expected value of a $e^{-2|x|}$ Let $f(x)=e^{-2|x|}$
Find: 


*

*$E(X)$

*$E(|X|)$

*$E(x')$ where $x'$ denotes the largest integer not greater than $x$.
I'm stuck on this question and am confused about how to use the modulus sign. I feel I need to show it separately for cases when $x\ge0$ and $x\le0$ but I'm not sure.
Any help is appreciated

*

*$f(x)=f(-x)$, so if $E[X]$ exists then $E[X]=0$ as a symmetric random variable.  It does exist, because $E[|X|]$ exists.

*If $Y=|X|$ then the density of $Y$ is $g(y)=2e^{-2y}$ for positive $y$ and $E[|X|]=E[Y]=\frac12$ as an exponential random variable

*If $Z=\lfloor X \rfloor $ then $\Pr(Z=n) = \Pr(Z=-(n+1)) = e^{-2n}(1-e^{-2})$ for non-negative integer $n$, so is should be easy to see $E[\lfloor X \rfloor]=E[Z]=-\frac12$. 
