# What are continuous distributions that are additive and have finite support

I'm wondering what are continuous distributions that are additive and have finite support. Joint normal distribution is continuous, and is additive in the sense that if $$X,Y$$ are joint normal, then $$X+Y$$ are still normal, but they have infinite support (on the real line). I want something that is continuous, additive and has finite support. Thanks!

• Your language is ambiguous because "distribution" in this context could be interpreted to mean a distribution or a set of distributions. Concerning the latter, you may begin with literally any distribution with finite support. Create the set of distributions arising from all finite sums of that distribution and voila, you have an additively closed family of such distributions. Is that the sort of thing you are looking for? – whuber Mar 5 '20 at 13:48
• Yes, you are right. I'm looking for additively closed family of such distributions. Thanks for pointing this out. Preferably this family also has a nice closed form pdf. – T34driver Mar 5 '20 at 19:46
• I meant preferably the pdf for the sum has the same form as the pdf for each individual random variable. Do you have any suggestions on what distributions to start with? – T34driver Mar 5 '20 at 20:12
• Please explain, then, what you mean by "same form"! – whuber Mar 5 '20 at 20:45
• Yeah, you are right, the phrase "same form" is not very rigorous. What I meant is something like the normal family, $X\sim N(u_{x},\sigma_{x})$, $X\sim N(u_{y},\sigma_{y})$ then $X+Y\sim N(u_{x}+u_{y},\sigma_{x+y})$, so that to evaluate the joint pdf I only need to change the mean and variance parameters. – T34driver Mar 6 '20 at 1:36

A minimal example is obtained by taking literally any distribution with compact support on the nonnegative reals. Letting $$F$$ be its cumulative distribution function, this means there exist $$0\le a\le b$$ for which $$F(a)=0$$ and $$F(b)=1.$$ Repeated convolution produces the sequence of distribution functions $$F_n,$$ $$n=1,2,3,\ldots,$$ for which $$F_1 = F$$ and for all $$n\ge 1,$$

$$F_{n+1}(x) = \int_{\mathbb{R}} F_n(x-y)\,\mathrm{d} F(y).$$

Because this corresponds to addition of random variables, the support of $$F_n$$ is contained in the interval $$[na,nb],$$ which is compact and non-negative.

When $$X_n$$ is a random variable with distribution $$F_n$$ and $$X_m$$ a random variable with distribution $$F_m,$$ this ensures that $$X_n+X_m$$ has distribution $$F_{n+m},$$ showing this family $$(F_n)$$ is closed under addition. Its parameter is $$n.$$ When $$F$$ is a continuous distribution, so is $$F_n.$$ Thus, $$(F_n)$$ satisfies all the requirements of the question.

At https://stats.stackexchange.com/a/43075/919 I describe such a family explicitly (and in great detail) where $$F$$ is the Uniform$$(0,1)$$ distribution. This shows that the construction of $$(F_n)$$ is not just an abstraction: it can lead to distributions with computable formulas and concrete applications.

It might be objected that $$n$$ can attain only integral values. We can try to fix that. If there were such a family associated with all positive real $$n$$ it would be infinitely divisible: for all $$n,$$ the distribution $$F_n$$ could be expressed as the convolution of $$F_{n/2}$$ with itself, or generally the $$k$$-fold convolution of $$F_{n/k}$$ for any whole number $$k.$$ The Wikipedia article on infinite divisibility asserts that apart from atomic distributions (which concentrate all probability on a single value), there exist no infinitely divisible family of distributions with "bounded (finite) support." It does not prove this, but refers to Sato, Ken-iti (1999), Lévy Processes and Infinitely Divisible Distributions for the details.

The minimal example can, however, be enlarged by adding in other distributions. For instance, by adding in any atomic distribution supported on a non-negative value $$\mu$$ we can introduce a location parameter. In effect, writing

$$F_{n;\mu}(x) = F_n(x-\mu)$$

we create a two-parameter family where the parameter $$n$$ must be a whole number and $$\mu$$ can be any non-negative real number. These distributions are supported on the intervals $$[na+\mu,nb+\mu]$$ which are non-negative and compact. For corresponding random variables $$X_{n;\nu}$$ and $$X_{m;\mu},$$ clearly $$X_{n;\nu}+X_{m;\mu}$$ has $$F_{m+n;\mu+\nu}$$ for its distribution. Thus, the family $$(F_{n;\mu}),\, n=1,2,3,\ldots;\, \mu \ge 0$$ satisfies the condition of the question.

We can generate larger families very generally by emulating this process of combining two families of distributions under addition. When $$(G_n)$$ is also an additively closed family of distributions with non-negative compact support, then for any pairs of whole numbers $$m$$ and $$n$$ let $$H_{m,n}$$ be the distribution of $$X_m+Y_n$$ where $$X_m\sim F_m$$ and $$Y_n\sim G_n.$$ The family $$(H_{m,n},\, m=1,2,3,\ldots;\, n=1,2,3,\ldots)$$ satisfies all the conditions of the question. This process can be repeated as many times as you like.

In a specific sense, all solutions to this problem arise in this way. When $$\mathcal F$$ is any set of distributions with compact non-negative support, define the "additive closure" of $$\mathcal F$$ to be the intersection of all sets of distributions that are additively closed and include $$\mathcal F.$$ This intersection exists because the set of all distributions is one such set. All additively closed families obviously equal their own additive closures. As we noted before, the only distributions in $$\mathcal F$$ that can be infinitely divisible would have to be atomic.

• Thanks, this is very helpful. – T34driver Mar 6 '20 at 19:56

Let $$X$$ and $$Y$$ are from same continues distribution with finite support $$(a,b)$$.

so $$a and $$a , in-hence $$2a. Since the support of $$X+Y$$ equal to $$(2a,2b)$$ so I think it can not be happen. at least one of $$a$$ , $$b$$ $$\rightarrow$$ $$\infty$$ and other should be zero. like Chi-square distribution.

• Thanks, this is helpful. This means we cannot have a distribution family that is closed under addition and is supported on the same interval. Maybe the statement of my problem is not rigorous enough, I actually do not require that the sum having the same support as the original two random variables, and I guess as long as we incorporate the support as part of the parameter for the distribution family (like uniform distribution which are characterized by support parameter) it will fix the problem. – T34driver Mar 5 '20 at 19:53
• You should look at en.wikipedia.org/wiki/… – Masoud Mar 5 '20 at 22:39
• This argument is incorrect: the fact that various distributions in the family must have increasingly large supports does not imply that any distribution in the family must be supported on $(0,\infty).$ – whuber Mar 6 '20 at 14:38