Menu
Home
People
Places
Arts
History
Plants & Animals
Science
Life & Culture
Technology
Reference.org
Indecomposable distribution
open-in-new
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
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 {\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
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[
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.