1
$\begingroup$

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?

Improve this question
$\endgroup$
1
  • $\begingroup$ Can you clarify what it is that we are given? We are told the $\mu_i$ and the $a_i$, but we are not told the joint distribution of the $X_i$, or even just the distribution of $S$ alone - right? $\endgroup$
    – James Martin
    Oct 19, 2020 at 9:08

1 Answer 1

Reset to default
1
$\begingroup$

(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 $S=1$.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Right. What is the result if we assume the distribution of the sum is also available? $\endgroup$
    – RyanChan
    Oct 19, 2020 at 23:35
  • $\begingroup$ Thank you for your remainder. I will rewrite a new question. $\endgroup$
    – RyanChan
    Oct 20, 2020 at 9:15
  • $\begingroup$ @RyanChen Thanks. If distribution of $S$ is known, I think it is possible. First, when the distribution of $S$ is discrete, then $\phi_i(S)=E(X_i|S)$ is a solution, and you can find it via linear programming (you are looking for a solution to a system of linear equations which is contained within a given simplex). If $S$ is not discrete, maybe there's a more sophisticated approach, but at least you could approximate the problem by discrete ones to get an answer to within whatever desired accuracy. $\endgroup$
    – James Martin
    Oct 20, 2020 at 9:24
  • $\begingroup$ Can you explain how to obtain $E(X_i|S)$ with more details? Some related papers are also useful. $\endgroup$
    – RyanChan
    Oct 20, 2020 at 9:48
  • $\begingroup$ I was a bit loose. You need a solution of a particular set of linear equations, which lies within a given simplex, and $E(X_i|S)$ is one such solution. You may not be able to determine precisely which one is $E(X_i|S)$ (but any such solution is good to give you your $\phi_i$). Suppose $S$ takes finitely many values and let $p_j=P(S=s_j)$. Then you can set $\phi_i(s_j)=e_{ij}$ where $(e_{ij})$ solves $\sum_j p_je_{ij}=\mu_i$ for each $i$, subject to $\sum_i e_{ij}=s_j$ for each $j$ and $e_{ij}\in[0,a_i]$ for each $i$, $j$. You know that $e_{ij}=E(X_i|S=s_j)$ gives one solution to this system. $\endgroup$
    – James Martin
    Oct 20, 2020 at 10:17