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)

I think that it is even not enough to demand absolute convergence of only the left hand side. All of the absolute expectations on the right have to converge, or one might have $\infty-\infty$. Also, absolute convergence of the right hand side is enough as we then can estimate the left hand side. So, a note might be added that the absolute expectation of the $X_i$ existing is a necessary and sufficient condition for the equality stated. --Lord_Farin 03:08, 27 October 2011 (CDT)