# Law of Iterated Logarithm when the mean does not exist

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 $$\alpha=\frac 12$$.

• 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? – Diger Nov 23 '18 at 23:41
• @Diger you are right, the condition should be $\alpha <1$ not $\alpha <2$. I will correct it. – Ali Khezeli Nov 24 '18 at 3:13
• $\alpha \leq 1$ but fair enough ;) – Diger Nov 24 '18 at 13:16