Ketaksamaan Chebysev

Teorema. Diberikan variabel acak X dengan E[X]=\mu dan Var(X)=\sigma^2, maka untuk setiap \lambda > 0 berlaku

P\left(|X-\mu| \ge \lambda \sigma \right) \le \dfrac{1}{\lambda^2}.

Bukti. Kita punyai

\sigma^2 =Var(X)=E[(X-\mu)^2]

=\underbrace{(X-\mu)^2P(|X-\mu| < \lambda \sigma)}_{\ge 0}+\underbrace{(X-\mu)^2}_{\ge \lambda^2 \sigma^2}P(|X-\mu| \ge \lambda \sigma)

\ge \lambda^2 \sigma^2 P(|X-\mu| \ge \lambda \sigma)

sehingga benar bahwa P(|X-\mu| \ge \lambda \sigma) \le \dfrac{1}{\lambda^2}. \blacksquare

 

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