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I obtained the following problem from an old probability problem-set. Let $X_1,X_2,...$ be uncorrelated random variables with $\mathbb EX_i=0$ for all $i$. Suppose that there exists $0<p\leq 2$ such that $$\sum_{i=1}^\infty \frac{Var(X_i)}{i^p}<\infty \tag{*}$$ Show that with $S_n = \sum_{i=1}^n X_i$, we have $$\limsup_{n\rightarrow \infty}\frac{S_n}{n^{p/2}\log^2 n} \leq 0 \text{ a.s.}$$

My attempt: By Chebychev's Inequality and uncorrelatedness of the $X_i$, we have $$P(|S_n| \geq \epsilon n^{p/2}\log^2n) \leq \epsilon^2 \frac{Var(S_n)}{n^{p}\log^4n} \rightarrow 0$$ since $ \frac{Var(S_n)}{n^p} = \frac{1}{n^p}\sum_{i=1}^n Var(X_i) \rightarrow 0 $ by Kronecker's Lemma applied to $(*)$. This means $\frac{S_n}{n^{p/2}\log^2 n} \xrightarrow{p} 0$. Since $$ \sum_{n=1}^\infty \frac{1}{\log^4 n} = \infty$$ Borel-Cantelli Lemma does not immediately apply...

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I belive you can even improve your result by a bit. Check hartman-winter theorem for a very similar theorem that we can apply to your problem. This theorem states that

$$ \limsup \frac{S_n}{\sqrt{loglog(n)\sum_{i=1}^nvar(X_i)}}\to const. $$

Notice that you want to prove something similar, so if we prove that your divisor is larger (in limit) than $\sqrt{loglog(n)\sum_{i=1}^nvar(X_i)}$, it will imply what you want. That is, we want to prove $ \frac{\sqrt{loglog(n)\sum_{i=1}^nvar(X_i)}}{n^{p/2}log^2n}\to 0. $ This is actually easy, do a square and you will get that $$ \frac{loglog(n)}{log^4n}\frac{\sum_{i=1}^nvar(X_i)}{n^{p}}\to 0, $$ where the first fraction is trivial that converge to 0 and the second also $\frac{\sum_{i=1}^nvar(X_i)}{n^{p}}\to 0$ from Kroneker lemma.

All together

$$ \limsup \frac{S_n}{n^{p/2}log^2n} = \frac{S_n}{\sqrt{loglog(n)\sum_{i=1}^nvar(X_i)}} \cdot o(1)\to const\cdot 0 =0. $$

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  • $\begingroup$ Very nice! Would it be possible to do even better and prove the limit exists and is actually zero? (that is that the liminf is nonnegative) $\endgroup$ Commented Jan 10, 2022 at 14:04
  • $\begingroup$ Well of course, it does converge to 0. Because something larger converge to a constant, you divide it by something large, surely it has to converge to 0 not only to something non-positive. So yes, you can prove stronger thing :) $\endgroup$ Commented Jan 10, 2022 at 14:11
  • $\begingroup$ Oh yes that's for limsup but I was asking about lim. Either way, would the law of iterated logarithm only apply when the random variables are independent? Here, we only assume uncorrelatedness. $\endgroup$ Commented Jan 10, 2022 at 15:23
  • $\begingroup$ If the variables were independent, the problem actually becomes very trivial. $\endgroup$ Commented Jan 10, 2022 at 15:24
  • $\begingroup$ Hmm, i see. Yes, LIL says that it requires independence. Intuitively, there exists some result using only uncorrelated r.v. for LIL, but you have to look i am not sure. $\endgroup$ Commented Jan 10, 2022 at 15:38

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