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Marcinkiewicz-Zygmund strong law of large numbers:

Let $X_1,X_2,···$ be i.i.d. with $E|X_1|^p< \infty$ for some $0< p <2$. Then $$ \begin{cases} \frac{S_n - nEX}{n^{1/p}} \to 0 \text{ a.s.} & \text{if } 1< p < 2\\ \frac{S_n}{n^{1/p}} \to 0 \text{ a.s.} & \text{if } 0<p<1\\ \end{cases} $$ where $S_n = X_1+X_2+\cdots +X_n$.

I tried to use Kronecker’s lemma, so we just need to show $$ \begin{cases} \sum_{n=1}^{\infty} \frac{X_n-EX}{n^{1/p}} \text{ converges a.s.} & \text{if } 1< p < 2\\ \sum_{n=1}^{\infty} \frac{X_n}{n^{1/p}} \text{ converges a.s.} & \text{if } 0<p<1\\ \end{cases} $$ Moreover, by Kolmogorov Three Series Theorem, it is equivalent to show $$ \begin{cases} \sum_n P(|X_n|>n^{1/p})<\infty \\ \sum_n E\left(\frac{X_n}{n^{1/p}} 1_{\{|X_n|<n^{1/p}\}} \right)\text{ converges} \\ \sum_n \text{Var}\left(\frac{X_n}{n^{1/p}} 1_{\{|X_n|<n^{1/p}\}} \right)<\infty \end{cases} $$ For the first one, I can use $E|X_n|^p< \infty$ to prove, but for the rest two I have no idea how to bound them.

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  • $\begingroup$ There is a difference between $\sum a_n <\infty$ and $\sum a_n$ convergent. $\endgroup$ Dec 4, 2023 at 11:46
  • $\begingroup$ Oh, I even don't realise this, I have updated, thanks $\endgroup$
    – Fireond
    Dec 4, 2023 at 12:05

1 Answer 1

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Here is a hint for the case $1<p<2$. Assume that $X_1$ is centered. Since $X_n$ has the same distribution as $X_1$, we have $$ \mathbb E\left(\frac{X_n}{n^{1/p}} 1_{\{|X_n|<n^{1/p}\}} \right)=\mathbb E\left(\frac{X_1}{n^{1/p}} 1_{\{|X_1|<n^{1/p}\}} \right). $$ Since $X_1$ is centered, we have $$ \mathbb E\left(\frac{X_1}{n^{1/p}} 1_{\{|X_1|<n^{1/p}\}} \right)=\mathbb E\left(\frac{X_1}{n^{1/p}} 1_{\{|X_1|\geqslant n^{1/p}\}} \right). $$ then use the same type of argument that are used to show the convergence of the first series.

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