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Kolmogorov's 3-series theorem states that if $X_k, k \geq 1$ are independent, $A > 0$ a threshold level, $Y_k = X_k 1(|X_k|\leq A)$,

then $\sum_k X_k$ converges iff all three series converge:

  1. $\sum_{k \geq 1} P[|X_k| > A] < \infty$
  2. $\sum_{k \geq 1} E[Y_k]$ converges
  3. $\sum_{k \geq 1} Var[Y_k] < \infty$

In the lecture it was mentioned that 1. and 3. are absolute convergence, whereas 2. is convergence of partial sums.

First of all: is this clear from notation? So if we write $< \infty$, then it is not clear that the sum converges? But since the terms all are non-negative, we have that the sequence must converge, and hence we also have absolute convergence? When it is written that a serie converges, then this means that it does not oscillate but goes to a fixed value $(< \infty)$? But it could be that some of the terms cancel, so that we would not have absolute convergence?

Why do we write $< \infty$ for absolute convergence and "converges" for convergence of partial sums? Or has this something to do with the nonnegativity? Would it be wrong to write in 2. $\sum_{k \geq 1} E[Y_k] < \infty$, because that could also imply that the series jumps between some values that are finite?

When can we do reordering of the terms? Whenever they converge or only with absolute convergence?

Sorry for the weird questions, but I do not study maths and only attended a very basic analysis and probability course. Thanks a lot!

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    $\begingroup$ You can only reorder terms for absolutely convergent sequences. The notation $\sum a_k <\infty $ is/should generally only be used if $a_k \geq 0$ for all $k $. In this case, it means absolute convergence, which (in this case) is the same as convergence of partial sums. $\endgroup$ – PhoemueX Dec 20 '15 at 11:57
  • $\begingroup$ Thanks for your answer. Can you then maybe help me explaining why we are allowed to do the following step: We assume 1,2,3 and want to show that $\sum X_k$ converges. We introduce $\widetilde{Y}_k:=Y_k -E[Y_k]$ and can show, with some other method, that $\sum_k \widetilde{Y}_k$ converges P-a.s. Then we write $\sum_k Y_k = \sum_k \widetilde{Y}_k + \sum_k E[Y_k]$ to show (together with assumption 2.) that $\sum_k Y_k$ converges P-a.s. So why are we allowed to do reordering here? Why is the convergence of $\sum \widetilde{Y}_k$ absolute? $\endgroup$ – user146358 Dec 20 '15 at 12:58
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    $\begingroup$ Reordering means to consider the new series $\sum a_{\pi(k)}$, where $\pi : \Bbb{N} \to \Bbb{N}$ is a bijection. What you want to do is from convergence of $\sum_k a_k$ and $\sum_k b_k$ to deduce convergence of $\sum_k (a_k -b_k)$. This is always justified, even without absolute convergence, simply by using the usual rules for limits (applied to the sequence of partial sums). $\endgroup$ – PhoemueX Dec 20 '15 at 13:28
  • $\begingroup$ Thanks a lot, I wasn't aware of this! This explains everything now. :-) $\endgroup$ – user146358 Dec 20 '15 at 13:37

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