Prove $\limsup n^{-1/\alpha} |X_1+\ldots+X_n|=\infty$ for an iid sequence $(X_n)_{n \in \mathbb{N}}$ with $\mathbb{E}(|X_1|^{\alpha})=\infty$

Question

Suppose that $X_1, X_2,...$ are i.i.d random variables with $E|X_1|^{\alpha}=\infty$ for some $\alpha>0$. Define $S_n=\sum_{k=1}^nX_k$. Prove that $$\limsup_{n\rightarrow\infty} n^{-1/\alpha}|S_n|=\infty$$ almost surely.

What I proved so far is: For $\beta>0$, one has $\sum_{k=1}^\infty P(|X_k|>\beta k^{1/\alpha})=\infty$, and also $\limsup_{n\rightarrow\infty}n^{-1/\alpha}|X_n|=\infty$. I m not sure how is this result useful to prove the above limit supremum is infinite. How are we going to use Borel Cantelli, or is there an easier way to do this without using BC?

Thanks

Answer 1

As you pointed out, it follows from the Borel-Cantelli lemma that

$$\mathbb{P} \left( \limsup_{n \to \infty} n^{-1/\alpha} |X_n| = \infty \right) = 1.$$

If we can show

$$A:=\left\{ \limsup_{n \to \infty} n^{-1/\alpha} |S_n|< \infty \right\} \subseteq \left\{ \limsup_{n \to \infty} n^{-1/\alpha} |X_n| < \infty \right\}=:B,$$

then we are done. To see this, fix $\omega \in A$. Since

$$X_n = S_n-S_{n-1}$$

we have

$$\limsup_{n \to \infty} n^{-1/\alpha} |X_n(\omega)| \leq \limsup_{n \to \infty} n^{-1/\alpha} |S_n(\omega)| + \limsup_{n \to \infty} n^{-1/\alpha} |S_{n-1}(\omega)| < \infty;$$ hence, $\omega \in B$. Here we have used the subadditivity of $\limsup$ and the fact that

$$\begin{align*} \limsup_{n \to \infty} n^{-1/\alpha} |S_{n-1}(\omega)| &= \limsup_{n \to \infty} \frac{n^{-1/\alpha}}{(n-1)^{-1/\alpha}} (n-1)^{\alpha} |S_{n-1}(\omega)| \\ &= \limsup_{n \to \infty} (n-1)^{-1/\alpha} |S_{n-1}(\omega)|<\infty \end{align*}$$

for any $\omega \in A$.

Remark: This result is, for $\alpha=1$, a converse to the strong law of large numbers.