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 &ne; 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.

Indecomposable

 * The simplest examples are Bernoulli-distributions: if


 * $$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 &ge; 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 distribution of the sum of two Bernoulli-distributed random variables) 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 p, q &isin; [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 (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 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


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


 * 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:


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


 * 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.


 * 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


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


 * 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.


 * On the other hand, let Dn be the nth binary digit of Y, for n &ge; 0. Then the Dn's are independent and


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


 * 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.