How to find the pdf of $x+y$, where $x$, $y$ are iid and both are uniform $(\theta, \theta +1)$? I don't know how to divide intervals. I know how to solve if uniform $(0,1)$, since I can draw a square and use areas to analyze it. But I don't know how to solve out uniform $(\theta, \theta +1)$. I can use convolution, but after that, I don't know how to discuss intervals.
My attempt: I denote $Z=X+Y$ their sum. Then we have $f_X(x)=f_Y(y)=1 $ if $\theta\le x \le \theta+1$ and $0$ otherwise.
and the density function for the sum is given by
$$f_Z(z)=\int_{-\infty}^\infty f_X(z-y)f_Y(y)\,dy.$$
Since $f_Y(y)=1$ if $\theta\le y \le \theta+1$ and 0 otherwise, this becomes
$$f_Z(z)=\int_{\theta}^{\theta+1} f_X(z-y)\,dy.$$
Now the integrand is 0 unless $\theta \le z-y \le \theta+1$, i.e., $z-\theta-1 \le y \le z-\theta$ and then it is 1 in this case.
I don't know how to do next.
CDF when $x$ and $y$ are uniform $(0,1).$ $$ P(Z\leq z)= \begin{cases} \frac{z^2}{2}& 0< z \leq 1\\ 1-\frac{(2-z)^2}{2}& 1<z<2. \end{cases} $$
Taking the derivative, the PDF is $$ f(z)= \begin{cases} z & 0<z\leq1\\ 2-z & 1<z<2. \end{cases} $$
If $X$ and $Y$ are uniform $(\theta,\theta+1),$ then $X-\theta$ and $Y-\theta$ are uniform $(0,1).$ Then $(X-\theta)+(Y-\theta)$ has the distribution given above. Then the PDF for $X+Y$ is $f$ translated by $2\theta.$ $$ f_{\theta}(z)=f(z-2\theta)= \begin{cases} (z-2\theta) &2\theta < z \leq 2\theta+1\\ 2-(z-2\theta)& 2\theta+1 < z < 2\theta+2. \end{cases} $$