Chebyshev's inequality

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

ℙ (| 𝑋 - 𝜇 | \geq 𝑎) \leq 𝜎 2 𝑎 2

Proof

ℙ ((𝑋 - 𝜇) 2 \geq 𝑎 2) = 𝔼 [(𝑋 - 𝜇) 2] 𝑎 2

as required.

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