📂Lemmas

Proof of Chebyshev's Inequality

Theorem ¹

If the variance $\sigma^2 < \infty$ of a random variable $X$ exists, then for $\mu := E(X)$ and any positive number $k>0$, $$ P(|X-\mu| \ge k\sigma) \le {1 \over k^2} $$

Explanation

Since its form is relatively simple, the expression is easy to manipulate, and the result is immediately clear, it is widely used as a lemma. However, compared to Markov’s inequality, it has one additional condition: the variance must exist.

Seeing that the condition requires the $2$nd moment to exist, one might regard it as an all-too-easy and obvious condition. Well, that is true to some extent, but if you are at least an undergraduate, you should at least know the fact that that existence is not entirely obvious.

Proof

Strategy: Start from Markov’s inequality and use the fact that an inequality involving a square remains the same when converted into an inequality about the absolute value. Note that since the variance exists by assumption, there is no need to prove the existence of the mean $\mu$.

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Let $u(X) : =(X-\mu)^2$.

Markov’s inequality $$ P(u(X) \ge c) \le {E(u(X)) \over c} $$

If we set $c:=k^2 \sigma^2$, then $$ P((X-\mu)^2 \ge k^2 \sigma ^2) \le {E((X-\mu)^2) \over {k^2 \sigma^2}} $$ Since $P((X-\mu)^2 \ge k^2 \sigma ^2) = P(|X-\mu| \ge k \sigma)$ and $E((X-\mu)^2)=\sigma^2$, $$ P(|X-\mu| \ge k \sigma) \le {1 \over k^2} $$

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