An upper bound for $\sum_{n=1}^{\infty} n^{r-2} P\{|\sum_{k=1}^{n} \sum_{i=-\infty}^{\infty} a_{i} X_{i+k}|>\varepsilon n\}$

Question

Let $1 \leq r<2$ and let $\left\{X, X_{i},-\infty<i<\infty\right\}$ be a sequence of pairwise i.i.d. random variables. Let $\left\{a_{i},-\infty<i<\infty\right\}$ be a sequence of real constants such that $\sum_{i=-\infty}^{\infty}\left|a_{i}\right|^{\theta}<\infty,$ where $\theta \in(0,1)$ if $r=1$ and $\theta=1$ if $1<r<2$. Assume additionally that $EX = 0$ and $E|X|^r<\infty$.

for every $\varepsilon > 0$, show that :

\begin{array}{c} \sum_{n=1}^{\infty} n^{r-2} P\left\{\left|\sum_{k=1}^{n} \sum_{i=-\infty}^{\infty} a_{i} X_{i+k}\right|>\varepsilon n\right\} \\ \leq \sum_{n=1}^{\infty} n^{r-2} P\{|\sum_{i=-\infty}^{\infty} a_{i} \sum_{j=i+1}^{i+n} X_{j} I(|X_{j}|>n)|>\varepsilon n / 2\} \\ +c \sum_{n=1}^{\infty} n^{r-2} P\left\{\left|\sum_{i=-\infty}^{\infty} a_{i} \sum_{j=i+1}^{i+n}\left(X_{j} I\left(\left|X_{j}\right| \leq n\right)\right.\right.\right. \left.\left.-E X_{j} I\left(\left|X_{j}\right| \leq n\right)\right) |>\varepsilon n / 4\right\} \end{array}

where $c$ is a constant.

In case it helps, I've already proven that : $$\sum_{k=1}^{n} \sum_{i=-\infty}^{\infty} a_{i} X_{i+k}=\sum_{i=-\infty}^{\infty} a_{i} \sum_{j=i+1}^{i+n} X_{j}$$

and that : $$n^{-1}\left|E \sum_{i=-\infty}^{\infty} a_{i} \sum_{j=i+1}^{i+n} X_{j} I\left(\left|X_{j}\right| \leq n\right)\right| \to 0$$

this inequality that I'm stuck with is used to proved a lemma that is helpful in proving a variant of the strong law of large numbers for pairwise i.i.d random variables.

any tips or comments will be greatly appreciated, thanks !

Edit : it's from this paper, page 507/7

Answer 1

After having looked at the paper, it seems that the authors are interested in showing the convergence of the series $$ \sum_{n=1}^{\infty} n^{r-2} P\left\{\left|\sum_{k=1}^{n} \sum_{i=-\infty}^{\infty} a_{i} X_{i+k}\right|>\varepsilon n\right\}.$$ Fix a positive $\varepsilon$; there exists an $N$ such that for $n\geqslant N$, $$\tag{*}n^{-1}\left|E \sum_{i=-\infty}^{\infty} a_{i} \sum_{j=i+1}^{i+n} X_{j} I\left(\left|X_{j}\right| \leq n\right)\right| \lt \varepsilon/4.$$ For $n\geqslant N$, the inclusion $$ \left\{\left|\sum_{k=1}^{n} \sum_{i=-\infty}^{\infty} a_{i} X_{i+k}\right|>\varepsilon n\right\}\subset \left\{\left|\sum_{i=-\infty}^{\infty} a_{i} \sum_{j=i+1}^{i+n} X_{j} I(|X_{j}|>n)\right|>\varepsilon n / 2\right\}\cup\left\{\left|\sum_{i=-\infty}^{\infty} a_{i} \sum_{j=i+1}^{i+n} X_{j} I(|X_{j}|\leqslant n)\right|>\varepsilon n / 2\right\} $$ holds and using (*) gives $$ \left\{\left|\sum_{i=-\infty}^{\infty} a_{i} \sum_{j=i+1}^{i+n} X_{j} I(|X_{j}|\leqslant n)\right|>\varepsilon n / 2\right\}\subset\left\{\left|\sum_{i=-\infty}^{\infty} a_{i} \sum_{j=i+1}^{i+n} X_{j} I(|X_{j}|\leqslant n)-\mathbb E\left[ X_{j} I(|X_{j}|\leqslant n)\right]\right|>\varepsilon n / 4\right\} $$ hence $$ \left\{\left|\sum_{k=1}^{n} \sum_{i=-\infty}^{\infty} a_{i} X_{i+k}\right|>\varepsilon n\right\}\subset \left\{\left|\sum_{i=-\infty}^{\infty} a_{i} \sum_{j=i+1}^{i+n} X_{j} I(|X_{j}|>n)\right|>\varepsilon n / 2\right\}\cup \left\{\left|\sum_{i=-\infty}^{\infty} a_{i} \sum_{j=i+1}^{i+n} X_{j} I(|X_{j}|\leqslant n)-\mathbb E\left[ X_{j} I(|X_{j}|\leqslant n)\right]\right|>\varepsilon n / 4\right\} $$ and it suffices to prove the convergence of the series $$\sum_{n=1}^{\infty}n^{r-2} P\left\{\left|\sum_{i=-\infty}^{\infty} a_{i} \sum_{j=i+1}^{i+n} X_{j} I(|X_{j}|>n)\right|>\varepsilon n / 2\right\}$$ and $$ \sum_{n=1}^{\infty} n^{r-2} P\left\{\left|\sum_{i=-\infty}^{\infty} a_{i} \sum_{j=i+1}^{i+n}\left(X_{j} I\left(\left|X_{j}\right| \leq n\right) -E X_{j} I\left(\left|X_{j}\right| \leq n\right)\right) \right|>\varepsilon n / 4\right\}. $$

Answer 2

I don't think that this inequality is true as written. Consider taking n = 2, and $X_1 = -100$ w.p. $1/100$ and $100/99$ w.p. 99/100. Choose $\epsilon = 1/2$, and note that $P(|X_1| > 1) =1$. However, $P(X_1 1(|X_1| > 2)) = 1/100$, and $P(|X_1 1(|X_1| < 2) - E[X_1 1(|X_1| < 2)]| > 1/4 ) = P(|X_1 1(|X_1| < 2) - 1| > 1/4) = 0$.

If you did not subtract off that expectation inside the probability in the second term, it would be true by a simple union bound. However, as it stands, I do not think it is correct. If you can show the source material, people might be more helpful.

(P.S. I checked whether Etemadi's proof of SLLN which uses pairwise independence has this step, and it does not since it deals with finite variance case. It seems you are interested in the case where there isn't a finite variance?)