| Line 30: | Line 30: | ||
===== <math>\color{blue}\text{Solution 2:}</math> ===== | ===== <math>\color{blue}\text{Solution 2:}</math> ===== | ||
Discrete Case:<br> | Discrete Case:<br> | ||
| + | Let <math class="inline">p_{X}(x)</math> be the pmf of X. The probability that <math class="inline">X</math> differs from <math class="inline">\mu</math> by at least <math class="inline">\varepsilon </math> is <br> | ||
| + | <math> P(|X-\mu| \geq \varepsilon)= \sum_{|X-\mu| \geq \varepsilon}p_{X}(x)</math><br> | ||
| + | Based on the definition of the variance, we have<br> | ||
| + | <math> \sigma^2 = \sum_{x}(x-\mu)^2 p_{X}(x)</math><br> | ||
| + | Let a set <math class="inline">A= \{ x|\,|x-\mu| \geq \varepsilon \}</math>. We have<br> | ||
| + | <math> \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)</math><br> | ||
| + | <math> \Rightarrow\sigma^2 \geq \sum_{x \in A}(x-\mu)^2 p_{X}(x)</math><br> | ||
| + | |||
Continuous Case:<br> | Continuous Case:<br> | ||
Revision as of 19:53, 25 January 2014
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2012
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:
