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

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

$V(|X|)>V(X)$

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)$. Is my approach right? • You're right. You need to replace inequality with non-strict inequality though. Jan 14 '18 at 10:44 • Also, just add one more change. A symmetric distribution can be symmetric around values other than zero, so it is not true that a symmetric distribution necessarily has$\mathbb{E}(X) = 0\$. So what you should say instead is that since you are only concerned with the variance, without loss of generality, you can assume a symmetric distribution around zero.
– Ben
Jan 19 '18 at 6:03