Talk:Sum of Expectations of Independent Trials

When splitting the summation, I use that both of the expressions exist (which they do by the assumption I added - when they exist). This should be mentioned... --LF
 * Doesn't that just come under distributive law? Not sure whether we've proved it for infinite sets, but it's definitely there for finite ones. -- PM
 * No. Consider $\sum\frac1n$ and $\sum\frac{-1}n$, and their sum. Generally, the convergence of the right implies that on the left, I think... By the way, the countable case of this theorem relates to the infinite cartesian product I just posted about (for the associated sigma-algebra), and so depends on AoC. --LF
 * Now you mention it, I seem to remember something about this: doesn't the sum have to be absolutely convergent? In this case yes I concur, we do need a result that can be cited. Might be in $\S 1.2.3$ of Knuth TAOCP. --prime mover 17:03, 26 October 2011 (CDT)
 * I'm pretty sure absolute convergence is the right condition; essentially without absolute convergence, you can't mess with the order/grouping of terms at all. I've actually had this issue come up twice in the past week - once in my Probability class and once when running a Calc help session (had to find the problem with $0 = 0 + 0 + \ldots = (1 - 1) + (1 - 1) + \ldots = 1 - (1 - 1) - (1 - 1) - \ldots = 1 - 0 - 0 - \ldots = 1$). --Alec  (talk) 23:52, 26 October 2011 (CDT)