Comparison between $\|X\|_2$ and $\|X\|_{2,1}$

Question

For any real random variable $X$, define $$\|X\|_{2,1}=\int_0^\infty \sqrt{\Pr(|X|>t)}dt.$$ This quantity (it is not a norm) appears in various problems, e.g. the multiplier central limit theorem (see, e.g., Section 2.9 in this book) or in L-statistics (see, e.g., this paper). Problem 2.9.1 of the book cited above mentions the inequality $\|X\|_{2,1}^2\ge E(X^2)/4$. I think we have actually better. For all $x\ge 0$, $$\|X\|_{2,1} \ge \int_0^x \sqrt{\Pr(|X|>t)}dt\ge x \sqrt{\Pr(|X|>x)},$$ which implies that $$\begin{array}{rcl} \|X\|_{2,1}^2 & = & \int_0^\infty \|X\|_{2,1} \sqrt{\Pr(|X|>x)}dx \\ & \ge & \int_0^\infty x \Pr(|X|>x)dx \\ & =& E[X^2]/2. \end{array}$$ My question is: is this bound sharp (I don't think it is)? If not, what is the best constant in the inequality?

Answer 1

$$\begin{aligned} \|X\|_{2,1}^2&=\int_0^\infty\int_0^\infty ds\,dt\,\sqrt{P(|X|>s)}\sqrt{P(|X|>t)} \\ &\ge\int_0^\infty\int_0^\infty ds\,dt\,P(|X|>\max(s,t)) \\ &=E\int_0^\infty\int_0^\infty ds\,dt\,1(|X|>\max(s,t)) \\ &=E|X|^2=EX^2. \end{aligned}$$ So, we have an improvement of your bound. Moreover, the lower bound $EX^2$ on $\|X\|_{2,1}^2$ is exact: it is attained when $P(|X|=c)=1$ for some real $c$.