1
$\begingroup$

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!

$\endgroup$

1 Answer 1

0
$\begingroup$

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, $$\frac{S_{n}}{\sqrt[p]{n}}=\frac{S_{n}}{n}\cdot \frac{n}{\sqrt[p]{n}}\overset{a.s}{\to} 0\cdot c =0$$

$\endgroup$
1
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.