Comparison between variance of $|x|$ and $x$ for the symmetric distribution

by RAHUl JHa   Last Updated January 14, 2018 11:19 AM

For a symmetric distribution, how the following inequality holds which is given by my teacher:


What I think is that it should be opposite since for a symmetric distribution the mean is zero. Also, $E(|X|^2)=E(X^2)=V(X). Then, the

$V(|X|)=V(X)-[E(|X|)]^2$ which implies that $V(X)=V(|X|)+[E(|X|)]^2$ which is obviously greater than $V(|X|)$ since $[E(|X|)]>0$. Hence , it should be $V(|X|)<V(X)$ not $V(|X|)>V(X)$. Is my approach right?

Related Questions

regarding symmetric distributions

Updated February 26, 2017 16:19 PM

About symmetric distributions

Updated January 18, 2018 00:19 AM

Identity related to symmetric distribution

Updated July 09, 2017 19:19 PM