Let $(E,\mathcal{A},P)$ be a probability space, and $X$ a random variable on $E$. Markov's inequality states that for any $\epsilon>0$,
$$P(\mid X \mid \geq \epsilon)\leq \frac{1}{\epsilon} \int \mid X \mid dP.$$
Since $w\in\{\mid X \mid\geq \epsilon\} \iff w\in\{\mid X \mid^r\geq \epsilon^r\}, r\in \mathbb{N} $, it implies that
$$P(\mid X \mid \geq \epsilon)\leq \frac{1}{\epsilon^r} \int \mid X \mid^r dP,$$ for any natural $r$.
In terms of moments, it would be a strong result.
My question: Does it really holds?
Best Answer
Yes, this fact is true and well-known. It's often called "Chebyshev's inequality". See https://en.wikipedia.org/wiki/Chebyshev%27s_inequality#Measure-theoretic_statement