# Indecomposable distribution

(Redirected from Decomposable distributions)

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} \le 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} \le 1. \$
Thus, for example, the discrete uniform distribution on the set {0, 1, 2} is indecomposable, but the binomial distribution assigning respective probabilities 1/4, 1/2, 1/4 is decomposable.
$f(x) = {1 \over \sqrt{2\pi\,}} x^2 e^{-x^2/2}$
is indecomposable.

### Decomposable

• 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 Xn are each equal to 0 or 1 with equal probabilities – this is a Bernoulli trial of each digit of the binary expansion.
$\Pr(Y = y) = (1-p)^n p\,$
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
$Y = \sum_{n=1}^\infty {D_n \over 2^n},$
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.