Let $X_1,X_2,\ldots$ be an i.i.d. sequence of random variables such that $X_1\geq 0$ a.s. and $\mathbb P[X_1>x]\sim x^{-\alpha}$, where $\alpha<1$. This implies that $X_1$ does not have finite mean. Does the following law of iterated logarithm hold? $$ \limsup_{n\to\infty} \frac{X_1+\cdots+X_n}{n^{\alpha}(\log\log n)^{1-\alpha}}=c,\quad \text{a.s.}, $$ where $c$ is a constant depending on the distribution of $X_1$. A continuous-time version holds (see here). Also, a special case of the claim exists here (Theorem 3) regarding zeros of the simple random walk, in which $\alpha=\frac 12$.
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$\begingroup$ When $\alpha=3/2$ in $\mathbb P[X_1>x]\sim x^{-\alpha}$ say, then the distribution must fall off like $x^{-5/2}$ at $\infty$, so the mean should lead to the order $x\cdot x^{-5/2}=x^{-3/2}$ and thus the integral is convergent, or? $\endgroup$ – Diger Nov 23 '18 at 23:41
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$\begingroup$ @Diger you are right, the condition should be $\alpha <1$ not $\alpha <2$. I will correct it. $\endgroup$ – Ali Khezeli Nov 24 '18 at 3:13
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$\begingroup$ $\alpha \leq 1$ but fair enough ;) $\endgroup$ – Diger Nov 24 '18 at 13:16
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Yes, it holds. Thanks to Jean Bertoin for telling me the answer, the problem is solved in Theorem 4 of this paper.
