Past exam of the graph Theory and Probabilistic Combinatorics course of the University of Oxford / 2009 / B / 4 / a

Chebyshev's inequality states that if $X$ has finite mean $\mu$ and variance ${\sigma }^2$, then for $t>0$,Indeed $(X-\mu {)}^2\ge t^2{1}_{\{|X-\mu |\ge t\}}$ pointwise. Taking expectations proves the result, including the zero-variance case.

The assumptions implyEventually ${\mu }_n>0$, and $X_n=0$ implies $|X_n-{\mu }_n|\ge {\mu }_n$. Thus Chebyshev's inequality givesNo independence between the different $X_n$ is needed; this is a second moment method.