Suppose $X_i$ are i.i.d., $\mathbb{E}X_1 = 0$, $\mathbb{E}X^2_1<\infty$ and $S_n = \sum_{i=1}^n X_i$ I am to calculate three following limits:
- $\liminf_{n \to \infty} \frac{S_n}{ \sqrt{n \ln ( \ln(n))}}$
- $\lim_{n \to \infty} \frac{S_n}{n^{\alpha}}$ where $\alpha > \frac{1}{2}$
- $\lim_{n \to \infty} \frac{-S_n}{\sqrt{n}lnn}$
The first one was relatively easy, and I got $-\sqrt{2 \mathbb{E}X^2_1}$ as result (using symmetry and LIL Hartman's-Winter). I got stuck on the next two however. What can be done here?
Write $$\frac{S_n}{n^\alpha} = \frac{S_n}{\sqrt{n \ln \ln(n)}} \frac{\sqrt{\ln \ln (n)}}{n^{\alpha-1/2}}.$$ The first factor is bounded below while the second one goes to $0$. Thus $$\liminf_n \frac{S_n}{n^\alpha} = 0.$$ Now applying this result to the random walk $(-S_n)_n$ gives $$0=\liminf_n \frac{-S_n}{n^\alpha} = -\limsup_n \frac{S_n}{n^\alpha}.$$ This proves that $\lim_{n\to \infty}\frac{S_n}{n^\alpha} = 0$. The third limit is similar.