If $E(|X|)$ is finite, is $\lim_{n\to\infty} nP(|X|>n)=0$? For a continuous random variable $X$, if $E(|X|)$ is finite, is $\lim_{n\to\infty}n P(|X|>n)=0$?
This is a problem I found on the internet, but I'm not sure whether it holds or not.
I know that $n P(|X|>n)<E(|X|)$ holds by Markov inequality, but I can't show that it goes to 0 as $n$ goes to infinity.
 A: I can provide an answer for a continuous random variable (there is surely a more general answer). Let $Y=|X|$:
$$\mathbb{E}[Y]=\int_0^\infty yf_Y(y)\text{d}y=\int_0^n yf_Y(y)\text{d}y+\int_n^\infty yf_Y(y)\text{d}y\ge\int_0^n yf_Y(y)\text{d}y+n\int_n^\infty f_Y(y)\text{d}y=\dots+n\left(F_Y(\infty)-F_Y(n)\right)=\dots+n(1-F_Y(n))=\int_0^n yf_Y(y)\text{d}y+nP(Y\gt n)$$
Thus
$$0\leq nP(Y\gt n)\le\left(\mathbb{E}[Y]-\int_0^n yf_Y(y)\text{d}y\right)$$
Now,since by hypothesis $\mathbb{E}[Y]$ is finite, we have that
$$\lim_{n\to \infty}\left(\mathbb{E}[Y]-\int_0^n yf_Y(y)\text{d}y\right)=\mathbb{E}[Y]-\lim_{n\to \infty}\int_0^n yf_Y(y)\text{d}y=\mathbb{E}[Y]-\mathbb{E}[Y]=0$$
Then
$$\lim_{n\to \infty}nP(Y\gt n)=0$$
by the sandwich theorem.
A: Look at the sequence of random variables $\{Y_n\}$ defined by retaining only large values of $|X|$: $$Y_n:=|X|I(|X|>n).$$ It's clear that $Y_n\ge nI(|X|>n)$, so $$E(Y_n)\ge nP(|X|>n).\tag1$$ Note that $Y_n\to0$ and $|Y_n|\le |X|$ for each $n$. So the LHS of (1) tends to zero  by dominated convergence.
A: $E\left | X \right |< \infty \Leftrightarrow  E\left | X \right |\mathbb{I}_{\left | X \right |>n}\rightarrow 0$ (uniformly integrable)
$E\left | X \right |=E\left | X \right |\mathbb{I}_{\left | X \right |>n}+E\left | X \right |\mathbb{I}_{\left | X \right |\leq n}$
$E\left | X \right |\mathbb{I}_{\left | X \right |>n}\leq E\left | X \right |< \infty $
$E\left | X \right |\mathbb{I}_{\left | X \right |>n}\geq  nE\mathbb{I}_{\left | X \right |>n}=nP\left (  \left | X \right |>n\right )$
$E\left | X \right |\mathbb{I}_{\left | X \right |>n} \rightarrow 0 \Rightarrow nP\left (  \left | X \right |>n\right )\rightarrow 0 \Rightarrow  P\left (  \left | X \right |>n\right )\rightarrow 0$
i.e. $ \underset{n\rightarrow \infty}{\lim} P\left (  \left | X \right |>n\right )=0$
