1 Jensen Inequality
Jensen’s inequality gives a lower bound on expectations of convex functions (such as linear functions, quadratic functions, exponential functions, modulus etc.).
For convex function \(\varphi\) and random variable \(X\) on a probability space \((\Omega, \mathcal{F}, \mathbb{P})\) we have \[ \varphi(\mathbb{E} X)\leq\mathbb{E}\varphi(X)~. \] Rewriting the expectation in its integral form we get the probability measure theoretic form of the result \[ \varphi\left(\int_{\Omega}Xd\mathbb{P}\right)\leq \int_{\Omega}\varphi(X)d\mathbb{P}. \]
PROOF: Since \(\varphi\) is convex, at each real number \(x\) we have a nonempty set of sub-derivatives (which may be thought of as lines touching the graph of \(\varphi\) at \(x\), but which are below the graph of \(\varphi\) at all points (support lines of the graph). If we define \[ x_0:=\int_{\Omega}X~d\mathbb{P} \] then from the existence of sub-derivatives for convex functions we may choose \(a\) and \(b\) such that \[ ax+b\leq \varphi(x) \] for all \(x\in\mathbb{R}\) and \(ax_0+b=\varphi(x_0)\). Then we have that \[ \varphi(X)\geq aX+b \] for almost all \(\omega\in\Omega\). From the monotonicity of the integral with respect to the probability measure we obtain the result since \[ \mathbb{E}\varphi(X)\geq a\mathbb{E}X+b=\varphi(\mathbb{E}X). \] \(\square\)