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?

  • $\begingroup$ 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

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 $0$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.