Proof that "symmetrized" sequence of random variables is independent. While studying the following theorem from Loeve's book on probability:

Let $X_n$ be uniformly bounded random variables. If $\sum X_n$
  converges a.s. then $\sum \sigma^{2}X_n$ and $\sum E X_n$ converges

He writes the proof as follows:

To the random variables $X_n$ we associate random variables $X_n'$
  such that $x_n$ and $X_n'$ are identically distributed for every n and
  $X_1,X_1',X_2,X_2'...$ is a sequence of independent random variables.
  We form the "symmetrized" sequence $X_{n}^{s} = X_n - X_n' $ of
  independent random variables.

So I checked the section on symmetrization but I cannot figure out the following:
-How can such a $X_n'$ be built in general? (they need to be independent and identically distributed)
-How can I proof that the "symmetrized" sequence is made up with independent random variables?
 A: Not a mathematician so please let me know if this is completely off base, and I will delete this answer. The way I understand symmetrization, as used in the proof, is the subtraction, from each $X_n$, of an independent RV $X_n'$ with the same distribution as $X_n$. Assuming $X_n$ were independent (not sure if this was supposed to be the case, but is suggested by the proof wherein the sequence $X_n$,$X_n'$ is independent), $X_n'$ will also be independent because they are identically distributed. By the same reason, for all $n$ and $m$, $X_n'$ will be independent of $X_m$, because otherwise $X_n$ would not be independent of $X_m$. Independence of $X_n^s$ follows from 3 independent RVs A,B,C satisfying
$$
P_{A-B|C}(a-b<x) = \int_{-\infty}^\infty dP_{B|C}(b) P_{A|C}(a<x+b) = \int_{-\infty}^\infty dP_{B}(b) P_{A}(a<x+b) = P_{A-B}(a-b<x)
$$
Then if we have another independent RV D, we have both C and D independent of A-B, so C-D is independent of A-B. This is a special case of this general proof.
To use this symmetrized series to show $\sum\sigma^2X_n$ and $\sum EX_n$ converge, note convergence of $\sum\sigma^2X_n$ implies convergence of the latter, and $\sum\sigma^2X_n=1/2\sum\sigma^2X_n^s$, so it suffices to show convergence of $\sum\sigma^2X_n^s$. I will omit this complicated proof, as it seems beyond the scope of the question.
