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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0Hi, your notation was quite ambiguous. I have tried to put the text into LaTeX for you but if anything look wrong just shout and I'll edit it again (or you could take the opportunity to learn some LaTeX yourself :) ) – 2012-11-23
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0The LaTeX text is fine! Maybe the symbol "{" and "}" is missing in the set of the indicator function. Some hint for this? Jimmy – 2012-11-24
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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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3Why the downvote? – 2012-12-02
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0Did E[|X_1|^r . I_{|X_1| > b_n}] have any asymptotic behaviour (with convergence and limit zero) too? – 2012-12-10
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0??? Since the random variables (X_i) are i.i.d., this is the same as the last expectation written in my post. – 2012-12-10
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0Yes it is, but is something hard to believe since X_1 does not depend on n. An example or reference? – 2012-12-15
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0Reference of what? The random variables $X_1$ and $X_n$ are different but their distributions are the same hence $E(u(X_n))=E(u(X_1))$ for every (nonnegative) function $u$. Considering $u:x\mapsto|x|^r\cdot\mathbf 1_{|x|\gt b_n}$ solves your puzzlement. – 2012-12-15
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0These discussions must be constructive. Your function u(x) depends on n. Let's look only on E[1_{|X1|>b_n}] (put r = 0). This is 1 - F_|X1|(b_n). So, it would be to expect that if b_n change, then the asymptotic behavior of E[1_{|X1|>b_n}] change too, or not?! That is, 1 - F_|X1|(n^n^n^n) has not the same asymptotic behavior of 1 - F_|X1|(log(log(log(log(n))))). – 2012-12-16
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0Everything you write in your last comment is true and I completely fail to see your point... Once again, what part of your puzzlement is not solved by the last sentence of my post? (By the way, for every `X` there exists some `Y` such that `P(|X|>n^n^n^n) = P(|Y|>log(log(log(log(n)))))` for every `n`.) – 2012-12-16
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0Under the assumptions of my question, I presume that E[|X_1|^r . 1_{|X_1| > b_n}] has an asymptotic behaviour which depends on b_n. As you well observe in your first post, if we consider E[|X_1|^r . 1_{|X_2| > b_n}] then independence do wonderful things and such asymptotic behavior comes up. In the case E[|X_1|^r . 1_{|X_1| > b_n}], I presume that some asymptotic behavior exists, but maybe I'm wrong... – 2012-12-16
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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