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).