Let $X_1, X_2, ...$ be a sequence of i.i.d. random variables such that $\mathbb{E}|X_1|^r < \infty$ for some $r > 1$. If b_n is a sequence such that $b_n \to \infty$ as $n \to \infty$ then $\mathbb{E} [|X_1|^r . I_{|X_n| > b_n}] \to 0$ as $n \to \infty$. Does there exist any result which gives the asymptotics for $\mathbb{E} [|X_1|^r . I_{|X_n| > b_n}]$, that is, what is the speed of convergence of $\mathbb{E} [|X_1|^r . I_{|X_n| > b_n}]$?
Asymptotics for \mathbb{E} [|X_1|^r . I_{|X_n| > b_n}]
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probability-theory
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0If you go to edit the post you should just put \{ and \} around the set. The backslash tells the Latex interpreter to treat the brace as a brace and not as a special character. – 2012-11-25
1 Answers
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For every $n\ne1$, $\mathbb{E} [|X_1|^r \cdot\mathbf 1_{|X_n| > b_n}]=\mathbb E[|X_1|^r]\cdot\mathbb P[|X_n| > b_n]\leqslant\mathbb E[|X_1|^r]\cdot\mathbb E[|X_n|^r]\cdot b_n^{-r}$, thus $\mathbb{E} [|X_1|^r \cdot\mathbf 1_{|X_n| > b_n}]\leqslant C\cdot b_n^{-r}$ with $C=\mathbb E[|X_1|^r]^2$.
If, as is most probable, the question is in fact to bound $\mathbb{E} [|X_n|^r \cdot\mathbf 1_{|X_n| > b_n}]$, then any behaviour (with convergence and limit zero) is possible.
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0Yes, convergence to zero, nothing more. Give me some $(b_n)$ and $(p_n)$ such that $b_n\to\infty$ and $p_n\to0$--and there shall exist an i.i.d. sequence $(X_n)$ in $L^r$ such that $\mathbb E[|X_n|^r\,;\,|X_n|\gt b_n]\gg p_n$. – 2012-12-16