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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 [[statistical independence|independent]] [[random variable]]s: ''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|independent ''identically'' distributed]] random variables, then it is '''divisible:''' ''Z'' = ''X''1 + ''X''2.

==Examples==
=== Indecomposable ===
* The simplest examples are [[Bernoulli distribution]]s: if

::<math>X = \begin{cases}
1 & \text{with probability } p, \\
0 & \text{with probability } 1-p,
\end{cases}
</math>

: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

::<math>X = \begin{cases}
2 & \text{with probability } a, \\
1 & \text{with probability } b, \\
0 & \text{with probability } c.
\end{cases}
</math>

:This probability distribution is decomposable (as the sum of two Bernoulli distributions) if

::<math>\sqrt{a} + \sqrt{c} \le 1 \ </math>

: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

::<math>
\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}
</math>

: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

::<math>a = pq, \, </math>
::<math>c = (1-p)(1-q), \, </math>
::<math>b = 1 - a - c. \, </math>

:This system of two quadratic equations in two variables ''p'' and ''q'' has a solution (''p'', ''q'') ∈ [0, 1]2 if and only if

::<math>\sqrt{a} + \sqrt{c} \le 1. \ </math>

: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 [[absolute continuity|absolutely continuous]] indecomposable distribution. It can be shown that the distribution whose [[probability density function|density function]] is

::<math>f(x) = {1 \over \sqrt{2\pi\,}} x^2 e^{-x^2/2}</math>

:is indecomposable.

=== Decomposable ===
* All [[infinite divisibility (probability)|infinitely divisible]] distributions are a fortiori decomposable; in particular, this includes the [[stable distribution]]s, such as the [[normal distribution]].

* The [[uniform distribution (continuous)|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:

::<math> \sum_{n=1}^\infty {X_n \over 2^n }, </math>

: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 necessarily decomposable (as it is a sum), and in fact may a fortiori be an [[infinitely divisible distribution]] (not just decomposable as the given sum). Suppose a random variable ''Y'' has a [[geometric distribution]]

::<math>\Pr(Y = y) = (1-p)^n p\, </math>

:on {0, 1, 2, ...}. For any positive integer ''k'', there is a sequence of [[negative binomial distribution|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. But now let ''D''''n'' be the ''n''th binary digit of ''Y'', for ''n'' ≥ 0. Then the ''D''s are independent and

::<math> Y = \sum_{n=1}^\infty {D_n \over 2^n}, </math>

:and each term in this sum is indecomposable.

== Related concepts ==
At the other extreme from indecomposability is [[Infinite divisibility (probability)|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 decompositions of a sum of squares of normal random variables into sums of squares of linear combinations of these variables are always independent [[chi-squared distribution]]s.

== See also ==
* [[Cramér's theorem]]
* [[Cochran's theorem]]
* [[Infinite divisibility (probability)]]

==References==

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

[[Category:Probability theory]]
[[Category:Types of probability distributions]]

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