Let $X_k$ be independent random variable s.t. $\sum_{k=1}^nX_k\rightarrow_{a.s.} X$. So,
$$X=\sum_{k=1}^\infty (X_k^+ -X_k^-)$$
Is it true that $X=\sum_{k=1}^\infty (X_k^+)-\sum_{k=1}^\infty (X_k^-)$?
In particular, can we say that $E[X]=E[\sum_{k=1}^\infty (X_k^+)]-E[\sum_{k=1}^\infty (X_k^-)]$?
Thank's!
No, this is not even true for deterministic sequences. Just consider $$X_k := \frac{(-1)^k}{k},$$ then $\sum_{k=1}^n X_k \to 0$, but since both series $\sum_{k=1}^{\infty} X_k^+$ and $\sum_{k=1}^{\infty} X_k^-$ are divergent, the equality $$X = \sum_{k=1}^{\infty} X_k^+ - \sum_{k=1}^{\infty} X_k^-$$ does not hold.