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Suppose $\xi_1, \xi_2, \ldots$ are random variables that satisfy

$$E\left[\left|\sum_{i < l \leq j} \xi_l \right|^\gamma\right] \leq \left(\sum_{i < l \leq j} u_l\right)^\alpha$$

for $\gamma \geq 0$, $\alpha > 1$, $u_l$ non-negative, and $\sum u_l < \infty$. I want to show $\sum \xi_l$ converges almost surely.

Let $S_m = \sum_{i = 1}^m \xi_i$ and $M_m = \max_{1 \leq i \leq m} |S_i|$. This criterion on the expectation can be used to show

$$P(|S_j - S_i| \geq \lambda) \leq \frac{1}{\lambda^\gamma} \left(\sum_{i < l \leq j} u_l\right)^\alpha$$

Because of this we can use a theorem from Billingsley's book Convergence of Probability Measures (first edition, 1968) to say

$$P(M_m \geq \lambda) \leq \frac{K}{\lambda^\gamma} (u_1 + \ldots + u_m)^\alpha$$

(Billingsley says to use that theorem to reach the desired result.) $K$ depends only on $\gamma$ and $\alpha$.

After this I'm not sure how to proceed. Sure, one could take $m \to \infty$ and then see that the sum is bounded almost surely but I don't see how that can show that the sum is convergent. And I'm not seeing where to judiciously use the Borel-Cantelli lemma.

Basically, I don't see how the bound on $P(M_m \geq \lambda)$ is useful for proving almost-sure convergence but supposedly it can (and in fact needs to be) used to get the desired result.

So what should I do next?

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  • $\begingroup$ Are you sure there is no independence assumption here? The theorems you are quoting are not true without independence. $\endgroup$ – Kavi Rama Murthy Jan 7 at 7:49
  • $\begingroup$ There is an obvious counterexample with $\pm 1$ valued random variables $\xi_i$. $\endgroup$ – Kavi Rama Murthy Jan 7 at 7:50
  • $\begingroup$ @KaviRamaMurthy Billingsley explicitly notes that the variables need not be independent, and the counterexample you mention does not satisfy the assumption mentioned; the tightest $u_l$ you could have would not be summable. $\endgroup$ – cgmil Jan 10 at 22:44
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I guess we need some kind of weak dependence assumption on $(\xi_j)$, but anyway, this is only about how we get the convergence result using the maximal inequality about $M_n$. First, we note that we can bound $$ P(\max_{n\le i\le N}|S_i-S_n|>\lambda)\le\frac{K}{\lambda^\gamma}(u_{n+1} +\cdots u_N)^\alpha $$ and hence $$ P(\max_{n\le i, j\le N}|S_i-S_j|>\lambda)\le\frac{K'}{\lambda^\gamma}(u_{n+1} +\cdots u_N)^\alpha. $$ (We can regard $\xi_j$ as starting from index $j=n+1$.) To show that $S_n$ converges, id est, that $\limsup_{n\to\infty} S_n = \liminf_{n\to\infty} S_n$, it seems natural for us to control the oscillation defined by $$ \limsup_{i,j\to \infty}|S_i-S_j| =\lim_{n\to\infty}\sup_{i,j\ge n} |S_i-S_j|. $$ Let $W_{n,N} = \sup_{n\le i,j\le N} |S_i-S_j| $ and $W_n =\sup_{ i,j\ge n} |S_i-S_j|=\lim_{N\to\infty}W_{n,N}$. We have $$ P(W_n>\lambda) =\lim_{N\to\infty}P(W_{n,N}>\lambda) $$by monotonic convergence of $W_{n,N}$. Hence we get the bound $$ P(W_n>\lambda)\le\frac{K'}{\lambda^\gamma}\left(\sum_{j>n}u_{j} \right)^\alpha.\tag{*} $$ Finally, we observe $$ \limsup_{n\to\infty} S_n -\liminf_{n\to\infty} S_n =\lim_{n\to\infty} W_n=:W $$ and $$ \{S_n\text{ diverges}\} = \{\limsup_{n\to\infty} S_n -\liminf_{n\to\infty} S_n>0\} =\bigcup_{j\in\mathbb{N}}\{W\ge 1/j\}. $$ Since it holds that $$ P(W\ge \lambda) =\lim_{n\to\infty} P(W_n\ge \lambda)= 0 $$ for all $\lambda>0$ by the above estimate $(*)$, it follows that $\{S_n\text{ diverges}\}$ is a countable union of null probability events, and hence has probability $0$.

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  • $\begingroup$ Thank you for your answer. It was very helpful! I have a follow-up question of a similar nature; perhaps you could answer it too? math.stackexchange.com/q/3069263/360447 $\endgroup$ – cgmil Jan 10 at 22:40

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