I'm trying to find the joint density of $Z=X+Y$ where $X\in U(0,1), Y\in U(0,\alpha)$
Here $U$ is the uniform distribution.
The method I use i to introduce an auxilary variable $W=X$ and then use the transformation theorem according to which we have $f_{ZV}(z,v) = f_{XY}(v,z-v)|J| = \alpha^{-1}$
Now if I just could integrate with respect to $w$ I would be done, but I don't know which regions to integrate over. At least $v$ shold range from $0$ to $1$, and $z$ from $0$ to $1+\alpha$, but which is the third? How do I picture this? Also I think I may need to take into account whether $\alpha$ is greater or less than $1$?
Thanks in advance!
Suppose $\alpha< 1$ (for $\alpha>1$, it is similar.)
To calculate the probability (density) of a number $t$ with $0<t<1+\alpha$, we have to evaluate $$ \int_{x=0}^{t}P_1(x)P_2(t-x)\,dx $$ where $P_1$ is the uniform distribution $(0,1)$ and $P_2$ is the uniform distribution on $(0,\alpha)$. When $0< t a$, we get $$ \int_{x=0}^{t}P_1(x)P_2(t-x)\,dx=\int_{x=0}^{t}1\cdot \frac 1\alpha\,dx=\frac t\alpha $$ When $\alpha<t<1 $, we get \begin{align} \int_{x=0}^{t}P_1(x)P_2(t-x)\,dx&= \int_{x=0}^{t-\alpha}P_1(x)P_2(t-x)\,dx+\int_{x=t-\alpha}^{t}P_1(x)P_2(t-x)\,dx\\ &= \int_{x=0}^{t-\alpha}1\cdot 0\,dx+\int_{x=t-\alpha}^{t}1\cdot \frac 1\alpha\,dx\\ &=0+\alpha\frac 1\alpha=1 \end{align} When $1<t<1+\alpha$, we get \begin{align} &\int_{x=0}^{t}P_1(x)P_2(t-x)\,dx=\int_{x=0}^{t-\alpha}P_1(x)P_2(t-x)\,dx\\ &+\int_{x=t-\alpha}^{1}P_1(x)P_2(t-x)\,dx+\int_{x=1}^{t}P_1(x)P_2(t-x)\,dx\\ &=\int_{x=0}^{t-\alpha}1\cdot0\,dx+\int_{x=t-\alpha}^{1}1\cdot\frac 1\alpha \,dx+\int_{x=1}^{t}0\cdot\frac1\alpha\,dx\\ &=0+(1-(t-\alpha))\frac 1\alpha+0=\frac{1-t+\alpha}{\alpha} \end{align} We have to integrate over three regions here, because there are several critical points: $P_1(x)$ is discontinuous at $x=0$ and $x=1$ and $P_2(x)$ at $0$ and $\alpha$. Because we have $P_x(t-x)$, they are at $x=t$ and $x=t-\alpha$. Now, we sort them: $$ 0<t-\alpha<1<t $$ Because of the discontinuities, we have to integrate each of the three parts of the interval $[0,t]$ on its own. We can plot the result: