Proof that $E(X)<\infty$ entails $\lim_{n\to\infty}n\Pr(X\ge n) = 0$?

Question

As the title says. I think this should follow straightforwardly but I can't find a proof.

My random variable of interest $X$ takes values in the non-negative integers. The only other assumption on its distribution is that $E(X)<\infty$. I want to prove: $$\lim_{n\to\infty}n\Pr(X\ge n) = 0.$$ The fact that this should follow is referenced e.g. by DeGroot (2004) "Optimal Statistical Decisions" p. 295, but no proof is given.

All I have right now is that without the constant $n$ it is easy to prove using Markov's inequality: $$\Pr(X\ge n) \le \frac{1}{n}E(X) \to 0.$$ I appreciate any help in figuring this out.

Answer 1

This fact should be true for any monotonically decreasing sequence $a_n$ with $\sum_{i=1}^\infty a_i<\infty$. Recall the Cauchy Condensation test which says that $\sum_{i=1}^\infty a_i<\infty$ converges iff $\sum_{i=1}^\infty 2^i a_{2^i}<\infty$ converges, so that we have $2^na_{2^n}\rightarrow 0$ and by monotonicity if we let $k(n):=\log_2(n)$ then $0\leq na_n \leq 2^{k(n)}a_{2^{k(n)}}$ which implies $na_n\rightarrow 0$.

Now use the fact that $\sum_{i=1}^\infty P(X\geq i)=E(X)$ for nonnegative integer random variables.

Answer 2

Note that for each $N$, $$\sum_{j=1}^Nj\mu(j\leqslant X\leqslant j+1)\leqslant EX,$$ hence $$\sum_{1\leqslant i\leqslant j\leqslant N}\mu(j\leqslant X\leqslant j+1)\leqslant EX,$$ that is, $$\sum_{i=1}^N\mu(i\leqslant X\leqslant N+1)\leqslant EX.$$ This gives that the series $\sum_{j\geqslant 0}\mu(j\leqslant X)$ is convergent. As the sequence $\left(\mu(j\leqslant X),j\in\Bbb N\right)$ is non-negative and non-increasing, this gives the result.

Answer 3

Notice that $|X| \geq n 1_{|X|\geq n}$ for every $n$ and that as $n \to \infty$ the right hand side tends to zero pointwise almost surely since $E[X] < \infty$ implies that the measure of the set where $X = \infty$ is zero. The result follows from the dominated convergence theorem.