# If $E[|\sum_{i < l \leq j} \xi_l |^\gamma] \leq (\sum_{i < l \leq j} u_l)^\alpha$ and $\sum u_l < \infty$ then $\sum \xi_l$ converges almost surely

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?

• Are you sure there is no independence assumption here? The theorems you are quoting are not true without independence. – Kavi Rama Murthy Jan 7 at 7:49
• There is an obvious counterexample with $\pm 1$ valued random variables $\xi_i$. – Kavi Rama Murthy Jan 7 at 7:50
• @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. – cgmil Jan 10 at 22:44

## 1 Answer

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 $$0$$.

• 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 – cgmil Jan 10 at 22:40