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?
Best Answer
Hint:
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