# arrow_back Prove Chebyschev’s inequality

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Prove Chebyschev&rsquo;s inequality

(Chebyschev's inequality). Let $X$ be a random variable. Then
$P(|X-E X|>\epsilon) \leq \frac{\operatorname{Var}(X)}{\epsilon^{2}}$

## 1 Answer

Best answer

Best answer
Proof.

This is marked as a corollary because we simply apply Markov's inequality to the nonnegative random variable $(X-E X)^{2}$. We then have
$\begin{array}{rlr}P(|X-E X|>\epsilon) & =P\left((X-E X)^{2}>\epsilon^{2}\right) & (\text { statements are equivalent }) \\ & \leq \frac{E\left[(X-E X)^{2}\right]}{e^{2}} & \text { (Markov's inequality) } \\ & =\frac{\operatorname{Var}(X)}{\epsilon^{2}} & \text { (definition of variance) }\end{array}$
(definition of variance)
by Platinum
(106,844 points)

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