Decomposition of the sum of nonnegative random variables Non-necessarily independent random variables $X_1,~X_2,~\cdots,~X_n$ are supported on $[0,a_1],~[0,a_2],~\cdots,[0,a_n]$ and with mean values $\mu_1,~\cdots,~\mu_n$ respectively, where all $a_i$ and $\mu_i$ are positive real numbers.
Assume that some system can only observe the sum of the above random variables, i.e., $S=\sum_{i=1}^{n}X_i$.
How to decompose $S$ such that $S=\sum_{i=1}^n V_i$ and for any $i\in\{1,~2,~\cdots,n\}$:

*

*$V_i$ is supported on $[0,a_i]$;

*The mean value $\mathbb{E}\left[V_i \right]=\mu_i$
An additional question is whether such the decomposition can be achieved by a series functions (i.e. $V_i=\phi_i(S))$.

The above problem is edited as follows.
Assume that we know that $S$ is the sum of $n$ unknown nonnegative random variables $X_1,~\cdots,~X_n$ with known peak values ($a_1,~\cdots,a_n$) and mean values ($\mu_1,~\cdots,~\mu_n$) and the distribution of $S$ is available.
How to decompose the random variable $S$ such that $V_1,~\cdots,~V_n$ satisfy the above-mentioned constraints?
 A: (The OP's first clarification of the question asked about the case where the distribution of $S$ is not given.)
If you are not told the distribution of $S$, then it is not possible in general.
For example, suppose you are told $n=2$, $\mu_1=\mu_2=1/2$, $a_1=1$, $a_2=2$.
What will you do if you observe $S=1$?
It could be that $S=1$ always, say for example $(X_1,X_2)=(1/2,1/2)$ with probability $1$. Then on average your response to $S=1$ needs to divide up the mass equally between $V_1$ and $V_2$.
On the other hand it could be that
\begin{equation}
(X_1,X_2)=\begin{cases}
(1,0)& \text{with probability }1/4,\\
(1,2)& \text{with probability }1/4,\\
(0,0)& \text{with probability }1/2.
\end{cases}.
\end{equation}
So you'll observe $S=1$ a quarter of the time, and $S=3$ a quarter of the time, and when you observe $S=1$ you need to respond with $V_1=1$, $V_2=0$.
So without some information on, for example, the distribution of $S$, you do not know how to respond appropriately when you observe $S=1$.
