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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 \begin{equation} \sum_{i=1}^n h_iX_i=S,~\text{and} \end{equation} \begin{equation} \mathbb{E}\left[X_i\right]=\alpha_i. \end{equation}

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 $\mathbb{P}\left[X_1=\cdots=X_n=1|S=1\right]=1$.
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    $\begingroup$ independent $X_i$? $\endgroup$ – Dieter Kadelka Nov 27 '20 at 8:00
  • $\begingroup$ 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. $\endgroup$ – Alexandre Eremenko Nov 27 '20 at 14:13
  • $\begingroup$ @ Dieter Kadelka Not necessarily independent. $\endgroup$ – Ryan Chen Nov 28 '20 at 0:55
  • $\begingroup$ @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 $\endgroup$ – Ryan Chen Nov 28 '20 at 1:09
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    $\begingroup$ @YemonChoi You are right. The vector $\mathbf{\alpha}$ and the weighted vector $\mathbf{h}$ are fixed at the start. $\endgroup$ – Ryan Chen Dec 6 '20 at 1:09
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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 $$\alpha_1 \cdot \max S \le \mathbb{E}\left[S\right] \,.$$

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  • $\begingroup$ 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. $\endgroup$ – Ryan Chen Nov 29 '20 at 1:19

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