Probability theory MOC Markov's inequality For any real random variable 𝑋 :𝜉 →ℝ and 𝑎 ≥0 we have #m/thm/prob ℙ(|𝑋|≥𝑎)≤𝔼[|𝑋|]𝑎 ProofSince |𝑋| ≥𝑎1{|𝑋|≥𝑎} and therefore𝔼[|𝑋|]≥𝑎𝔼[1{|𝑋|≥𝑎}]=𝑎ℙ(|𝑋|≥𝑎)as required. #state/tidy | #lang/en | #SemBr