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I'm was asking myself the following question. Consider a real-valued random variable $X \in L^2$, i.e. $\mathbb{E}[X^2] < \infty$.

Clearly, $\mathbb{E}[X^2 \mathbb{1}_{\vert X \vert > n}] \to 0$ for $n \to \infty$ or in other words $\mathbb{E}[X^2 \mathbb{1}_{\vert X \vert > n}] = o(1)$.

I am wondering if there is any chance, to find a stronger result, something like $\mathbb{E}[X^2 \mathbb{1}_{\vert X \vert > n}] = o(n^{-1})$.

Any thoughts or recommendations to related problems would be great. Thanks in advance.

student7481
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1 Answers1

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Let $\left(\delta_k\right)_{k\geqslant 1}$ be a decreasing sequence of positive numbers such that $\delta_k\to 0$ and let $X$ be a random variable taking the value $k$ with probability $\left(\delta_k-\delta_{k+1}\right)/(Ck^2)$, where $C=\sum_{k=1}^\infty\left(\delta_k-\delta_{k+1}\right)/k^2$. Then $$ \mathbb E\left[X^2\mathbf{1}\left\{X>n\right\}\right] =\sum_{k=n+1}^\infty k^2\mathbb P\left(X=k\right)=\sum_{k=n+1}^\infty k^2\left(\delta_k-\delta_{k+1}\right)/(Ck^2)=\delta_n/C $$ hence the decay can be as slow as you wish.

Davide Giraudo
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