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Let $p \in (0,2)$ and let $\xi_n, n \geq 1$, be iid random variables. Show that the following two conditions are equivalent:

  1. With probability one, the limit $$ \lim_{n \rightarrow \infty} \frac{1}{n^{1/p}} \sum_{k=1}^n \xi_k$$ exists and is finite.

  2. $\mathbb{E}\lvert \xi_i \rvert^p < \infty$ AND either $\mathbb{E} \xi = 0$ or $p \leq 1$.

I have tried using Holder's inequality but didn't really get anywhere. I don't really have any idea how to approach the problem for either direction...

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  • $\begingroup$ Try using Chebychev's inequality on $\dfrac{\sum_{k=1}^n |\xi_k|^p}{n}$, i.e what I'm thinking is in the lines of proving the WLLN $\endgroup$ Feb 12, 2015 at 7:22

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Here's a proof that (1)$\implies$ (2).

Letting $T_n=n^{-1/p}\sum_1^n \xi_n$, we have $$ \frac{\xi_n}{n^{1/p}}=T_n - T_{n-1}\cdot \frac{(n-1)^{1/p}}{n^{1/p}} $$ Letting $n\to\infty$ above, since $T_n\to T$ a.s, and $\frac{(n-1)^{1/p}}{n^{1/p}}\to 1$, we get $$\frac{\xi_n}{n^{1/p}}=T_n - T_{n-1}\cdot \frac{(n-1)^{1/p}}{n^{1/p}}\to T-T\cdot 1=0$$ so that $\xi_n/n^{1/p}\to 0$ a.s. This means $P(|\xi_n|/n^{1/p}>1\text{ i.o.})=P(|\xi_n|^p>n\text{ i.o.})=0$, so (using Borel Cantelli on the last inequality), $$ E|\xi|^p=\int_0^\infty P(|\xi|^p>t)\,dt\le \sum_{n\ge0} P(|\xi_n|^p>n)<\infty $$ proving $E|\xi|^p<\infty$. Now, suppose by way of contradiction that $p>1$ and $E\xi\neq0$. Using Jensen's, $E|\xi|^p<\infty$ allows you to show $E|\xi|<\infty$, so by SLLN, $$ \frac{\sum_{k=1}^n\xi_n}{n}\to E\xi\neq0 $$ almost surely as $n\to\infty$. We also have, since $p>1$, that $$ \frac1{n^{1/p-1}}\to \infty $$ Multiplying the two above limits implies that $$ \frac{\sum_{k=1}^n\xi_n}{n^{1/p}}=T_n\to\infty\qquad\text{a.s.} $$ contradicting that the limit was finite. Thus, we must have $p\le 1$ or $E\xi=0$.

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  • $\begingroup$ Also: for a proof that $(2)\implies (1)$ in the case where $p\le 1$, see this question for the special case $p=1/2$, the generalization is not bad. The case $p\in (1,2)$ and $E\xi_1=0$ is covered in Theorem 2.5.8 in Durrett's Probability text (4th edition). Both proofs are too long to include here. $\endgroup$ Feb 21, 2015 at 5:13

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