How to change variable of distribution from vector to angle from fixed point?

Question

I have a distribution of vector $\textbf{x}=\langle \sin{\phi_x}\cos{\theta_x}, \sin{\phi_x}\sin{\theta_x}, \cos{\phi_x} \rangle$ on the unit sphere (von Mises-Fisher):

\begin{align}f(\phi_x,\theta_x;\kappa)=C(\kappa)e^{\kappa \cos{\phi_x}}\sin{\phi_x},\end{align}

where $\phi_x$ and $\theta_x$ are the spherical coordinate angles ($\phi_x$: angle between $\textbf{x}$ and the $z$-axis; $\theta_x$: angle around the $z$-axis to the $x$-axis), $\kappa \geq 0$ is a scaling factor, and $C(\kappa)$ is a normalization factor. This distribution is radially symmetric.

Another vector $\textbf{v}=\langle \sin{\phi_v}, 0, \cos{\phi_v} \rangle$ is set to a fixed point on the sphere and on the $xz$-plane, where the angle between $\textbf{v}$ and $\textbf{x}$,

\begin{align}\psi(\textbf{v}, \textbf{x}) = \cos^{-1}{(\textbf{v}\cdot \textbf{x})}=\cos^{-1}{(\sin{\phi_v}\sin{\phi_x}\cos{\theta_x} + \cos{\phi_v}\cos{\phi_x})}.\end{align}

I am interested in the PDF $g(\psi)$, with a fixed $\textbf{v}$, which should only be a function of $\phi_v$ and $\kappa$. Unfortunately, $\psi(\textbf{v},\textbf{x})$is not a 1-to-1 function. How can I go about finding this PDF $g(\psi)$?

Edit: Added spherical area element.

Answer 1

Here's the answer I came up with. From the link @whuber posted (in my notation above), if $\textbf{v}$ is set to the $z$-axis, \begin{align}f(\psi, \epsilon \mid \phi, \kappa) = C(\kappa)e^{\kappa(\cos\psi \cos\phi + \sin\psi \sin\phi\cos(\epsilon-\beta))}\sin\psi),\end{align} where $\phi$ is the angle between $\textbf{v}$ and $\boldsymbol{\mu}$ (the center vector of $\textbf{x}$), $\psi$ is the angle between $\textbf{v}$ and $\textbf{x}$, $\epsilon$ is the angle between $\textbf{x}$ and the $x$-axis around $\textbf{v}$, and $\beta$ is the angle between $\boldsymbol{\mu}$ and the $x$-axis around $\textbf{v}$. Since we're interested in angles not vectors, we can set $\boldsymbol{\mu}$ to be on the $xz$-plane, so $\beta=0$. Then, since $\psi$ and $\epsilon$ are independent, we can integrate out $\epsilon$.

\begin{align} \begin{split} g(\psi \mid \phi,\kappa) & = \int_0^{2\pi}{f(\psi, \epsilon \mid \phi, \kappa) d\epsilon} \\ & = \int_0^{2\pi}{C(\kappa)e^{\kappa(\cos\psi \cos\phi + \sin\psi \sin\phi\cos\epsilon)}\sin\psi)d\epsilon} \\ & = C(\kappa)e^{\kappa\cos\psi \cos\phi}\sin\psi \int_0^{2\pi}{e^{\kappa \sin\psi \sin\phi\cos\epsilon}d\epsilon} \\ & = C(\kappa)e^{\kappa\cos\psi \cos\phi}\sin\psi (2\pi I_0(\kappa \sin\psi \sin\phi)), \end{split} \end{align}

where $I_0(z)$ is the modified Bessel function of the first kind and $C(\kappa)=\frac{\kappa}{4\pi \sinh \kappa}$. Simplifying,

\begin{align} g(\psi \mid \phi,\kappa) = \frac{\kappa}{2 \sinh\kappa} e^{\kappa\cos\psi \cos\phi} \sin\psi I_0(\kappa \sin\psi \sin\phi). \end{align}

As a quick check, if $\textbf{v}=\boldsymbol{\mu}$, then $\phi=0$, and \begin{align} g(\psi \mid \kappa) = \frac{\kappa}{2 \sinh\kappa} e^{\kappa\cos\psi} \sin\psi, \end{align} which is the expected form of the von Mises-Fisher distribution.