# Prove $P(X > C) = 0$ if $E[|X|^n] \leq aC^n$

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.

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

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