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For example, given a positive r.v. $X$, if we know $\mathbb{E}(X)^{n} \leq f(n)$, where $f$ is a function of $n$, and we know the fact $\mathbb{E}(X^{n}) = \int_{0}^{\infty}n x^{n-1} \mathbb{P}(X > x)dx$. Then can we show $\mathbb{P}(X > x) = O(\frac{f(n)}{nx^{n-1}})$? It is quite intuitive but I don't know if it is true.

Thanks for reading!

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  • $\begingroup$ I suspect one might be able to make a counterexample by choosing an $X$ that's supported on an incredibly sparse sequence of positive integers: then $P(X>x)$ is constant for very long stretches, which tends to mess up decay statements. $\endgroup$
    – Greg Martin
    Nov 4, 2022 at 15:50
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    $\begingroup$ Assuming that $\mathbb{E}[X^n] \leq f(n)$ (it is not clear where the exponent is atm), then Markov inequality tells us $\Pr(X > x) = \Pr(X^n > x^n ) \leq x^{-n} \mathbb{E}[X^n] \leq x^{-n}f(n)$. Does that help? $\endgroup$
    – sudeep5221
    Nov 4, 2022 at 20:04
  • $\begingroup$ @sudeep5221 Thank you! This really helps $\endgroup$
    – Tom
    Nov 4, 2022 at 20:47

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