm{d}x; 0) = \delta_0(\mathrm{d}x)$. Then $\mathcal{M}_{\text{Gamma}}$ satisfies item 1 of property P, but it does not satisfy item 2.
Indeed, let $Z$ have an exponential distribution with unit rate, and let $X$ and $Y$ by
$$ X = \lfloor Z \rfloor \qquad\text{and} \qquad Y = Z - \lfloor Z \rfloor = Z \text{ mod } 1. $$
Then for any $k \in \mathbb{N}_0$ and $0 < y < 1$,
\begin{align*} \mathbf{P}(Y > y \mid X = k) &= \frac{\mathbf{P}(k+y \leq Z < k+1)}{\mathbf{P}(k \leq Z < k+1)} \\ &= \frac{e^{-k-y} - e^{-k-1}}{e^{-k} - e^{-k-1}} \\ &= \frac{e^{-y} - e^{-1}}{1 - e^{-1}} \end{align*}
does not depend on $k$, and so, $Y$ is independent of $X$. (Or, you can utilize the memorylessness property of $Z$.) Hence, $Z$ is decomposed into the sum $X+Y$ of independent RVS, but neither of the distribution of $X$ nor $Y$ lies in $\mathcal{M}_{\text{Gamma}}$.
This non-example shows that mere infinite-divisibility is not enough for property P.
