Use chebyshev inequality to find the probability $P[|X-E[X]| \ge k\sigma]$

Question



For an arbitrary random variable $X$, use the Chebyshev inequality to show that the probability that $X$ is more than $k$ standard deviations from its expected value $E[X]$ satisfies $$P[|X-E[X]| \ge k\sigma] \le \frac1{k^2}$$

Hi every one, I found this problem in the book any idea?

Answer 1

When you use Chebyshev's inequality, you get that $$ P\left( \lvert X - \mathbb{E}X \rvert \geq k \sigma \right) \leq \frac{\mathbb{E} \lvert X - \mathbb{E}X \rvert^2}{k^2 \sigma^2}.$$ What is the quantity in the numerator?