Indecomposable distribution

<h2 id="examples">Examples</h2>
<h3>Indecomposable</h3>
<ul><li>The simplest examples are <a href="/facts/Bernoulli_distribution/ChCtYyvs">Bernoulli-distributions</a>: if</li></ul>

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.
<ul><li>Suppose a + b + c = 1, a, b, c ≥ 0, and</li></ul>

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 <a href="/facts/Bernoulli_distribution/ChCtYyvs">Bernoulli-distributed</a> 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 <a href="/facts/Discrete_uniform_distribution/M52SQP5n">discrete uniform distribution</a> on the set {0, 1, 2} is indecomposable, but the <a href="/facts/Binomial_distribution/UMoFMjDj">binomial distribution</a> for two trials each having probabilities 1/2, thus giving respective probabilities a, b, c as 1/4, 1/2, 1/4, is decomposable.
<ul><li>An <a href="/facts/Absolute_continuity/NPKDt9Qn">absolutely continuous</a> indecomposable distribution. It can be shown that the distribution whose <a href="/facts/Probability_density_function/zvfybna4">density function</a> is</li></ul>

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.
<h3>Decomposable</h3>
<ul><li>All <a href="/facts/Infinite_divisibility_(probability)/XsDLIfs2">infinitely divisible</a> distributions are <a href="/facts/Argumentum_a_fortiori/XGR8N1j0">a fortiori</a> decomposable; in particular, this includes the <a href="/facts/Stable_distribution/rvtRmhlz">stable distributions</a>, such as the <a href="/facts/Normal_distribution/UapjjPyQ">normal distribution</a>.</li>
<li>The <a href="/facts/Uniform_distribution_(continuous)/XbnlVljT">uniform distribution</a> 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:</li></ul>

∑
          
            n
            =
            1
          
          
            ∞
          
        
        
          
            
              X
              
                n
              
            
            
              2
              
                n
              
            
          
        
        ,
      
    
    {\displaystyle \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.
<ul><li>A sum of indecomposable random variables is decomposable into the original summands. But it may turn out to be <a href="/facts/Infinitely_divisible_distribution/XsDLIfs2">infinitely divisible</a>. Suppose a random variable Y has a <a href="/facts/Geometric_distribution/c8wjfaoV">geometric distribution</a></li></ul>

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 <a href="/facts/Negative_binomial_distribution/znzj365Y">negative-binomially distributed</a> random variables Yj, j = 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 Dn'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.
<h2 id="related-concepts">Related concepts</h2>
At the other extreme from indecomposability is <a href="/facts/Infinite_divisibility_(probability)/XsDLIfs2">infinite divisibility</a>.

<ul><li><a href="/facts/Cram%25C3%25A9r%25E2%2580%2599s_decomposition_theorem/rYff3Awg">Cramér's theorem</a> shows that while the normal distribution is infinitely divisible, it can only be decomposed into normal distributions.</li>
<li><a href="/facts/Cochran%2527s_theorem/maUMfYzP">Cochran's theorem</a> 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 <a href="/facts/Chi-squared_distribution/wYnWWgUe">chi-squared distributions</a>.</li></ul>
<h2 id="see-also">See also</h2>
<ul><li><a href="/facts/Cram%25C3%25A9r%25E2%2580%2599s_decomposition_theorem/rYff3Awg">Cramér's theorem</a></li>
<li><a href="/facts/Cochran%2527s_theorem/maUMfYzP">Cochran's theorem</a></li>
<li><a href="/facts/Infinite_divisibility_(probability)/XsDLIfs2">Infinite divisibility (probability)</a></li>
<li><a href="/facts/Khinchin%2527s_theorem_on_the_factorization_of_distributions/tHCovCFg">Khinchin's theorem on the factorization of distributions</a></li></ul>

<ul><li>Linnik, Yu. V. and Ostrovskii, I. V. Decomposition of random variables and vectors, Amer. Math. Soc., Providence RI, 1977.</li></ul>
<ul><li>Lukacs, Eugene, Characteristic Functions, New York, Hafner Publishing Company, 1970.</li></ul>

Indecomposable distribution open-in-new

Indecomposable distribution