Is $\sum_{n=0}^\infty (n+1)P(|X|\ge n) \ge E(|X|^2)$?

Question

The question is in the title. I am doing a problem and wanted to know if this inequality holds: $\sum_{n=0}^\infty (n+1)P(|X|\ge n) \ge E(|X|^2)$.

Answer 1

By the fundamental theorem of calculus,

$$|X|^2 = 2 \int_0^{|X|} y \, dy, \tag{1}$$

which implies, by Fubini's theorem,

$$\begin{align*} \mathbb{E}(|X|^2) &\stackrel{(1)}{=} 2\int_{\Omega} \int_0^{|X|} y \, dy \, d\mathbb{P} \\ &= 2 \int_{\Omega} \int_0^{\infty} 1_{[y,\infty)}(|X|) \, dy \, d\mathbb{P} \\ &\stackrel{\text{Fub}}{=} 2 \int_0^{\infty} y \mathbb{P}(|X| \geq y) \, dy. \end{align*}$$

Since $[0,\infty) \ni y \mapsto y$ is increasing and $[0,\infty) \ni y \mapsto \mathbb{P}(|X| \geq y)$ decreasing, we get

$$\begin{align*} \mathbb{E}(|X|^2) &= 2 \sum_{n \in \mathbb{N}_0} \int_{n}^{n+1} y \mathbb{P}(|X| \geq y) \, dy \\ &\leq 2 \sum_{n \in \mathbb{N}_0} (n+1) \mathbb{P}(X| \geq n). \end{align*}$$