Convolution of uniform random variables I've 2 continuous random variables $X$ and $Z$ and I'm asked the p.d.f of $Y = X + Z$. $X$ is uniform over the interval $-\frac{1}{2} \leq x \leq \frac{1}{2}$ and $Z$ is uniform over $-\frac{a}{2} \leq z \leq \frac{a}{2}$.
I'm not sure how to break the intervals here. Cleary, $a \leq 1$ and $a > 1$ should be considered seperately but then I'm not sure how to proceed.
 A: $Y$ will lie within the range $-\frac{a+1}{2}\leq Y\leq \frac{a+1}{2}$.
Then by convolution:
$$\begin{align}f_Y(t)&=\int_\Bbb R f_{X,Z}(t-z,z)\operatorname d z~\cdot \mathbf 1_{-(a+1)/2\leqslant t\leqslant (a+1)/2}\\[2ex] & = \int_{\substack -1/2\leqslant t-z\leqslant 1/2\\ -a/2\leqslant z\leqslant a/2}\operatorname d z~\cdot \mathbf 1_{-(a+1)\leqslant 2t\leqslant (a+1)} \\[2ex] & = \int_{\max\{2t-1,-a\}}^{\min\{2t+1,a\}}\tfrac 12\operatorname d u~\cdot \mathbf 1_{-(a+1)\leqslant 2t\leqslant (a+1)} & z\gets u/2 \\[2ex] & = \int_{\max\{2t,1-a)\}-1}^{\min\{2t,a-1\}+1}\tfrac 12\operatorname d u~\cdot \mathbf 1_{-(a+1)\leqslant 2t\leqslant (a+1)}\end{align}$$
From here you partition into three subintervals for $2t$.
$$\begin{align} & ~\{-1-a\leqslant 2t\leqslant \min\{1-a,a-1\}\}\\ \cup &~ (\{1-a\leqslant 2t\leqslant a-1\}\cap\{1<a\}) \\ \cup &~ \{\max\{1-a,a-1\}\leqslant 2t\leqslant a+1\}\end{align}$$
A: Hint:
Can you compute $\mathbb{P}(Y \leq x)$? From this you get the cdf of $Y$, and then the pdf is the derivative of this.
