Indecomposable distribution

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.

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 &ge; 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 &isin; [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) &isin; [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.


 * 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 Xn 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 Yj, j = 1, ..., k, such that Y1 + ... + Yk has this geometric distibution.  Therefore, this distribution is infinitely divisible.  But now let Dn be the nth binary digit of Y, for n &ge; 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.