Let $X$ be a random variable such that $E|X|^m \leq A \cdot B^{m}$ for some positive number $A$ and $B$ and for all $m \geq 0$. Can we obtain the conclusion that $P(|X| > B) = 0$?
Doest polynomial growth of moments imply boundedness of random variables?
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0Do you mean $E|X|^m\le A\cdot B^m$? – 2017-02-21
1 Answers
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Assuming you mean $\mathbb{E}[|X|^m]\leq AB^m$ for all $m$, then yes. Proceed by contrapositive argument. Say $\mathbb{P}[|X|>B]>0$. Then $\mathbb{P}[|X|>B+\epsilon]>0$ for sufficiently small $\epsilon>0$. Then $\mathbb{E}[|X|^m]\geq (B+\epsilon)^m\mathbb{P}[|X|>B+\epsilon]$. No matter what $A$ you choose, this will be larger than $AB^m$ for sufficiently large $m$.