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

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.

**Chebyshev's inequality: if $X$ has finite mean $\mu$ and variance ${\sigma }^2$, then for $a>0$,Apply Markov's inequality to the nonnegative random variable $Y=(X-\mu {)}^2$ with threshold $a^2$. Since $\mathbb{E}[Y]={\sigma }^2$ and $\{|X-\mu |\ge a\}=\{Y\ge a^2\}$, the desired statement follows immediately.