Indecomposable distribution

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In probability theory, an indecomposable distribution is any probability distribution that cannot be represented as the distribution of the sum of two or more non-constant independent random variables.

[edit] Examples

The simplest examples are Bernoulli distributions: if

$X = \left\{ \begin{matrix} 1 & \mathrm{with}\ \mathrm{probability} & p, \\ 0 & \mathrm{with}\ \mathrm{probability} & 1-p, \end{matrix} \right.$

then the probability distribution of X is indecomposable.

Suppose a + b + c = 1, a, b, c ≥ 0, and

$X = \left\{\begin{matrix} 2 & \mathrm{with}\ \mathrm{probability}\ a, \\ 1 & \mathrm{with}\ \mathrm{probability}\ b, \\ 0 & \mathrm{with}\ \mathrm{probability}\ c. \end{matrix} \right.$

This probability distribution is decomposable 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

$\begin{matrix} U = \left\{ \begin{matrix} 1 & \mathrm{with}\ \mathrm{probability} & p, \\ 0 & \mathrm{with}\ \mathrm{probability} & 1 - p, \end{matrix} \right. & \mbox{and} & V = \left\{ \begin{matrix} 1 & \mathrm{with}\ \mathrm{probability} & q, \\ 0 & \mathrm{with}\ \mathrm{probability} & 1 - q, \end{matrix} \right. \end{matrix}$

for some p, q ∈ [0, 1]. 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} \le 1. \$

Thus, for example, the discrete uniform distribution on the set {0, 1, 2} is indecomposable, but the binomial distribution assigneing respective probabilities 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.

The uniform distribution on the interval [0, 1] is decomposable, since it is the probability distribution of the random variable

$\sum_{n=1}^\infty {X_n \over 2^n },$

where the independent random variables X_n are each equal to 0 or 1 with equal probabilities.

This example shows that a random variable whose probability distribution is infinitely divisible can also be a sum of indecomposable distributions. Suppose a random variable Y has a geometric distribution

$\Pr(Y = y) = (1-p)^n p\,$

on {0, 1, 2, ...}. For any positive integer k, there is a sequence of negative-binomially distributed random variables Y_j, j = 1, ..., k, such that Y₁ + ... + Y_k has this geometric distibution. Therefore, this distribution is infinitely divisible. But now let D_n be the nth binary digit of Y, for n ≥ 0. Then the Ds are independent and

$Y = \sum_{n=1}^\infty {D_n \over 2^n},$