Chebyshev Inequality

Says that the proability that X is far away from its mean, is bounded by quantity that increases as Var(X) increases.

Suppose, we have sequence of random variables with Var(X), then:

$$P(|X - E(X)| \ge t) \le {Var(X) \over t^2}\text{, where t > 0}$$

Proof: Suppose, \(Y = (X - E(X))^2\), \(P(Y \ge 0) = 1\) and E(Y) = Var(X); from Markov Inequality we have:

$$P(|X - E(X)| \ge t) = P(Y \ge t^2) \le {Var(X) \over t^2}$$

It can be seen from this proof that the Chebyshev inequality is simply a special case of the Markov inequality, therefore, the comments that were given following the proof of the Markov inequality can be applied as well to the Chebyshev inequality.