# 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 ProofOfJensensInequality 2013-03-22 12:45:15 2013-03-22 12:45:15 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 6 Andrea Ambrosio (7332) Proof msc 26D15 msc 39B62