# How can the var(|X|) be defined?

I know that $$var(|X|) = E[|X|^2] - (E[|X|])^2 = E[X^2] - (E[|X|])^2$$.

However, I don’t know if (E[|X|])^2 can be simplified in terms of E[X] or something similar that has no absolute value.

Generally speaking, you have $$\begin{split} \mathbb{E}[X] &= \int_\mathbb{R} x f(x) dx\\ \mathbb{E}[|X|] &= \int_\mathbb{R} |x| f(x) dx = \int_0^\infty x f(x) dx - \int_{-\infty}^0 x f(x) dx\\ &= \mathbb{E}[X] - 2\int_{-\infty}^0 x f(x) dx \end{split}$$
Let $$X_+=\max(X,0)=X{\bf 1}_{\{X>0\}},~X_-=\max(-X,0)=-X{\bf 1}_{\{X<0\}}$$.
Then $$X=X_+-X_-$$, and $$|X|=X_++X_-$$. Also, $$X_+,X_-\ge 0$$ and $$X_+X_-=0$$.
$$\mbox{Var}(|X|) = \mbox{Var}(X)-4 E[X_+]E[X_-]=\mbox{Var}(X) + 4 E[X{\bf 1}_{\{X>0\}}]E[X{\bf 1}_{\{X<0\}}].$$
So Variance of $$|X|$$ is always less than or equal to variance of $$X$$ with equality if and only if $$|X|=X$$ with probability $$1$$.