The <Cauchy-Schwarz inequality> is $|\mathbb E(XY)|^2\leq\mathbb E(X^2)\mathbb E(Y^2)$. <Markov inequality> says that for $X\geq0$ and $a>0$, $\mathbb P(X\geq a)\leq\mathbb EX/a$.

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

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 2 / 4F / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ia/paper-2/4f/a/solution