The family of Poisson distributions (parametrised by its mean) has an interesting 'linearity' property:
$$\mathrm{Poisson}(x) + \mathrm{Poisson(y)} \sim \mathrm{Poisson(x+y)}$$
…meaning that the sum of two independent random variables drawn from poisson distributions with means $x$ and $y$ is itself distributed as a Poisson variable, with mean $(x+y)$.
- Do any other probability distribution families have the same property?
- Are there similar distribution families $D(a)$ on other binary operations $\odot$?
$$\text{i.e.} \qquad D(a) \odot D(b) \sim D(a \odot b)$$
Best Answer
Regaring question 1), there is the class infinitely divisible distributions (for a textbook reference taking this further to stochastic processes, see Sato 1999). However, the parameter in question is not necessarily the mean like for the Poisson distribution.
The Gamma distribution, for example, belongs to this class. The sum of independent Gamma RVs all having equal scale but potentially different shape parameters is again Gamma with the same scale and the shape parameter given by the sum of the individual shapes. The chi-square distribution is another example. And all stable distributions belong to this class (see also this thread for infinitely divisible but not stable). The normal distribution is a stable distribution.
However, the class of infinitely divisible distributions does not fully characterize the class satisfying the "linearity property". Rather, it is a subclass because two independent binomial distributions with the same success probability, for example, also satisfy the "linearity property" (proof by multiplication of their moment generating functions) but a binomial distribution is not infinitely divisible (as a binomial cannot be expressed as a sum of an arbitrary number of binomials, whereby I exclude degenerate binomials in the sense of having a success probability of zero).
Regarding question 2), as maybe a "cheap" example I consider multiplication as the operator and the exp of an infinitely divisible RV as distribution:
$exp(X) * exp(Y) = exp(X+Y). $
If $X$ and $Y$ are normals, then we get the log-normal example, mentioned in Bryden C's answer.
EDIT-log: I added the paragraph "However, the class of..." and a reference to Sato.