# Generalization of Markov's inequality

$$X$$ is a random variable of any sign, such that $$\mathbb{E}(|X|^k)$$ exists for $$k$$ positive integer.

My question is: how to show that Markov's Markov inequality can be generalized to the following form

$$P(|X>\epsilon |) \leq \frac{\mathbb{E}(|X|^k)}{\epsilon^k}$$

• What have you tried? Also, do you mean $P(|X|>\epsilon)$? Commented Sep 29, 2022 at 8:44
• $\begin{array}{ll} \mathbb{E}(|X|^k) &\leq \mathbb{E}\left(|X|^k \mathbb{1}_{\{|X|^k\geq \epsilon^k\}}\right)\\ &\leq \mathbb{E}\left(\epsilon^k \mathbb{1}_{\{ |X|^k \geq \epsilon^k \}}\right)\\ &\leq \epsilon^k P\left(|X|^k\geq \epsilon^k\right) \end{array}$ @jakobdt Commented Sep 29, 2022 at 9:52
• The first two inequalities should be $\geq$ rather than $\leq$. The last one is simply an equality due to linearity of the expectation. Then you are done. Commented Sep 29, 2022 at 10:21

For positive numbers, is $$a>b$$ equivalent to $$a^{p}>b^{p}$$ ? If yes then what can you say about the event $$\{|X|>c\}$$ and $$\{|X|^{p}>c^{p}\}$$ ? . What can you then say about the probabilities of the above events? What can you then do with Markov's inequality ?
• $\begin{array}{ll} \mathbb{E}(|X|^k) &\leq \mathbb{E}\left(|X|^k \mathbb{1}_{\{|X|^k\geq \epsilon^k\}}\right)\\ &\leq \mathbb{E}\left(\epsilon^k \mathbb{1}_{\{ |X|^k \geq \epsilon^k \}}\right)\\ &\leq \epsilon^k P\left(|X|^k\geq \epsilon^k\right) \end{array}$ @Mr.Gandalf Sauron Commented Sep 29, 2022 at 9:53
• You are just reproving Markov's Inequality. What hint did I give? . $|X|>c\iff |X|^{k}>c^{k}$ . So $P(|X|>c)=P(|X|^{k}>c^{k})\leq \frac{E(|X|^{k})}{c^{k}}$ . Commented Sep 29, 2022 at 12:04