# 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 = X1 + X2.

## Examples

### Indecomposable

$X={\begin{cases}1&{\text{with probability }}p,\\0&{\text{with probability }}1-p,\end{cases}}$ then the probability distribution of X is indecomposable.
Proof: Given non-constant distributions U and V, so that U assumes at least two values ab and V assumes two values cd, 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, abc ≥ 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 sum of two Bernoulli distributions) 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 pq ∈ [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 (pq) ∈ [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 three trials each having probabilities 1/2, 1/2, thus giving respective probabilities a, b, c as 1/4, 1/2, 1/4, is decomposable.
$f(x)={1 \over {\sqrt {2\pi \,}}}x^{2}e^{-x^{2}/2}$ is indecomposable.

### Decomposable

$\sum _{n=1}^{\infty }{X_{n} \over 2^{n}},$ where the independent random variables Xn are each equal to 0 or 1 with equal probabilities – this is a Bernoulli trial of each digit of the binary expansion.
$\Pr(Y=n)=(1-p)^{n}p\,$ on {0, 1, 2, ...}. For any positive integer k, there is a sequence of negative-binomially distributed random variables Yj, j = 1, ..., k, such that Y1 + ... + Yk has this geometric distribution. Therefore, this distribution is infinitely divisible. But now let Dn be the nth binary digit of Y, for n ≥ 0. Then the Ds are independent and
$Y=\sum _{n=1}^{\infty }{D_{n} \over 2^{n}},$ [clarification needed]
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.