Say that I have a probability density function defined on the unit circle, $f_{\Theta}(\theta)$, with $\theta \in \left[0,2\pi\right)$. I have a joint pdf, assuming independence, of $f_{\Theta_1,\Theta_2,\Theta_3}(\theta_1,\theta_2,\theta_3)=f_{\Theta}(\theta_1) f_{\Theta}(\theta_2) f_{\Theta}(\theta_3)$. I also have the following transformations: $\theta_{12}=\theta_2-\theta_1$, and $\theta_{23}=\theta_3-\theta_2$. Using these transformations, I rewrite the joint pdf as:
$$f_{\Theta_{12},\Theta_2,\Theta_{23}}(\theta_{12},\theta_2,\theta_{23})=f_{\Theta}(\theta_2-\theta_{12}) f_{\Theta}(\theta_2) f_{\Theta}(\theta_2+\theta_{23})$$
I integrate out the dependence on $\theta_2$ to arrive at:
$$f_{\Theta_{12},\Theta_{23}}(\theta_{12},\theta_{23})=\int_{0}^{2\pi} f_{\Theta}(\theta_2-\theta_{12}) f_{\Theta}(\theta_2) f_{\Theta}(\theta_2+\theta_{23}) \text{d}\theta_2$$
This is a sort of double convolution, the original function is being convolved with two copies of itself each shifted in opposite directions by independent amounts ($\theta_{12}$ and $\theta_{23}$ respectively).
Here is my question: I have $f_{\Theta_{12},\Theta_{23}}(\theta_{12},\theta_{23})$ and would like to recover $f_{\Theta}(\theta)$, how can I do this?