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


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

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?

  • $\begingroup$ 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 $\endgroup$
    – user190080
    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.


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