Distribution of sum of 2 circular uniform random variables I hope you can help me resolve the following problem:
Let $\Phi_1$ and $\Phi_2$ circular uniform random variables such that $0\leq\Phi_i\leq 2\pi$ (with $i=1,2$). Then the probability density function (pdf) and the cumulative distribution function (cdf) are given by
\begin{align}
f_{\Phi_i}(\phi_i) &= \frac{1}{2\pi}
\\
F_{\Phi_i}(\phi_i) &= \frac{\phi_i}{2\pi}.
\end{align}
I'd like to find the distribution of sum $\Phi=\Phi_1+\Phi_2$ so that I've calculated the cdf of $\Phi$ as follows:
\begin{align}
F_{\Phi}(\phi) = Pr\{\Phi_1+\Phi_2\leq\phi\} 
&= \int_0^{2\pi} Pr\{\Phi_1\leq\phi-\phi_2\bigl|\Phi_2=\phi_2\bigr.\} f_{\Phi_2}(\phi_2) d\phi_2
\\
&= \int_0^{2\pi} F_{\Phi_1}\left(\Phi_1\leq\phi-\phi_2\right) f_{\Phi_2}(\phi_2) d\phi_2
\\
&= \int_0^{2\pi} \frac{(\phi-\phi_2)}{2\pi} \frac{1}{2\pi} d\phi_2
\\
&= \frac{\phi}{2\pi} - \frac{1}{2}.
\end{align}
Due to $\Phi=\Phi_1+\Phi_2$, then $0\leq\Phi\leq 4\pi$. Unfortunately, we have
\begin{align}
F_{\Phi}(4\pi) = \frac{4\pi}{2\pi} - \frac{1}{2} = \frac{3}{2} > 1.
\end{align}
This is really unreasonable! Can anyone help me find the exact distribution of $\Phi$, please?

The immediate problem I see is the expression of the conditional probability in the integrand in the first step.  To help you understand why it is not correct, consider the case where $\Phi_1 = 3\pi/2$, $\Phi_2 = \pi$.  Then their sum is $$\Phi = \Phi_1 + \Phi_2 = 5\pi/2 \equiv \pi/2$$ where in the last step we have to reduce the angle modulo $2\pi$.  So you cannot write the conditional probability this way:  $$\Pr[\Phi \le \phi \mid \Phi_2 = \phi_2] \ne \Pr[\Phi_1 \le \phi - \phi_2].$$  You have to take into consideration the modulo operation.
