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'' ≠ ''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'' = ''X''₁ + ''X''₂.

Examples

Indecomposable

* The simplest examples are Bernoulli-distributeds: if

::$X = \begin 1 & \text p, \\ 0 & \text 1-p, \end$

: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

::$X = \begin 2 & \text a, \\ 1 & \text b, \\ 0 & \text c. \end$

:This probability distribution is decomposable (as the distribution of the sum of two Bernoulli-distributed random variables) if

::$\sqrt + \sqrt \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

::$\begin U = \begin 1 & \text p, \\ 0 & \text 1 - p, \end & \mbox & V = \begin 1 & \text q, \\ 0 & \text 1 - q, \end \end$

:for some ''p'', ''q'' ∈  , 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'') ∈  , 1sup>2 if and only if

::$\sqrt + \sqrt \le 1. \$

:Thus, for example, the discrete uniform distribution on the set 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

::$f(x) = x^2 e^$

: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 , 1is decomposable, since it is the sum of the Bernoulli variable that assumes 0 or 1/2 with equal probabilities and the uniform distribution on , 1/2 Iterating this yields the infinite decomposition:

::$\sum_^\infty ,$

: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 .

:For any positive integer ''k'', there is a sequence of negative-binomially distributed random variables ''Y''_''j'', ''j'' = 1, ..., ''k'', such that ''Y''₁ + ... + ''Y''_''k'' has this geometric distribution. Therefore, this distribution is infinitely divisible.

:On the other hand, let ''D''_''n'' be the ''n''th 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.

See also

* Cramér's theorem
* Cochran's theorem
* Infinite divisibility (probability)
* Khinchin's theorem on the factorization of 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.

Types of probability distributions