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 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 ∈ [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) ∈ [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, each with parameters
p and non-integer
r = 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 ''D'
ns are independent and :: Y = \sum_{n=1}^\infty 2^n D_n, :and each term in this sum is indecomposable. == Related concepts ==