Given a random variable $X$ with $\mathsf{supp}\, X \subseteq [0,1]$ and $n$ positive numbers $h_1,\cdots,h_n$ with $\sum_{i=1}^n h_i=1$, I want to know some sufficient conditions for decomposing $X$ into $X=\sum_{i=1}^nh_iW_i$, where components $W_i$ are mutually independent and satisfy $\mathsf{supp} \, W_i \subseteq [0,1]$.

A particular appearling case is decomposition of a uniform distribution on $[0,1]$.

Some personal remarks may be meaningful.

  1. For a Bernoulli distribution $B(p)$ with any $p\in (0,1)$, there does not exist such a decomposition.
  2. If $n=2$, for any pair $(h_1,h_2)$, we can decompose a random variable $X$ uniformly distributed on $[0,1]$ as follows:

we first calculate the binary respresentation $\{b_i\}_{i=1}^{+\infty}$ of $h_1$, i.e., $h_1=\sum_{i=1}^{+\infty} b_i\cdot 2^{-i}$. Then let $$ W_1=\sum_{i=1}^{+\infty}\frac{b_i}{h_1} \cdot 2^{-i} \cdot B_i, $$ and $$ W_2=\sum_{i=1}^{+\infty}\frac{(1-b_i)}{h_2} \cdot 2^{-i} \cdot B_i, $$ where $B_i$'s are independent copies of a Bernoulli random variable $B(\frac{1}{2})$. It is clear that $W_1$ is independent of $W_2$ and $X=h_1W_1+h_2W_2$ in distribution.

  • $\begingroup$ In your second example, in general $X$ will not equal $h_1W_1+h_2W_2$ in distribution. Indeed, for almost all $h_1$, the distribution will have an infinitely smooth density. $\endgroup$ Oct 21 '21 at 12:31
  • $\begingroup$ @losif Pinelis I have corrected my previous post. $\endgroup$
    – Ryan Chen
    Oct 21 '21 at 13:05

On your general question, there are several books which address it:

Yu. V. Linnik and I. V. Ostrovskii, Decompositions of random variables and vectors,

Lukacz, Characteristic functions,

Ramachandran, Advanced theory of characteristic functions.


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