I'm studying probability theory, especially about limit theorem. And I got some trouble in moving forward.

The problem is:

Let $(X_n)$ be i.i.d random variables with $E(X_1)=0$ and $E(|X_1|^p)<1$ for some $1<p<2$. Show that $\displaystyle \frac{S_n}{n^{1/p}}$ converges to $0$ almost surely, where $S_k=X_1+...+X_k$.

At first, I tried with Strong Laws of Large Numbers, but it doesn't help. I think the condition that $E(|X_1|^p)<1$ for some $1<p<2$ is crucial, but I don't figure out what it means.

Please give some advice!


1 Answer 1


First note that $$\frac{S_{n}}{n^{\frac{1}{p}}}=\frac{S_{n}}{\sqrt[p]{n}}=\frac{S_{n}}{n}\cdot \frac{n}{\sqrt[p]{n}}$$ Now, since that $(X_{n})$ are identically distributed, and since that $$\mathbb{E}[|X_{1}-0|^{p}]<1<\infty \implies \text{absolutely integrable:} \quad \mathbb{E}[X]<\infty$$ So, by The Strong Law of Large Number (SLLN), we can conclude that $$\frac{S_{n}}{n}\overset{a.s}{\to}\mathbb{E}[X]=0$$ So, $$\frac{S_{n}}{\sqrt[p]{n}}=\frac{S_{n}}{n}\cdot \frac{n}{\sqrt[p]{n}}\overset{a.s}{\to} 0\cdot c =0$$

  • 1
    $\begingroup$ In the last line, how can $\frac{n}{\sqrt[p]{n}} \to $ a constant almost surely ? $\endgroup$
    – Kolmogorov
    Dec 14, 2020 at 23:59

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