How might I show that if a random variable (call it Z) is such that EZ (expectation of Z) is finite (i.e. it is Lebesgue integrable), then nP(|Z|>n) tends to 0?
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Note that $nP(|Z|\gt n)=E(Z_n)$ with $Z_n=n\mathbf 1_{|Z|\gt n}$ and that $Z_n\to0$ almost surely. Hence all that is needed is to ensure that the integral of the limit is the limit of the integrals. Lebesgue dominated theorem tells you that, if $|Z_n|\leqslant Y$ uniformly over $n$, for some integrable $Y$, everything works fine. Surely you have some idea about a candidate for $Y$... |
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As in the de la Vallée-Poussin theorem, let $G$ be an increasing function on $[0,\infty)$ so that $G(x)/x\to\infty$ but so slowly that $E(G(|Z|))<\infty$. Then by Markov's inequality, $$n\,P(|Z|>n)=n\,P(G(|Z|)>G(n))\leq {n\, E(G(|Z|))\over G(n)}\to0.$$ |
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