For i.i.d. random variables $X_1,\dots, X_n$ with $E(|X_1|)<\infty$. Does the following equation hold? $$ \left|\frac{1}{n}\sum_{i=1}^n X_i-E(X_1)\right|=O_P\left(\frac{1}{\sqrt{n}}\right) $$ I know that if $\operatorname{var}(X_1)<\infty$, the above equation holds due to the central limit theorem. But now I only have $E(|X_1|)<\infty$ available. Besides, under $E(|X_1|)<\infty$, the strong law of large numbers does not give the rate $\frac{1}{\sqrt{n}}$.
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1$\begingroup$ What does $O_P$ mean? $\endgroup$– Fedor PetrovCommented Jun 12, 2021 at 17:23
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$\begingroup$ @FedorPetrov, big O in the probability sense. See arxiv.org/pdf/1108.3924.pdf $\endgroup$– JohnCommented Jun 12, 2021 at 17:27
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5$\begingroup$ No, of course not. Just consider $X_n$ symmetric stable random variables with index $\alpha \in (1, 2)$: then $\tfrac1n \sum_{i=1}^nX_i$ has the same distribution as $n^{-1+1/\alpha}X_1$, and so you get $O_P(n^{-1+1/\alpha})$ (if I guess the meaning of $O_P$ correctly). $\endgroup$– Mateusz KwaśnickiCommented Jun 12, 2021 at 18:08
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1$\begingroup$ @MateuszKwaśnicki Yes, for stable distribution, the rate doesn't hold. thanks. $\endgroup$– JohnCommented Jun 28, 2021 at 23:24
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