1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

Generally speaking, you have $$ \begin{split} \mathbb{E}[X] &= \int_\mathbb{R} x f(x) dx\\ \mathbb{E}[|X|] &= \int_\mathbb{R} |x| f(x) dx = \int_0^\infty x f(x) dx - \int_{-\infty}^0 x f(x) dx\\ &= \mathbb{E}[X] - 2\int_{-\infty}^0 x f(x) dx \end{split} $$

$\endgroup$
0
$\begingroup$

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 $X_+X_-=0$.

Using these,

$$\mbox{Var}(|X|) = \mbox{Var}(X)-4 E[X_+]E[X_-]=\mbox{Var}(X) + 4 E[X{\bf 1}_{\{X>0\}}]E[X{\bf 1}_{\{X<0\}}].$$

So Variance of $|X|$ is always less than or equal to variance of $X$ with equality if and only if $|X|=X$ with probability $1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.