# Entropy of probability distributions on a sphere

I have a family of probability distributions on the $$n$$-dimensional sphere $$\mathbb S^n \subset \mathbb R^{n+1}$$ defined in the following way:

$$D_0$$ is the uniform distribution, which is constructed by sampling $$k$$ points $$z_i \in \mathbb R^{n+1}$$ from a normal distribution of mean $$m_0 = (0,0, \dots, 0)$$ and covariance matrix $$\Sigma = diag(1,1, \dots,1)$$. In order to transform this Gaussian distribution into a uniform distribution on the sphere, we normalize every $$z_i$$. Our points will then be $$x_i = \frac{z_i}{\| z_i \|}$$.

The other distributions $$D_t$$ are constructed in the same way, the only difference is that they have mean $$m_t = (t,0, \dots, 0)$$. The covariance matrix remains constant.

How can I compute the entropy of these distributions?

I know that the entropy $$E$$ has a maximum under the uniform distribution, and intuitively I can see that $$E(D_t) > E(D_s)$$ if $$t, but I don't know how to show it formally.

Thank you very much!

Distributions on the sphere is studied and used in directional statistics, see for instance the text Directional Statistics by Kanti Mardia & Peter Jupp. The entropy of the uniform distribution on the $$n$$-sphere $$\mathbb S^n \subset \mathbb R^{n+1}$$ is by my calculations $$\log\{\frac{2\cdot \pi^{(n+1)/2}}{\Gamma((n+1)/2)}\}$$, which simply uses the surface area of the sphere$$^\dagger$$.
For the non-uniform distribution given, by the sampling process given the radius squared $$R^2=\| x\|^2$$ will have a non-central chisquared distribution woth non-centrality parameter $$\lambda=t^2$$. Then by calculating the conditional density given $$R^2=1$$ we can find the density on the sphere (details not given) as proportional to $$\exp(t x_1)$$, which by results in chapter 9 of the Mardia&Jupp book proves this is a von Mises-Fisher distribution. By the same reference, there is a maximum entropy characterization of the von Mises-Fisher distribution with given expectation, see also. With this results it shouldn't be difficult to calculate the entropy directly. And, indeed it is calculated here.
$$^\dagger$$The surface area of the $$n$$-sphere is $$\frac{2\cdot \pi^{(n+1)/2}}{\Gamma((n+1)/2)}$$ and is obtained by integrating the constant 1 with respect to the surface measure on the sphere. So to get the density of the uniform probability measure on the sphere we have to invert that, giving the (constant) density $$f(x)=\frac{\Gamma((n+1)/2)}{2 \cdot\pi^{(n+1)/2}}\cdot \mathbb{I}_{(\|x\|=1)}$$ and now you can compute the entropy.