I am considering the following question: Let $\theta_1$ and $\theta_2$ be the rest-angels observed in two consecutive spins of a roulette wheel, and suppose that expectations on the two-dimensional sample space thus generated are given by $$E[X(\theta_1, \theta_2)] = \frac{1}{4\pi^2} \int_{0}^{2\pi} \int_{0}^{2\pi} X(\theta_1, \theta_2) d\theta_1 d\theta_2.$$ Show that $P(\theta_1 \in A_1, \theta_2 \in A_2) = P(\theta_1 \in A_1)P(\theta_2 \in A_2).$ If $X = \theta_1 + \theta_2 - 2\pi$, then show either from the relation $f(x) = \frac{\partial F(x)}{\partial x} = \frac{\partial}{\partial x}P(X \leq x)$ or by calculating $E[H(X)]$ for arbitrary $H$, that $X$ is continuously distributed over the interval $[-2\pi, 2\pi]$ with density $(2\pi - |x|)/4\pi^2.$
I have shown that $P(\theta_1 \in A_1, \theta_2 \in A_2) = P(\theta_1 \in A_1)P(\theta_2 \in A_2)$, in the following way: $\begin{align} P(\theta_1 \in A_1, \theta_2 \in A_2) &= E(I(\theta_1 \in A_1, \theta_2 \in A_2)) \\\\ &= \frac{1}{4\pi^2} \int_{0}^{2\pi} \int_{0}^{2\pi} I(\theta_1 \in A_1, \theta_2 \in A_2) d\theta_1 d\theta_2 \\\\ &= \frac{1}{4\pi^2} \int_{0}^{2\pi} \int_{0}^{2\pi} I(\theta_1 \in A_1) I(\theta_2 \in A_2) d\theta_1 d\theta_2 \\\\ &= \frac{1}{2\pi} \int_{A_1} d\theta_1 \frac{1}{2\pi} \int_{A_2} d\theta_2 \\\\ &= P(\theta_1 \in A_1)P(\theta_2 \in A_2) \end{align}$
However, I'm uncertain how to show the second result. With the first method proposed, I get $\begin{align} F(x) = P(X \leq x) = E(I(X \leq x)) &= \frac{1}{4\pi^2} \int_{0}^{2\pi} \int_{0}^{2\pi} I(\theta_1 + \theta_2 - 2\pi \leq x) d\theta_1 d\theta_2 \\\\ &= \frac{1}{4\pi^2} \int_{0}^{2\pi + x} \int_{0}^{2\pi + x - \theta_1} d\theta_2 d\theta_1 \\\\ ... \\\\ &= \frac{1}{4\pi^2}\left(\frac{1}{2}(x+2\pi)^2\right), \end{align}$
so that $f(x) = \frac{1}{4\pi^2}(x+2\pi)$, which is not quite the desired answer. Where might I have gone awry?