# Indecomposable distribution

In probability theory, an indecomposable distribution is a probability distribution that cannot be represented as the distribution of the sum of two or more non-constant independent random variables: Z ≠ X + Y. If it can be so expressed, it is decomposable: Z = X + Y. If, further, it can be expressed as the distribution of the sum of two or more independent identically distributed random variables, then it is divisible: Z = X$_1$ + X$_2$.

## Examples
### Indecomposable

* The simplest examples are Bernoulli-distributions: if
       $X={\begin{cases}1&{\text{with probability }}p,\\0&{\text{with probability }}1-p,\end{cases}}$
    then the probability distribution of X is indecomposable.
    Proof: Given non-constant distributions U and V, so that U assumes at least two values a, b and V assumes two values c, d, with a < b and c < d, then U + V assumes at least three distinct values: a + c, a + d, b + d (b + c may be equal to a + d, for example if one uses 0, 1 and 0, 1). Thus the sum of non-constant distributions assumes at least three values, so the Bernoulli distribution is not the sum of non-constant distributions.
* Suppose a + b + c = 1, a, b, c ≥ 0, and
       $X={\begin{cases}2&{\text{with probability }}a,\\1&{\text{with probability }}b,\\0&{\text{with probability }}c.\end{cases}}$
    This probability distribution is decomposable (as the distribution of the sum of two Bernoulli-distributed random variables) if
       ${\sqrt {a}}+{\sqrt {c}}\leq 1\$
    and otherwise indecomposable. To see, this, suppose U and V are independent random variables and U + V has this probability distribution. Then we must have
       ${\begin{matrix}U={\begin{cases}1&{\text{with probability }}p,\\0&{\text{with probability }}1-p,\end{cases}}&{\mbox{and}}&V={\begin{cases}1&{\text{with probability }}q,\\0&{\text{with probability }}1-q,\end{cases}}\end{matrix}}$
    for some p, q ∈ [0, 1], by similar reasoning to the Bernoulli case (otherwise the sum U + V will assume more than three values). It follows that
       $a=pq,\,$
       $c=(1-p)(1-q),\,$
       $b=1-a-c.\,$
    This system of two quadratic equations in two variables p and q has a solution (p, q) ∈ [0, 1]$^2$ if and only if
       ${\sqrt {a}}+{\sqrt {c}}\leq 1.\$
    Thus, for example, the discrete uniform distribution on the set {0, 1, 2} is indecomposable, but the binomial distribution for two trials each having probabilities 1/2, thus giving respective probabilities a, b, c as 1/4, 1/2, 1/4, is decomposable.
* An absolutely continuous indecomposable distribution. It can be shown that the distribution whose density function is
       $f(x)={1 \over {\sqrt {2\pi \,}}}x^{2}e^{-x^{2}/2}$
    is indecomposable.

### Decomposable

* All infinitely divisible distributions are a fortiori decomposable; in particular, this includes the stable distributions, such as the normal distribution.
* The uniform distribution on the interval [0, 1] is decomposable, since it is the sum of the Bernoulli variable that assumes 0 or 1/2 with equal probabilities and the uniform distribution on [0, 1/2]. Iterating this yields the infinite decomposition:
       $\sum _{n=1}^{\infty }{X_{n} \over 2^{n}},$
    where the independent random variables X$_n$ are each equal to 0 or 1 with equal probabilities – this is a Bernoulli trial of each digit of the binary expansion.
* A sum of indecomposable random variables is decomposable into the original summands. But it may turn out to be infinitely divisible. Suppose a random variable Y has a geometric distribution
       $\Pr(Y=n)=(1-p)^{n}p\,$
    on {0, 1, 2, ...}.
    For any positive integer k, there is a sequence of negative-binomially distributed random variables Y$_j$, j = 1, ..., k, such that Y$_1$ + ... + Y$_k$ has this geometric distribution. Therefore, this distribution is infinitely divisible.
    On the other hand, let D$_n$ be the nth binary digit of Y, for n ≥ 0. Then the D$_n$'s are independent and
       $Y=\sum _{n=1}^{\infty }2^{n}D_{n},$
    and each term in this sum is indecomposable.

## Related concepts

At the other extreme from indecomposability is infinite divisibility.
* Cramér's theorem shows that while the normal distribution is infinitely divisible, it can only be decomposed into normal distributions.
* Cochran's theorem shows that the terms in a decomposition of a sum of squares of normal random variables into sums of squares of linear combinations of these variables always have independent chi-squared distributions.