Divergence of general random series and a special case

Is there any sufficient condition in terms of moments under which $$\sum_{n=1}^{\infty} X_n$$ diverges a.s.?Here $X_n$ are not independent

I am given that $\sum_n E[X_n]$ diverges. Actually, I am interested in the following special case:

Let $\{F_n\}$ be a filtration, and $\{A_n\}$ be a sequence of events such that $A_n \in F_n$. Suppose $P(A_n) > 0 \forall n$. Then what is the weakest additional condition that will ensure the following

$$\sum_{n=1}^{\infty} P(A_n | F_{n-1}) = \infty \mbox{ a.s.}$$

One condition is $P(A_n i.o) =1$, but I want something verifiable.

• $E(X_n)$ = constant $\neq 0$ – Carlo Beenakker Feb 4 '16 at 10:37
• Probably not. It's easy to make it converge by making it almost always 0, say, multiply anything bi independent bernouillis with $\sum p_i < \infty$, but that leaves you a lot of room to pick the moments. – user83457 Feb 4 '16 at 11:58
• Do you mean that the sum diverges almost surely? If so, then having constant expectation is not enough. – Ori Gurel-Gurevich Feb 4 '16 at 13:02
• Certainly not! If the $X_n$ take a value $A_n$ with probability $2^{-n}$ and 0 otherwise, then the series is convergent by Borel-Cantelli, no matter what the dependence structure is. At the same time, by building large $A_n$'s, you can exceed any desired moment bounds. – Anthony Quas Feb 4 '16 at 16:13
• @AnthonyQuas You can exceed lower bounds, but not all conditions are lower bounds. – Robert Israel Feb 4 '16 at 22:31

3 Answers

Here is one simple sufficient condition: $\sum \mathsf{E} X_n$ diverges, and $\sum (\mathsf{Var} X_n)^{1/2} < \infty$.

Indeed, the variance condition ensures that $\sum (X_n - \mathsf{E} X_n)$ converges in $L^2$, so if $\sum \mathsf{E} X_n$ diverges, so does $\sum X_n$ (say, in probability).

• Can you explain the 2nd and 3rd line a bit more. Thanks. – Sosha Feb 5 '16 at 2:08

Let $\mu_n$ and $\sigma_n$ be the mean and standard deviation of $X_n$. Suppose there is a sequence of positive numbers $k_n$ such that $\sum_n 1/k_n^2 < \infty$ while $\sum_n (\mu_n - k_n \sigma_n) = + \infty$. By Chebyshev's inequality, $\mathbb P(X_n \ge \mu_n - k_n \sigma_n) > 1 - 1/k_n^2$, so $\sum_n 1/k_n^2 < \infty$ implies a.s. $X_n \ge \mu_n - k_n \sigma_n$ for all sufficiently large $n$, and thus a.s. $\sum_n X_n = +\infty$.

Analogous conditions could be formulated in terms of other moments.

The key observation is that even if the sum of expectations of individual $X_n$ diverges, if correlations among the $\{X_{k_1}, X_{k_2}\ldots X_{k_m}\}$ with $k_m = \sup (k_i)$ do not fall off sufficiently fast as $k_m \to \infty$, then it is possible to "save" the convergence, that is, to have a finite probability of that the sum does not diverge.

With that concern in mind, here is a sufficient condition (though I would not say this is anything like a necessary condition) that $\sum_1^\infty X_n$ diverges a.s., expressed in terms of the moments (including of course the correlation expectations):

If for some $C \in \Bbb{R}, C > 0$ $$\lim_{n\to \infty} n E(X_n) = C$$ and for all partitions of integers $k = \sum_{i=1}^m k_i$ with $k > 1$, and $k_i \geq 0$ for all $i$, $$\lim_{n\to \infty} n^{k+1} \left[\prod_{i=1}^k E(X_i^{k_i})- E\left((\prod_{i=1}^k X_i^{k_i}\right)\right]=0$$ then $\sum_{n=1}^\infty X_n$ diverges a.s.