ProofOfChebyshevsInequality 2002-06-17 04:24:21 2008-02-15 18:37:25 Proof Chebyshev's inequality 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newcommand{\Prob}[2]{\mathbb{P}_{#1}\left\{#2\right\}} \newcommand{\Expect}{\mathbb{E}} \newcommand{\norm}[1]{\left\|#1\right\|} % Some sets \newcommand{\Nats}{\mathbb{N}} \newcommand{\Ints}{\mathbb{Z}} \newcommand{\Reals}{\mathbb{R}} \newcommand{\Complex}{\mathbb{C}} %%%%%% END OF SAVED PREAMBLE %%%%%% The proof of Chebyshev's inequality follows from the application of Markov's inequality. Define $Y = (X - \mu)^2$. Then $Y \ge 0$ is a random variable, and $$\Expect[Y] = \operatorname{Var}[X] = \sigma^2.$$ Applying Markov's inequality to $Y$, we see that $$\Prob{}{\left|X - \mu \right| \ge t} = \Prob{}{Y \ge t^2} \le \frac{1}{t^2}\Expect[Y] = \frac{\sigma^2}{t^2}.$$