Indecomposable distribution

Examples

Indecomposable

• The simplest examples are Bernoulli-distributions: if

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

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

\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

: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 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

: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 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

::\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, 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 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.

References

• Linnik, Yu. V. and Ostrovskii, I. V. Decomposition of random variables and vectors, Amer. Math. Soc., Providence RI, 1977.

• Lukacs, Eugene, Characteristic Functions, New York, Hafner Publishing Company, 1970.