ProofOfJensensInequality 2002-06-06 10:18:48 2006-11-05 02:58:13 Proof Jensen'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}{\operatorname{\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 %%%%%% We prove an equivalent, more convenient formulation: Let $X$ be some random variable, and let $f(x)$ be a convex function (defined at least on a segment containing the range of $X$). Then the expected value of $f(X)$ is at least the value of $f$ at the mean of $X$: $$ \Expect[f(X)] \ge f(\Expect [X]). $$ Indeed, let $c=\Expect [X]$. Since $f(x)$ is convex, there exists a supporting line for $f(x)$ at $c$: $$ \varphi(x)=\alpha (x-c) + f(c) $$ for some $\alpha$, and $\varphi(x)\le f(x)$. Then $$ \Expect[f(X)] \ge \Expect[\varphi(X)] = \Expect[\alpha (X-c) + f(c)] = f(c) $$ as claimed.