This is a cross-posting of this math SE question.
I want to compute or approximate the following expected value with some analytic expression:
$\mathbb{E}\left( \frac{X}{||X||} \right)$ , where $X \in \mathbb{R}^n$ is a multivariate non-centered gaussian $X\sim N(\mu, \Sigma)$.
For a general $X\sim N(\mu, \Sigma)$ Gaussian, the variable $\frac{X}{||X||}$ is called the general projected normal distribution. This is a complicated distribution, whose moments don't seem to have closed form formulas.
However, I'm wondering whether there is either a useful approximation, or some formula for the case with diagonal/identity covariance. This question comes close to what I need, asking for the expectation $\mathbb{E} \left( \frac{1}{1+\|X\|^2} \right)$ and someone posted an approximation.
One solution I thought of is using a Taylor approximation, in which: $\mathbb{E}\left( \frac{A}{B} \right) \approx \frac{\mathbb{E}(A)}{\mathbb{E}(B)} - \frac{Cov(A,B)}{\mathbb{E}(B)^2} + \frac{\mathbb{E}(A)var(B)}{\mathbb{E}(B)^3}$, where I would have $A = X$ and $B = ||X||$ in the description above. However, I'm not sure there's an expression for $Cov(X, ||X||)$, or whether there's an analytic expression for the variance of the non-centered Chi distribution (i.e. $var(||X||)$).
So, I'm wondering, does the approach above seem valid? Is there expressions for $Cov(X, ||X||)$ and $var(||X||)$ where $X\sim N(\mu, \Sigma)$? Is there some expression or simplification if we assume that $Sigma$ is diagonal or the identity?