Show that $\mathbb{E}[X^2]< \infty \Longleftrightarrow \sum_{n=1}^{\infty}n\mathbb{P}[|X|>n]< \infty$

Question

Show that $\mathbb{E}[X^2]< \infty \Longleftrightarrow \sum_{n=1}^{\infty}n\mathbb{P}[|X|>n]< \infty$

I try the use the next inequality: if $Y$ is a non negative random variable then $$ \sum_{n=1}^{\infty} \mathbb{P}[Y>n] \leq \mathbb{E}[Y] \leq 1+\sum_{n=1}^{\infty} \mathbb{P}[Y>n]$$ Taking $Y:= X^2$ i get $$ \sum_{n=1}^{\infty} \mathbb{P}[X^2>n] \leq \mathbb{E}[X^2] \leq 1+\sum_{n=1}^{\infty} \mathbb{P}[X^2>n]$$ but I'm not sure how to manipulate the sum $\sum_{n=1}^{\infty} \mathbb{P}[X^2>n]$ to somehow make it appear $\sum_{n=1}^{\infty}n\mathbb{P}[|X|>n]$

Any hint or help i will be very grateful.

Answer 1

I suggest summation by parts. $$2\sum_{n=1}^N n\mathbb{P}[|X|>n]\le \sum_{n=1}^N [(n+1)^2-n^2]\mathbb{P}[|X|>n]\\ = \sum_{n=2}^{N+1}n^2\mathbb{P}[|X|>n-1]- \sum_{n=1}^{N}n^2\mathbb{P}[|X|>n]\\ = \sum_{n=2}^{N+1}n^2\mathbb{P}[n-1<|X|\le n]+(N+1)^2\mathbb{P}[|X|>N+1]\\ \le 4\sum_{n=2}^{N+1}(n-1)^2\mathbb{P}[n-1<|X|\le n]+\mathbb{E}[X^2]\\ \le 4\mathbb{E}[X^2]+\mathbb{E}[X^2]$$ Since $N$ is arbitrary we get $$\sum_{n=1}^\infty n\mathbb{P}[|X|>n]\le {5\over 2}\mathbb{E}[X^2]$$

Answer 2

Note that $$ \sum_{k=0}^\infty k P(X\gt k) = E\left(\sum_{k=0}^\infty k I(X\gt k)\right) = E\sum_{k=0}^{\lfloor X\rfloor-1} k = E\frac{\lfloor X\rfloor (\lfloor X\rfloor-1)}{2} $$ whence $$ EX^2\lt\infty \iff \sum_{k=0}^\infty k P(X\gt k)\lt\infty $$