I have this question-
Let $X \sim Uniform(0,1)$ . $f_Y(y)= 1-|1-y|$ when $0 < y < 2$ and $0$ otherwise. It is also given that $X$ and $Y$ are independent. Then find the distribution of $(X+Y)$.
I have been trying to solve this using transformation technique,i.e $(X,Y) \to (U=X+Y,V=Y)$ and after finding the joint pdf of $(U,V)$ I have to integrate out $v$ to get the pdf of $(X+Y)$. But I am having a hard time finding the region of integration because I cannot separate the cases.
Any help would be very much appreciated!
Note that $$ f_X(x) = \begin{cases} 1 & x\in[0,1], \\ 0 & \text{else,} \end{cases} \qquad f_Y(y) = \begin{cases} y & y\in[0,1], \\ 2-y & y\in(1,2], \\ 0 & \text{else.} \end{cases} $$ The density of $Z$ is given by the convolution of $f_Y$ and $f_X$ (alternatively, you can interchange $f_Y$ and $f_X$ in the next formula, but for me it is easier to think this way since $f_Y$ is more "complicated"): $$ f_Z(z) = \int_{-\infty}^{\infty} f_Y(y)f_X(z-y)\, \mathrm dy $$ Now, being very careful for which $y$ and $z$ the densities $f_Y(y)$ and $f_X(z-y)$ take which values (and being non-zero in particular), we obtain: \begin{align*} \text{for }z\in[0,1]: \qquad & f_Z(z) = \int_{0}^{z} y\cdot 1\mathrm dy \\ \text{for }z\in(1,2]: \qquad & f_Z(z) = \int_{z-1}^{1} y\cdot 1\mathrm dy + \int_{1}^{z} (2-y)\cdot 1\mathrm dy \\ \text{for }z\in(2,3]: \qquad & f_Z(z) = \int_{z-1}^{2} (2-y)\cdot 1\mathrm dy \\ \text{for all other }z: \qquad & f_Z(z) = 0 \end{align*} I assume you can figure out the integrals by yourself. I hope I didn't make any typos, which is best verified by checking whether $\rho_Z(z)$ integrates to one.