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$X$ 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

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

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    $\begingroup$ What have you tried? Also, do you mean $P(|X|>\epsilon)$? $\endgroup$
    – jakobdt
    Commented Sep 29, 2022 at 8:44
  • $\begingroup$ $\begin{array}{ll} \mathbb{E}(|X|^k) &\leq \mathbb{E}\left(|X|^k \mathbb{1}_{\{|X|^k\geq \epsilon^k\}}\right)\\ &\leq \mathbb{E}\left(\epsilon^k \mathbb{1}_{\{ |X|^k \geq \epsilon^k \}}\right)\\ &\leq \epsilon^k P\left(|X|^k\geq \epsilon^k\right) \end{array}$ @jakobdt $\endgroup$ Commented Sep 29, 2022 at 9:52
  • $\begingroup$ 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. $\endgroup$
    – jakobdt
    Commented Sep 29, 2022 at 10:21

1 Answer 1

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Hint:-

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 ?

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  • $\begingroup$ $\begin{array}{ll} \mathbb{E}(|X|^k) &\leq \mathbb{E}\left(|X|^k \mathbb{1}_{\{|X|^k\geq \epsilon^k\}}\right)\\ &\leq \mathbb{E}\left(\epsilon^k \mathbb{1}_{\{ |X|^k \geq \epsilon^k \}}\right)\\ &\leq \epsilon^k P\left(|X|^k\geq \epsilon^k\right) \end{array}$ @Mr.Gandalf Sauron $\endgroup$ Commented Sep 29, 2022 at 9:53
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    $\begingroup$ You are just reproving Markov's Inequality. What hint did I give? . $|X|>c\iff |X|^{k}>c^{k}$ . So $P(|X|>c)=P(|X|^{k}>c^{k})\leq \frac{E(|X|^{k})}{c^{k}}$ . $\endgroup$ Commented Sep 29, 2022 at 12:04
  • $\begingroup$ thank you for your help @Mr.Gandalf Sauron $\endgroup$ Commented Oct 5, 2022 at 15:39
  • $\begingroup$ @learnermathematics You can mark as answer then. It is important to accept answers as otherwise this question would remain in the unanswered tab. $\endgroup$ Commented Oct 5, 2022 at 16:19

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