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Suppose $X$ is a random variable where for all integers $n \geq 0$, $E[|X|^n] \leq aC^n$ for some positive number $a$ and $C$. Prove that $P(X > C) = 0$.

My thought is to use Markov's inequality and notice that $a$ is fixed for all $n$, so if we take a $n$-th root of $aC^n$, we can bound it by some value. But I don't know what does the $n$-th root of $E[|X|^n]$ and that is where I got stuck.

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  • $\begingroup$ What are your thoughts on this, and what have you tried? $\endgroup$ – Clement C. Oct 16 '18 at 19:10
  • $\begingroup$ I updated the question $\endgroup$ – Cassie Liu Oct 16 '18 at 19:20
  • $\begingroup$ Note that you just need a bound, not the exact value. $\endgroup$ – user10354138 Oct 16 '18 at 19:36
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For a hint, consider the event:

$E_k:=\Big\{ \vert X\vert \geq C+\frac{1}{k} \Big\} $

Then:

$\mathbb{E}\Big[ \frac{\vert X\vert^n }{C^n} \Big]\geq \mathbb{E}\Big[ \frac{\vert X\vert^n }{C^n} \cdot 1_{E_k} \Big]\geq \frac{\Big(C+\frac{1}{k} \Big)^n}{C^n}\cdot P(E_k)$

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    $\begingroup$ You can rewrite this directly using Markov's inequality, which may be more intuitive to the OP: for all $k\geq 1$, for all $n$, $$\mathbb{P}\{X > C+\frac{1}{k}\} = \mathbb{P}\!\left\{X^n > \left(C+\frac{1}{k}\right)^n\right\} \leq \frac{\mathbb{E}[|X|^n]}{\left(C+\frac{1}{k}\right)^n} \leq \frac{a}{\left(1+\frac{1}{Ck}\right)^n} \xrightarrow[n\to\infty]{} 0$$ $\endgroup$ – Clement C. Oct 16 '18 at 20:17

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