Probability Proof with E(x)

Status
Not open for further replies.
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.
 
Last edited by a moderator:
Status
Not open for further replies.
Cookies are required to use this site. You must accept them to continue using the site. Learn more…