Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability / 9 / a / i

Markov's inequality: if $X\ge 0$ and $a>0$, then $\mathbb{P}(X\ge a)\le \mathbb{E}[X]/a$. Pointwise,Taking expected values, using their monotonicity and the identity $\mathbb{E}[1_E]=\mathbb{P}(E)$, gives $\mathbb{E}[X]\ge a\mathbb{P}(X\ge a)$. Division by the positive threshold proves the inequality. If the expectation is infinite, the extended-valued bound remains true but uninformative.