proof of Jensen’s inequality

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 $X$ :

$\operatorname{\mathbb{E}}[f(X)]\geq f(\operatorname{\mathbb{E}}[X]).$

Indeed, let $c=\operatorname{\mathbb{E}}[X]$ . 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 $\varphi(x)\leq f(x)$ . Then

$\operatorname{\mathbb{E}}[f(X)]\geq\operatorname{\mathbb{E}}[\varphi(X)]=% \operatorname{\mathbb{E}}[\alpha(X-c)+f(c)]=f(c)$

as claimed.

Title	proof of Jensen’s inequality
Canonical name	ProofOfJensensInequality
Date of creation	2013-03-22 12:45:15
Last modified on	2013-03-22 12:45:15
Owner	Andrea Ambrosio (7332)
Last modified by	Andrea Ambrosio (7332)
Numerical id	6
Author	Andrea Ambrosio (7332)
Entry type	Proof
Classification	msc 26D15
Classification	msc 39B62