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?
 A: 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.
A: The way that you're doing it so far, you're treating $\Phi_1$ and $\Phi_2$ as ordinary uniformly distributed random variables.  That is, as heropup states, you are not taking the result of the addition modulo $2\pi$.
Even setting that aside, however, you have errors:
\begin{align}
F_{\Phi}(\phi)
    & = P(\Phi_1+\Phi_2\leq\phi) \\
    & = \int_0^{2\pi} P(\Phi_1\leq\phi-\phi_2\bigl|\Phi_2=\phi_2)
        f_{\Phi_2}(\phi_2) d\phi_2 \\
    & = \int_0^{2\pi} F_{\Phi_1}\left(\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 \cdots
\end{align}
You cannot expand $F_{\Phi_1}$ that way.  On one end, the reason is that if $\phi = \pi$, let's say, it cannot be the case that $\Phi_2 > \pi$.  The usual way to take that into account is to observe that $F_{\Phi_1}(\phi_1) = 0$ when $\phi_1 < 0$.  Unfortunately, you have just taken the expression for $F_{\Phi_1}(\phi_1)$ for the domain $0 \leq \phi_1 \leq 2\pi$ and rather carelessly expanded it to "the left" of that domain.  You didn't evaluate it at any point less than $2\pi$, however, so that wasn't your particular problem here.  You still have to fix it, though.
The other end is the problem you actually encountered.  The problem is analogous.  You've taken your linear expression for $F_{\Phi_1}(\phi_1)$ and continued it to "the right" of its proper domain.  Remember that the actual expression is
$$
F_{\Phi_1}(\phi_1) = \begin{cases}
    0 & \phi_1 < 0 \\
    \frac{\phi_1}{2\pi} & 0 \leq \phi_1 \leq 2\pi \\
    1 & \phi_1 > 2\pi
\end{cases}
$$
Try again, taking care with the domains, and see what you get.  You should end up with
$$
F_\Phi(\phi) = \begin{cases}
    0 & \phi < 0 \\
    \frac{\phi^2}{8\pi^2} & 0 \leq \phi \leq 2\pi \\
    1-\frac{(\phi-4\pi)^2}{8\pi^2} & 2\pi \leq \phi \leq 4\pi \\
    1 & \phi > 4\pi
\end{cases}
$$
Incidentally, if you do take the result modulo $2\pi$, you will find that the asymmetries cancel out, and you end up with just another uniformly distributed random variable between $0$ and $2\pi$.
