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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?

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    $\begingroup$ You're right. You need to replace inequality with non-strict inequality though. $\endgroup$ Jan 14 '18 at 10:44
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    $\begingroup$ 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. $\endgroup$
    – Ben
    Jan 19 '18 at 6:03

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