Expected value of $Z=2X-Y$

Question

Let $X\perp Y$ two exponential random variables with mean $1$.

1. Find the density of $Z=2X-Y$.

$\rightarrow f_Z(z)=\frac{1}{3}e^{-\frac{z}{2}}$.

2. $\mathbb{E}[Z],\mathbb{E}[|Z|],\mathbb{E}[Z^2]$.

$\rightarrow \mathbb{E}[Z]=1,\mathbb{E}[|Z|]=3,\mathbb{E}[Z^2]=6$

3. Say if $Z\perp X$.

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Hoping that's all correct, I've only a doubt on $\mathbb{E}[|Z|]=\mathbb{E}[|2X-Y|]=2\mathbb{E}[|X|]-\mathbb{E}[|Y|]$. Am I wrong? Thanks in advance.

Answer 1

The density of $\ Z\ $ is just the derivative of its cumulative distribution function. To get its cumulative distribution function you either have to integrate the joint density of $\ X,Y\ $ over the region $\ \left\{(x,y)\in\left.\mathbb{R}^2\,\right|2x-y\le z\right\}\ $, or (equivalently, since $\ X\ $ and $\ Y\ $ are independent) compute the convolution of the cumulative distribution functions of $\ X\ $ and $\ -2Y\ $. The following calculation takes the first approach \begin{align} P(Z\le z)&= P(2X-Y\le z)\\ &=\int_0^\infty\int_{\max(0,2x-z)}^\infty e^{-x}e^{-y}dydx\\ &=\cases{\displaystyle \int_0^\infty e^{-x}\int_{2x-z}^\infty e^{-y}dydx&if $\ z<0$\\ \displaystyle \int_0^\frac{z}{2}e^{-x}\int_0^\infty e^{-y}dydx+\int_\frac{z}{2}^\infty e^{-x}\int_{2x-z}^\infty e^{-y}dydx&if $\ z\ge0\ $. }\\ &=\cases{\displaystyle\frac{e^z}{3}&if $\ z<0$\\ \displaystyle 1-\frac{2e^{-\frac{z}{2}}}{3}& if $\ z\ge0\ $. } \end{align} Differentiating this with respect to $\ z\ $ gives $$ f_Z(z)=\cases{\displaystyle \frac{e^z}{3}&if $\ z<0$\\ \displaystyle \frac{ e^{-\frac{z}{2}}}{3}&if $\ z\ge0\ $.} $$