Indecomposable

• The simplest examples are Bernoulli-distributions: if

${\displaystyle X={\begin{cases}1&{\text{with probability }}p,\\0&{\text{with probability }}1-p,\end{cases}}}$

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

${\displaystyle X={\begin{cases}2&{\text{with probability }}a,\\1&{\text{with probability }}b,\\0&{\text{with probability }}c.\end{cases}}}$

This probability distribution is decomposable (as the distribution of the sum of two Bernoulli-distributed random variables) if

${\displaystyle {\sqrt {a}}+{\sqrt {c}}\leq 1\ }$

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

${\displaystyle {\begin{matrix}U={\begin{cases}1&{\text{with probability }}p,\\0&{\text{with probability }}1-p,\end{cases}}&{\mbox{and}}&V={\begin{cases}1&{\text{with probability }}q,\\0&{\text{with probability }}1-q,\end{cases}}\end{matrix}}}$

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

${\displaystyle a=pq,\,}$

${\displaystyle c=(1-p)(1-q),\,}$

${\displaystyle b=1-a-c.\,}$

This system of two quadratic equations in two variables p and q has a solution (p, q) ∈ [0, 1]² if and only if

${\displaystyle {\sqrt {a}}+{\sqrt {c}}\leq 1.\ }$

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

${\displaystyle f(x)={1 \over {\sqrt {2\pi \,}}}x^{2}e^{-x^{2}/2}}$

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:

${\displaystyle \sum _{n=1}^{\infty }{X_{n} \over 2^{n}},}$

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

${\displaystyle \Pr(Y=n)=(1-p)^{n}p\,}$

on {0, 1, 2, ...}.

For any positive integer k, there is a sequence of negative-binomially distributed random variables Y_j, j = 1, ..., k, each with parameters p and non-integer r = 1/k, such that Y₁ + ... + 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^{[why?][citation needed]} and

${\displaystyle Y=\sum _{n=1}^{\infty }2^{n}D_{n},}$

and each term in this sum is indecomposable.