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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).

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    $\begingroup$ Yes for your last question. Sum of independent random variables with Poisson distribution is poisson distributed with parameter the sum of the parameters. $\endgroup$ – Shashi Dec 18 '17 at 16:54
  • $\begingroup$ @Shashi, It is better to formulate this way: if the finite sum of independent non-negative random variables has a Poisson distribution, then the summands themselves must have the Poisson distribution. That is so called Raikov's theorem. $\endgroup$ – Rodvi Dec 18 '17 at 17:08
  • $\begingroup$ Well, there is a similar Cramer's theorem for the normal distribution: if the finite sum of independent random variables has a normal distribution, then the summands themselves must have the normal distribution. $\endgroup$ – Rodvi Dec 18 '17 at 17:18
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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.

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  • $\begingroup$ Just one question. If we use infinite divisibility of Poisson distribution to decompose the Poisson r.v. into the sum of i.i.d. r.vs and then want to apply Raikov's theorem we need to ensure that all the identically distributed summands are non-negative (according to Wikipedia). The last thing does not follow directly from the definition of infinite divisible distribution. Can we just say that the non-negativity of the summands follows from their identical distribution and the non-negativity of the their sum? $\endgroup$ – Rodvi Dec 18 '17 at 19:50
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    $\begingroup$ I'm using a weaker result than Raikov's theorem, which is a standard textbook exercise. Fairly easy computation (eg. manipulating the formulae for the distribution functions) using Poisson variables should convince you that of the fact I state above: $\text{Poi}(\lambda) + \text{Poi}(\mu) = \text{Poi}(\lambda + \mu)$. This is weaker than Raikov, but sufficient to prove infinite divisibility of a Poisson distribution (using Poisson rate $\lambda/N$). $\endgroup$ – owen88 Dec 18 '17 at 20:17
  • $\begingroup$ Ok, now I understand your idea. Unfortunately, this way doesn't show the uniqueness of the Poisson distribution for the summands in the property of the infinite divisibility. Nevertheless, you have a simple and good result. $\endgroup$ – Rodvi Dec 18 '17 at 20:48

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