# Sufficient conditions for decomposition of a bounded random variable into several small pieces

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.

• 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. Oct 21 '21 at 12:31
• @losif Pinelis I have corrected my previous post. Oct 21 '21 at 13:05