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Examples

Indecomposable

• The simplest examples are Bernoulli-distributions: if

X = { 1 with probability  p , 0 with probability  1 − p , {\displaystyle 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 ≥ 0, and

X = { 2 with probability  a , 1 with probability  b , 0 with probability  c . {\displaystyle 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 a + c ≤ 1   {\displaystyle {\sqrt {a}}+{\sqrt {c}}\leq 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 U = { 1 with probability  p , 0 with probability  1 − p , and V = { 1 with probability  q , 0 with probability  1 − q , {\displaystyle {\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 a = p q , {\displaystyle a=pq,\,} c = ( 1 − p ) ( 1 − q ) , {\displaystyle c=(1-p)(1-q),\,} b = 1 − a − c . {\displaystyle 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 a + c ≤ 1.   {\displaystyle {\sqrt {a}}+{\sqrt {c}}\leq 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 2 π x 2 e − x 2 / 2 {\displaystyle 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:

∑ n = 1 ∞ X n 2 n , {\displaystyle \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 {\displaystyle \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 ≥ 0. Then the Dn's are independent[why?] and Y = ∑ n = 1 ∞ 2 n D n , {\displaystyle 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.

See also

• Cramér's theorem
• Cochran's theorem
• Infinite divisibility (probability)
• Khinchin's theorem on the factorization of distributions

• 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.