For any real random variable $X$, define $$||X||_{2,1}=\int_0^\infty \sqrt{\Pr(|X|>t)}dt.$$$$\|X\|_{2,1}=\int_0^\infty \sqrt{\Pr(|X|>t)}dt.$$ 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$$\|X\|_{2,1}^2\ge E(X^2)/4$. I think we have actually better. For all $x\ge 0$, $$||X||_{2,1} \ge \int_0^x \sqrt{\Pr(|X|>t)}dt\ge x \sqrt{\Pr(|X|>x)},$$$$\|X\|_{2,1} \ge \int_0^x \sqrt{\Pr(|X|>t)}dt\ge x \sqrt{\Pr(|X|>x)},$$ which implies that $$\begin{array}{rcl} ||X||_{2,1}^2 & = & \int_0^\infty ||X||_{2,1} \sqrt{\Pr(|X|>x)}dx \\ & \ge & \int_0^\infty x \Pr(|X|>x)dx \\ & =& E[X^2]/2. \end{array}$$$$\begin{array}{rcl} \|X\|_{2,1}^2 & = & \int_0^\infty \|X\|_{2,1} \sqrt{\Pr(|X|>x)}dx \\ & \ge & \int_0^\infty x \Pr(|X|>x)dx \\ & =& E[X^2]/2. \end{array}$$ My question is: is this bound sharp (I don't think it is)? If not, what is the best constant in the inequality?
