Chebyshev Inequality

The probability of being far from the mean is very low.

Inequality

For any RV $X$, probability of both tails is bounded by Variance. $P(|X-\mu |\ge a)\le \frac{V(X)}{a^2}$

Proof

1. Apply Markov Inequality to $g(x)=(x-\mu {)}^2>0$
2. Then, $P(g(X)\ge a^2)\le \frac{E(g(x))}{a^2}$
3. $⟺P((x-\mu {)}^2\ge a^2)\le \frac{E((X-\mu {)}^2}{a}$
4. $⟺P(|X-\mu )|\ge a)\le \frac{V(X)}{a^2}$