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I searched for a proof for the generalisation about the indepence of sums of independent random variables, but I can't find the result anywhere and I've also got quite some difficulty with proving the result myself, can anyone give me some good hints on how to proof this, or a link to the proof of the result itsself? What I'm trying to proof is the following:

Let $X_1, X_2, \dots X_n$ be independent discrete random variables all with the same image, then for all partitions $P_1, P_2, \dots P_j$ of the index set $\{1, 2, \dots, n\}$ the random variables $R_1 = \sum\limits_{i \in P_1} X_i $$,$ $R_2 = \sum\limits_{i \in P_2} X_i $ $,\dots,$ $R_j = \sum\limits_{i \in P_j} X_i $ are independent.

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  • $\begingroup$ What is your working definition of (mutually) independent random variables? $\endgroup$ – Erick Wong Sep 21 '16 at 19:56
  • $\begingroup$ @ErickWong I call $X_1, X_2, \dots X_n$ independent if $P(X_1 = x_1, X_2 = x_2, \dots X_n = x_n) = \prod_{i=1}^{n} P(X_i = x_i)$ $\endgroup$ – pokemonfan Sep 21 '16 at 21:10
  • $\begingroup$ Ok cool, then you just need to show that $P(R_1 = r_1, R_2 = r_2, \ldots, R_j = r_j) = \prod_{i=1}^j P(R_i = r_i)$. The idea is that $P(R_1 = r_1)$ can be expressed as a (possibly infinite) summation of values involving $X_i$ for $i \in P_j$. But the notation might get pretty hairy. So one suggestion would be to break it down into smaller steps. If you can first prove this for just the "finest possible" partition $\{1\}, \{2\}, \{3\}, \ldots, \{n-2\}, \{n-1,n\}$, then you can use induction (and relabelling) to prove it for all partitions. $\endgroup$ – Erick Wong Sep 21 '16 at 21:36

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