Truncated Von Mises-Fisher distribution I am putting a von Mises-Fisher prior on my data. The data does lie on a unit sphere, but the only problem is that my data is always positive. So I feel like I am wasting my prior on unnecessary negative space. This problem probably gets exaggerated once I go into high dimensions (~500) since I'm only using a single quadrant out of the possible $2^{500}$.
So my question is how do I work out the normalising constant:
It is known that $\int_{x\in S}\exp(\kappa\mu^Tx)dx$ where S is the surface of the unit sphere is $\frac{(2\pi)^{p/2} I_{p/2-1}(\kappa) }{\kappa^{p/2-1}}$ where $p$ is the number of dimensions and $I$ is the modified Bessel function of the first kind: http://en.wikipedia.org/wiki/Von_Mises%E2%80%93Fisher_distribution.
However, what would happen if I restrict this sphere to the first quadrant. i.e. $x_i>0$?
$$\int_{x\in S, x_i>0}\exp(\kappa\mu^Tx)dx=?$$
 A: Because the analysis should not be too sensitive to the prior, we should feel free to make minor modifications to the prior.  Instead of truncating it, why not reflect all the probability into the first hyperquadrant?  That is, continue to use a von Mises-Fisher prior for $(x_1,x_2,\ldots,x_{n})$ (with $n\approx 500$) but base your analysis on $(|x_1|,|x_2|,\ldots,|x_{n}|)$. That would not need any renormalization at all.
The objection immediately arises that the calculations would require a $500$-fold sum, amounting to $2^{500}$ terms, which is an impossible calculation.  Although that is true, an algebraic simplification makes it possible.  I am suggesting using a prior
$$f(\mathbf x; \mu, \kappa) = C(\mu, \kappa) \sum_{i\in \{-1,1\}^n} \exp\left(\kappa (i_1 \mu_1, i_2\mu_2, \ldots, i_n\mu_n) \cdot \mathbf x\right)$$
where $C(\mu,\kappa)$ is the normalizing constant for the von Mises-Fisher distribution with parameters $(\mu, \kappa)$, all the $x_i$ are non-negative (and, without any loss of generality, you may as well assume all the $\mu_i$ are non-negative, too).  But by separately performing the sum over the last component, the foregoing can be written
$$C(\mu, \kappa) \sum_{i\in \{-1,1\}^{n-1}} \left(\exp\left(\kappa (i_1 \mu_1, i_2\mu_2, \ldots,\mu_n) \cdot \mathbf x\right) + \exp\left(\kappa (i_1 \mu_1, i_2\mu_2, \ldots,-\mu_n) \cdot \mathbf x\right)\right) \\
= C(\mu, \kappa) 2\cosh(\kappa \mu_n x_n)\sum_{i\in \{-1,1\}^{n-1}} \exp\left(\kappa (i_1 \mu_1, i_2\mu_2, \ldots,i_{n-1}\mu_{n-1}) \cdot \mathbf x_{[-n]}\right)$$
where $\mathbf x_{[-n]} = (x_1, x_2, \ldots, x_{n-1})$.  Proceeding inductively on $n$ yields
$$f(\mathbf x; \mu, \kappa) = C(\mu, \kappa) 2^n \prod_{i=1}^n \cosh(\kappa \mu_i x_i)$$
which is quite tractable.  For $\kappa \gg 0$ (that is, as this prior grows a little less diffuse), $f(\mathbf x; \mu, \kappa)$ approaches the truncated von Mises-Fisher distribution.
