Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 2 / 4F (Probability)

The Cauchy-Schwarz inequality is $|\mathbb{E}(XY){|}^2\le \mathbb{E}(X^2)\mathbb{E}(Y^2)$. Markov inequality says that for $X\ge 0$ and $a>0$, $\mathbb{P}(X\ge a)\le \mathbb{E}X/a$.

Jensen inequality states that for convex $\varphi$,Let $\mu =\mathbb{E}X$. A supporting line at $\mu$ gives $\varphi (x)\ge \varphi (\mu )+c(x-\mu )$. Taking expectations makes the linear term vanish.

With $\mu =\mathbb{E}X$,Every summand is nonnegative. If the sum is zero, every value of positive probability equals $\mu$, so $\mathbb{P}(X=\mu )=1$.