# Sufficient and necessary conditions for decomposing the sum of random variables

Given two $$n$$-tuple vectors $$\vec{\alpha}=(\alpha_1,\cdots,\alpha_n)$$ and $$\vec{h}=(h_1,\cdots,h_n)$$, where $$h_i\ge0$$, $$\sum_{i=1}^nh_i=1$$, and $$\alpha_i\in(0,1)$$, we consider a random variable $$S$$ on the interval $$[0,1]$$. My question is how to find the sufficient and necessary conditions for decomposability of $$S$$ into $$n$$ random variables $$X_1,~X_2,~\cdots,X_n$$ on $$[0,1]$$ satisfying $$$$\sum_{i=1}^n h_iX_i=S,~\text{and}$$$$ $$$$\mathbb{E}\left[X_i\right]=\alpha_i.$$$$

Please note that the parameters $$h_i$$ and $$\alpha_i$$ are given at the start.

Two trivial necessary conditions are given as follows:

1. $$\mathbb{E}\left[S\right]=\sum_{i=1}^nh_i\alpha_i$$;
2. $$\mathbb{P}\left[X_1=\cdots=X_n=0|S=0\right]=1$$ and $$\mathbb{P}\left[X_1=\cdots=X_n=1|S=1\right]=1$$.
• independent $X_i$? – Dieter Kadelka Nov 27 '20 at 8:00
• The problem is meaningless, unless you add the condition that $X_j$ are independent. With this condition, there is an extensive theory: see Linnik, Ostrovskii, Decomposition of random variables and vectors, AMS 1977. – Alexandre Eremenko Nov 27 '20 at 14:13
• @ Dieter Kadelka Not necessarily independent. – Ryan Chen Nov 28 '20 at 0:55
• @Alexandre Eremenko: I think this question is not easy even without the assumption of independent $X_i$. I found the sufficient and necessary condition in the case of $n=2$, i.e. $\mathbb{E}\left[ (S-h_1)\mathbf{1}\left(S-h_1\right)\right]\le (1-h_1)\alpha_{2}$ (the ordering of descending $\alpha_i$ is adopted), while the case of $n\ge3$ seems intractable – Ryan Chen Nov 28 '20 at 1:09
• @YemonChoi You are right. The vector $\mathbf{\alpha}$ and the weighted vector $\mathbf{h}$ are fixed at the start. – Ryan Chen Dec 6 '20 at 1:09

## 1 Answer

As Alexandre Eremenko noted, the question is easy if you do not require independence. The necessary condition $$\mathbb{E}\left[S\right]=\sum_{i=1}^nh_i\alpha_i$$ noted by the OP is also sufficent (Edit: when one does not require an upper bound on the $$X_i$$.) If this condition holds, and $$\mathbb{E}\left[S\right]>0$$, then define $$X_i=\frac{\alpha_i S}{\mathbb{E}\left[S\right]}\,.$$ These variables will satisfy $$\sum_i h_iX_i=S$$.

Edit: If one requires an upper bound of 1 on the $$X_i$$, then a sufficient condition (when the $$\alpha_i$$ are decreasing) is to also require $$\alpha_1 \cdot \max S \le \mathbb{E}\left[S\right] \,.$$

• Note that all $X_i$ should be on the interval $[0,1]$. Since the case of identical $\alpha_i$ is trivial, we can assume that $\alpha_1> \alpha_2\ge \cdots \ge \alpha_n$ without loss of generality. In your answer, $X_1=\frac{\alpha_1S}{\sum_{i=1}^n h_i\alpha_i}>S$ because $\sum_{i=1}^n h_i=1$ and all $h_i>0$. This is not acceptable. – Ryan Chen Nov 29 '20 at 1:19