# Proof Chebyshev's Inequality

I want to proof Chebyshev's Inequality using Markov's inequality.

Cheb.Ineq:

$$P(|X-\mu| \ge a) \le \frac{Var(X)}{a}$$

So I'm starting with Markov's Inequality:

$$P(|X-\mu| \ge a) \le \frac{E(|X-\mu|)}{a}$$

I replace $|X-\mu|$ by $(X-\mu)$, and square both sides, which leads to:

$$P((X-\mu)^2 \ge a^2) \le \frac{E((X-\mu)^2)}{a^2}$$

The numerator of the right hand side is the variance of $X$. So we get:

$$P((X-\mu)^2 \ge a^2) \le \frac{Var(X)}{a^2}$$

So far so good. The proof that I read now just simply states that $P((X-\mu)^2 \ge a^2)$ is equal to $P(|X-\mu| \ge a)$ and this leads to

$$P(|X-\mu| \ge a) \le \frac{Var(X)}{a^2}=\frac {\sigma^2} {a^2}$$

I really don't get the last step. Why are $P((X-\mu)^2 \ge a^2)$ and $P(|X-\mu| \ge a)$ are equal?

• I corrected some mistakes in your inequalities, please make sure that those really just were typos...you can see what has been changed in the edit history of your question Sep 12 '15 at 19:18

think of what $P((X-\mu)^2 \ge a^2)$ represents? It represents (the measure) of all those values $X$ for which $(X-\mu)^2 \ge a^2$ which are the exact same values for $|X-\mu| \ge |a|$ (just take square roots), now since $a \ge 0$, $|a|=a$. And the measures $P(\cdot)$ therefore are equal
You just need to show that the sets are equal, so that for $a\geq0$ $$\left\{\omega:(X(\omega)-\mu)^2\geq a^2\right \} =\left\{\omega:|X(\omega)-\mu|\geq a\right \}$$ holds, for this we show $$\left\{\omega:(X(\omega)-\mu)^2\geq a^2\right \} \subset\left\{\omega:|X(\omega)-\mu|\geq a\right \}$$ which is clearly the case, just use the square root, so we get $$(X(\omega)-\mu)^2\geq a^2\Rightarrow |X(\omega)-\mu|\geq a$$ and we show $$\left\{\omega:|X(\omega)-\mu|\geq a\right \} \subset\left\{\omega:(X(\omega)-\mu)^2\geq a^2\right \}$$ which is again clear by squaring, so $$|X(\omega)-\mu|\geq a\Rightarrow (X(\omega)-\mu)^2\geq a^2$$ so since the sets are equal and so is the measure.