Probability theory MOC

Markov's inequality

For any real random variable 𝑋 :𝜉 and 𝑎 0 we have #m/thm/prob

(|𝑋|𝑎)𝔼[|𝑋|]𝑎
Proof

Since |𝑋| 𝑎1{|𝑋|𝑎} and therefore

𝔼[|𝑋|]𝑎𝔼[1{|𝑋|𝑎}]=𝑎(|𝑋|𝑎)

as required.


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