Sharp convergence rate of law of large numbers with infinite higher moments

Question

Let $X_1,X_2,...$ be i.i.d. with $\mathbb{E}[X_1]=0$ and $\mathbb{E}[|X_1|^a]<\infty$ if and only if $a=1$. By the law of large numbers $\frac{1}{n}\sum_{i=1}^n X_i \to \mathbb{E}[X_1]=0$. Is there a $\varepsilon>0$, such that, $\frac{1}{n^{1-\varepsilon}} \sum_{i=1}^n X_i \to 0$? I know that the answer would be yes, if there was an $a>1$ with $\mathbb{E}[|X_1|^a]<\infty$. Since this is not the case, I guess the answer is no.

Answer 1

Suppose that $(X_i)_{i\geqslant 1}$ is an i.i.d. sequence such that $\frac{1}{n^{1-\varepsilon}} \sum_{i=1}^n X_i \to 0$ in probability. Let $(X'_i)_{i\geqslant 1}$ be an independent copy of $(X_i)_{i\geqslant 1}$. Defining $Y_i=X_i-X'_i$, we get that $(Y_i)_{i\geqslant 1}$ is i.i.d., $Y_i$ has the same law as $-Y_i$ and $\frac{1}{n^{1-\varepsilon}} \sum_{i=1}^n Y_i \to 0$ in probability.

By Theorem 1 in Michael Klass. Henry Teicher. "Iterated Logarithm Laws for Asymmetric Random Variables Barely with or Without Finite Mean." Ann. Probab. 5 (6) 861 - 874, December, 1977. https://doi.org/10.1214/aop/1176995656,

we derive that $n\mathbb P\left(\lvert Y_1\rvert>n^{1-\varepsilon}\right)\to 0$. This implies that there is $a>1$ such that $\mathbb E\left[\lvert Y_1\rvert^a\right]<\infty$ hence $\mathbb E\left[\lvert X_1\rvert^a\right]<\infty$.