Assume that $X_{1}$ and $X_{2}$ are independent exponential random variables with parameter $\lambda$. Let $Y_{1} = X_{1} + X_{2}$ and $Y_{2} = X_{1} - X_{2}$. Determine
(a) Find the joint distribution of $Y_{1}$ and $Y_{2}$
(b) Find the marginal distribution of $Y_{1}$
(c) Find the marginal distribution of $Y_{2}$
d) Are $Y_{1}$ and $Y_{2}$ independent?
MY SOLUTION
(a) To begin with, I noticed that $X_{1} = (Y_{1} + Y_{2})/2$ and $X_{2} = (Y_{1} - Y_{2})/2$. Hence we get \begin{align*} f_{Y_{1},Y_{2}}(Y_{1},Y_{2}) & = f_{X_{1},X_{2}}\left(\frac{Y_{1}+Y_{2}}{2},\frac{Y_{1}-Y_{2}}{2}\right)|\det J(y_{1},y_{2})|\\\\ & = \frac{1}{2}f_{X_{1}}\left(\frac{Y_{1} + Y_{2}}{2}\right)f_{X_{2}}\left(\frac{Y_{1}-Y_{2}}{2}\right) = \frac{\lambda^{2}}{2}\exp(-\lambda y_{1}) \end{align*}
where the support of $Y$ is given by the intersection of $Y_{1} - Y_{2} \geq 0$ and $Y_{1} + Y_{2} \geq 0$.
(b) According to the definition of marginal distribution, we have \begin{align*} f_{Y_{1}}(y_{1}) = \int_{-\infty}^{+\infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})\mathrm{d}y_{2} = \int_{-y_{1}}^{y_{1}}\frac{\lambda^{2}}{2}\exp(-\lambda y_{1})\mathrm{d}y_{2} = \lambda^{2}y_{1}\exp(-\lambda y_{1}) \end{align*}
where $y_{1} \geq 0$.
(c) Analogously, we have \begin{align*} f_{Y_{2}}(y_{2}) = \int_{-\infty}^{+\infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})\mathrm{d}y_{1} = \int_{y_{2}}^{\infty}\frac{\lambda^{2}}{2}\exp(-\lambda y_{1})\mathrm{d}y_{1} = \frac{\lambda}{2}\exp(-\lambda y_{2}) \end{align*}
Then I get stuck, because I do not know how to describe the support of $Y_{2}$ neither if the integration limits are wrong. Could someone help me out?
you must have $Y_1 \ge Y_2$ AND $Y_1 \ge - Y_2$
meaning you must have $Y_1 \ge max(Y_2,-Y_2) = |Y_2| $
\begin{align*} f_{Y_{2}}(y_{2}) &= \int_{-\infty}^{+\infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})\mathrm{d}y_{1} \\ &= \int_{|y_2|}^{+\infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})\mathrm{d}y_{1} \\ &= \frac{\lambda}{2}\exp(-\lambda |y_{2}|) \end{align*}