Simple convolution problem Let $X$ be continuous uniform over $[0,2]$ and $Y$ be continuous uniform over $[3,4]$. Find and sketch the PDF of $Z = X + Y$, using convolutions.

So I have:
$$f(x) = 1/2, 0 \leq x \leq 2 $$
$$f(y) = 1,  \leq y \leq 4 $$
And my bounds of integration are between $[3,6]$(?)
I know I have to use the convolution formula, but I don't really understand how it all works out!
Can someone please take me through a step by step process for this type of question? I have some gaps in my knowledge that I can't seem to figure out.
 A: Suppose for the moment that we're working with uniform discrete random variables, $U\sim \mathcal{U}\{0..2\}$ and $V\sim\mathcal{U}\{3..4\}$
The probability mass of $W=U+V$ is calculated by summing the the joint probability mass $p_{\small U,V}(u,w-u)$ over all supported values of $u$:
$$\begin{align}
p_{\small W}(w) & = \sum_\overbrace{u \mid u\in\{0,1,2\}, w-u\in\{3,4\}}p_{\small U}(u)p_{\small V}(w-u)
\\[1ex] &=\begin{cases} p_{\small U}(0)p_{\small V}(3) & : w=3
\\ p_{\small U}(0)p_{\small V}(4)+p_{\small U}(1)p_{\small V}(3) & : w=4
\\ p_{\small U}(1)p_{\small V}(4)+p_{\small U}(2)p_{\small V}(3) & : w=5
\\ p_{\small U}(2)p_{\small V}(4) & :w=6
\\ 0 & :\text{elsewhere}
\end{cases}
\\[1ex] &=\begin{cases} 1/6 & : w=3
\\ 2/6 & : w=4 \vee w=5
\\ 1/6 & : w=6
\\ 0 & :\text{elsewhere}
\end{cases}
\end{align}$$
Can you follow the logic of that?  

Now let's extend this to our required continuous random variables.  It's just slightly more involved but still the same basic principle.  Instead of summing we integrate, and we have to pay attention to the support of the probability density functions.
Now for our continuous uniform random variables:
$$X\sim\mathcal{U}[0,2] \iff f_X(x) = \tfrac 1 2 \;\mathbf 1_{0\leq x\leq 2}
\\ Y\sim \mathcal{U}[3,4]\iff f_Y(y) = \mathbf 1_{3\leq y\leq 4}$$
Where the boldface $\bf 1$ is an indicator function, taking on the value of one when its subscript is true or zero when its subscript is not.
Now the sum of these random variables, $Z$, will have the support $[3..6]$
So putting it together we evaluate the convolution as:
$$\begin{align}
f_Z(z) & =\mathbf 1_{3\leq z\leq 6} \int_\Bbb R f_X(x)f_Y(z-x)\operatorname d x \tag{1}
\\[1ex] & = \mathbf 1_{3\leq z\leq 6}\int_\Bbb R \big(\tfrac 1 2\cdot \mathbf 1_{0\leq x\leq 2}\big)\big(\mathbf 1_{3\leq z-x\leq 4}\big)\operatorname d x \tag{2}
\\[1ex] & = \tfrac 1 2 \mathbf 1_{3\leq z\leq 6}\int_\Bbb R \mathbf 1_{\max(0,z-4)\leq x\leq \min(2,z-3)}\operatorname d x \tag{3}
\\[1ex] & =\tfrac 1 2 \mathbf 1_{3\leq z\leq 6}\int_{\max(0,z-4)}^{\min(2,z-3)}\operatorname d x \tag{4}
\\[1ex] & = \frac 1 2\left(\mathbf 1_{3\leq z\lt 4}\int_{0}^{z-3}\operatorname d x +\mathbf 1_{4\leq z\lt 5}\int_{z-4}^{z-3}\operatorname d x+ \mathbf 1_{5\leq z\leq 6}\int_{z-4}^{2}\operatorname d x \right)\tag {5}
\\[1ex] & = \frac 1 2\left((z-3)\mathbf 1_{3\leq z\lt 4} + \mathbf 1_{4\leq z\lt 5}+ (6-z)\mathbf 1_{5\leq z\leq 6} \right)\tag{6}
\\[3ex]
f_Z(z) &=\begin{cases}(z-3)/2 &:3\leq z\lt 4\\1/2 &: 4\leq z\lt 5 \\ (6-z)/2 &:5\leq z\leq 6\\ 0 & :\text{elsewhere} \end{cases} \tag{7}
\end{align}$$
A: $$f_{Z}\left(z\right)=\int f_{Y}\left(z-x\right)f_{X}\left(x\right)dx=\int_{z-4}^{z-3}f_{X}\left(x\right)dx$$
Now realize for what values $f_{X}\left(x\right)$ is zero or is not zero and in
that light (by calculating this integral) discern the cases:


*

*$z-3<0$

*$z-4<0\le z-3$

*$0\le z-4<1$

*$1\le z-4<2$

*$2\leq z-4$

