Expectation of von Mises Fisher Distribution The von Mises- Fisher distribution is defined as 
$$
\frac{\kappa^{p/2-1}}{2\pi I_{p/2-1}(\kappa)}\exp(\kappa \mu^Tx)
$$
It is defined over the unit sphere i.e. $||x||_2^2=1$. My question is what is $E(x)$. I got a feeling it's simply $\mu$ but how would you prove this?
I'm asking mostly because I do not know how to integrate over a sphere.
 A: You can derive the answer as follows.  Start with the definition of the normalizing constant:
$$
\int \exp(\kappa \mu^{T} x) dx = \frac{(2\pi)^{p/2-1} I_{p/2-1}(\kappa)}{\kappa^{p/2-1}}
$$
(Note I have corrected an error in the question.)  Let $y = \kappa \mu$ so that $y$ is an unconstrained vector with $\kappa = \sqrt{y^T y}$.  It is easy to show that $d\kappa/dy = \mu$.
$$
\int x \exp(y^T x) dx = \frac{d}{dy} \int \exp(y^T x) dx
\\
= \frac{d\kappa}{dy} \frac{d}{d\kappa} \int \exp(y^T x) dx
\\
= \mu \frac{d}{d\kappa} \frac{(2\pi)^{p/2-1} I_{p/2-1}(\kappa)}{\kappa^{p/2-1}}
\\
= \mu \left(\frac{I'_{p/2-1}(\kappa)}{I_{p/2-1}(\kappa)} - \frac{p/2-1}{\kappa}\right) \frac{(2\pi)^{p/2-1} I_{p/2-1}(\kappa)}{\kappa^{p/2-1}}
$$
$$
E(x) = \frac{\int x \exp(y^T x) dx}{\int \exp(y^T x) dx} = \mu \left(\frac{I'_{p/2-1}(\kappa)}{I_{p/2-1}(\kappa)} - \frac{p/2-1}{\kappa}\right)
$$
Note $I'$ can be written in terms of $I$, as explained in wikipedia.
A: This is more like an extended comment rather than a full answer. I am working on a problem that is, I think, somewhat related to the question so I share my thoughts.
In what sense are you interested in the expectation? For example, for $\kappa = 0$ I don't think you can define a sensible expectation which is location "on" the sphere.
Let
\begin{equation}
c_d(\kappa) = \frac{\kappa^{p/2-1}}{2\pi I_{p/2-1}(\kappa)} \text,
\end{equation}
so the normalization constant is easier to handle.
Note that the problem is symmetric for rotation, so you can take, say, $\mu = e_1 = (1, 0, 0, \ldots, 0)$. In this case the PDF is simply
\begin{equation}
P(x) = c_p(\kappa) \exp(\kappa x_1) \text.
\end{equation}
If you then take the expectation of $x$ componentwise, you get for each $x_i$, $i = 2, 3, \ldots, p$
\begin{equation}
\mathbb{E}[x_i] = \int_{x_i \in S^{p - 1}} x_i c_p(\kappa) \exp{\kappa x_1} \,dx_i \text, \tag{*}
\end{equation}
where $S^{p - 1} = \{x \in \mathbb{R}^p : |x| = 1\}$ the $(p - 1)$-sphere on which the $p$-variate vMF distribution is defined. Now divide $S^{p - 1}$ into two "hemispheres" $H_1$ and $H_2$,
\begin{align}
H_1 &= \{x \in S^{p - 1} : x_i \ge 0\} \text, \\
H_2 &= \{x \in S^{p - 1} : x_i < 0\} \text.
\end{align}
It doesn't really matter which inequality is strict, the integrand of (*) vanishes for $x_i = 0$ anyway.
Now let's write (*) using $H_1$ and $H_2$ (remember that $i \ne 1$, so $x_i$ will not appear in the argument of $\exp$),
\begin{align}
\mathbb{E}[x_i] &= \int_{x_i \in H_1 \cup H_2} x_i c_p(\kappa) \exp{\kappa x_1} \,dx_i \\
&= \int_{x_i \in H_1} x_i c_p(\kappa) \exp{\kappa x_1} + \int_{x_i \in H_2} x_i c_p(\kappa) \exp{\kappa x_1} \\
&= \int_{x_i \in H_1} x_i c_p(\kappa) \exp{\kappa x_1} + \int_{x_i \in H_1} -x_i c_p(\kappa) \exp{\kappa x_1} \\
&= 0 \text.
\end{align}
Well, the situation for $x_1$ is considerably harder, as evaluating $\mathbb{E}[x_1]$ would result in and ugly mess of Gamma and modified Bessel functions... For $\kappa = 0$, we have the uniform distribution on $S^{p - 1}$, so $\mathbb{E}[x_1] = 0$. For $\kappa > 0$, intuitively, there is more probability mass on the $x_1 \ge 0$ "hemisphere" than on the $x_1 < 0$ "hemisphere", so $\mathbb{E}[x_1]$ should be $> 0$.
If we accept the handwaving above (or do the integration for real), we get $\mathbb{E}[x] = \mathbb{E}[x_1] e_1 = \mathbb{E}[x_1] \mu$. Combined with the fact that $\mathbb{E}[x_1] > 0$, this is vector pointing to the same direction as $\mu$. To get a vector that lies in $S^{p - 1}$, we can normalize to get
\begin{equation}
\frac{\mathbb{E}[x]}{|\mathbb{E}[x]|} = \frac{\mathbb{E}[x_1]\mu}{|\mathbb{E}[x_1]|} = \mu \text.
\end{equation}
If you remove the middle part of the equality above, the result will hold for any $\mu$ by rotation.
In this sense, yes, the expectation of the vMF distribution on the unit sphre is $\mu$ for $\kappa > 0$. For $\kappa = 1$, $\mathbb{E}[x] = 0$ (in the Euclidean sense) and the normalization will fail.

The distribution of $x_i$ (with $\mu = e_1$) is interesting for another reason, only slightly related to your question. It is the same thing for cosine similarity (or after a bit of scaling and shifting, cosine distance) and the vMF distribution that the $\chi$-distribution is for the Euclidean distance and the Gaussian distribution. That is,
\begin{align}
\chi &= |x - \mu| && \text{for $x \sim \mathcal{N}(\mu, 1)$,} \\
\text{like } x_1 &= \mu^T x = \frac{\mu^T x}{|\mu| \cdot |x|} && \text{for $x \sim \mathrm{vMF}(\mu, \kappa)$.}
\end{align}
