Jump to Problem 2,3

Problem 2

Problem statement: Let $ X $ be a continuous or discrete random variable with mean $ \mu $ and variance $ \sigma^2 $. Then, $ \forall \varepsilon >0 $, we have
$ P(|X-\mu| \geq \varepsilon) \leq \frac{\sigma^2}{\varepsilon^2} $

$ \color{blue}\text{Solution 1:} $

$ \color{blue}\text{Solution 2:} $

Discrete Case:
Let $ p_{X}(x) $ be the pmf of X. The probability that $ X $ differs from $ \mu $ by at least $ \varepsilon $ is
$ P(|X-\mu| \geq \varepsilon)= \sum_{|X-\mu| \geq \varepsilon}p_{X}(x) $
Based on the definition of the variance, we have
$ \sigma^2 = \sum_{x}(x-\mu)^2 p_{X}(x) $
Let a set $ A= \{ x|\,|x-\mu| \geq \varepsilon \} $. We have
$ \sigma^2 = \sum_{x}(x-\mu)^2 p_{X}(x)= \sum_{x \in A}(x-\mu)^2 p_{X}(x)+\sum_{x \notin A}(x-\mu)^2 p_{X}(x) $
$ \Rightarrow\sigma^2 \geq \sum_{x \in A}(x-\mu)^2 p_{X}(x) $

Continuous Case: