Converse statements for convolutions of probability distributions

There is a list of convolutions of probability distributions on wikipedia. Using that list we can find convolution distribution of given independent random variables.

From this online tutorial I found that the converse statement is true for the binomial distribution, i.e. if $Y \sim \mathrm{Bin}(n,p)$ then there exist i.i.d $X_1, \ldots ,X_n \sim \mathrm{Bern}(p)$ such that $\displaystyle Y = \sum_{i=1}^n X_i$.

So, are there similar converse statements for other convolutions from the above-mentioned list (excluding the last one)?

For example consider r.v. $Y \sim \mathrm{Pois}(\lambda), \, \lambda \gt 0$. Can we claim that there exist independent random variables $X_1 \sim \mathrm{Pois}(\lambda_1), X_2 \sim \mathrm{Pois}(\lambda_2), X_3 \sim \mathrm{Pois}(\lambda_3)$ such that $\lambda_1 + \lambda_2 + \lambda_3 = \lambda$ and $Y = X_1 + X_2 + X_3$?

I know that there is a list of infinitely divisible distributions.

And some distributions present in both above-mentioned lists. Unfortunately, the property of infinite divisibility doesn't indicate distribution law of the summands (it only says that the summands exist and they are i.i.d).

Answer

A slight tangent, but perhaps an angle you may find interesting is to look into Infinitely Divisible distributions.

A probably distribution $X$ is said to be infinititely divisible if for any $N \geq 1$ there exists a collection $(Y_n)_{n = 1}^N$ of independent identically distributed random variables such that:

$$ X = \sum_{n = 1}^N Y_n,$$

where the equality stands for equality in distribution.

As an example, from the fact that $\text{Poi}(\lambda) + \text{Poi}(\mu) = \text{Poi}(\mu + \lambda)$, from the comments above, then we can see that in fact for any fixed $N \geq 1$, and $\lambda > 0$:

$$ \text{Poi}(\lambda) = \sum_{n=1}^N \text{Poi}\left( \frac{\lambda}{N} \right),$$

so that Poisson variables are infinitely divisible. Similar proofs work Normal distributions, as well as Gamma distributions.

A simple example of a non-infinitely divisible distribution would be the Bernoulli distribution (see proof below). Similarly the Uniform distribution is not infinitely divisible.

Bernoulli is not Infinitely Divisible
Consider the case of the Bernoulli distribution, $X$, with $P(X = 1) = p$, with $0 < p < 1$ (the cases $p = 0,1$ are trivially infinitely divisible).

Fix $N = 2$, then if $X$ were infinitely divisible there would exists $Y_1, \,Y_2$ independent and identically distributed such that

$$X = Y_1 + Y_2$$

Since $X \in \{0,1\}$ the i.i.d. assumption forces $Y_n \in \left\{0, \frac12\right\}$.

But then the support of $Y_1 + Y_2$ is $\{0,\frac12, 1\}$; since $p \neq 0,1$ we have

$$P\left(Y_1+Y_2 = \frac12\right) > 0$$

whereas $P(X = \frac12) = 0$, so we have a contradiction.