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Probability Proof with E(x)

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me2please

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It's just Markov's ineq

\[X \ge 0, \epsilon > 0, P(X \ge \epsilon) \le \frac{E(X)}{\epsilon}\]
\[P(X \ge \sqrt \eta) \le \frac{\eta}{\sqrt \eta}\]

proof of Markov's ineq can be found in almost all probability books.
 
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