Chebyshev’s inequality

Let $𝑋 : 𝜉 \to ℝ$ be a real random variable with mean $𝜇$ and variance $𝜎 2$ . Then for any $𝑎 > 0$ prob

ℙ (| 𝑋 - 𝜇 | \geq 𝑎) \leq 𝜎 2 𝑎 2

Proof

By Markov’s inequality
$ℙ ((𝑋 - 𝜇) 2 \geq 𝑎 2) = 𝔼 [(𝑋 - 𝜇) 2] 𝑎 2$
as required.