Indecomposable distribution

From Wikipedia, the free encyclopedia
  (Redirected from Decomposable distribution)
Jump to: navigation, search

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 = X1 + X2.

Examples[edit]

Indecomposable[edit]

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 ab and V assumes two values cd, 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, abc ≥ 0, and
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 sum of two Bernoulli distributions) 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 pq ∈ [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 (pq) ∈ [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 assigning respective probabilities 1/4, 1/2, 1/4 is decomposable.
f(x) = {1 \over \sqrt{2\pi\,}} x^2 e^{-x^2/2}
is indecomposable.

Decomposable[edit]

  • 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.
\Pr(Y = y) = (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. But now let Dn be the nth binary digit of Y, for n ≥ 0. Then the Ds are independent and
 Y = \sum_{n=1}^\infty {D_n \over 2^n},
and each term in this sum is indecomposable.

Related concepts[edit]

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[edit]

References[edit]

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