Jensen's Inequality — Jemoka Knowledge Base

linear edition if f is convex, then for x,y \in \text{dom }f, 0 \leq \theta \leq 1, then:

\begin{equation} f\left(\theta x + \left(1-\theta\right) y\right) \leq \theta f\left(x\right) + \left(1-\theta\right) f\left(y\right) \end{equation}

probabilistic extension Let f be a convex function; that is, f’’\left(x\right) \geq 0; let x be a random variable. Then, f\left(\mathbb{E}[x]\right) \leq \mathbb{E}\left[f\left(x\right)\right]. Further, if f is strictly convex, that is f’’\left(x\right) > 0, then \mathbb{E}\left[f\left(x\right)\right] = f\left(\mathbb{E}[x]\right), that is, x is constant. the basic case is thus \begin{equation} P\left(z= x\right) = \theta , P\left(z=y\right) = 1-\theta \end{equation} concave edition Let f be a concave function; that is, f’’\left(x\right) \leq 0; let x be a random variable. Then, f\left(\mathbb{E}[x]\right) \geq \mathbb{E}\left[f\left(x\right)\right]. Further, if f is strictly concave, that is f’’\left(x\right) < 0, then \mathbb{E}\left[f\left(x\right)\right] = f\left(\mathbb{E}[x]\right), that is, x is constant.