Formulas or approximations for $\mathbb{E}\left( \frac{X}{\|X\|} \right)$ or $\mathbb{E}\left( \frac{X}{\|X\|^2} \right) $, $X\sim N(\mu, Id)$?

Question

I want to compute or approximate the following expected values with some analytic expression:

$\mathbb{E}\left( \frac{X}{\|X\|} \right)$ and $\mathbb{E}\left( \frac{X}{\|X\|^2} \right) $, where $X \in \mathbb{R}^n$ is a multivariate isotropic, non-centered gaussian $X\sim N(\mu, Id)$.

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, for the simpler case where the covariance is the identity matrix, I wondered whether there's a way to obtain or approximate these expected values. 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. This other question asks for the second expectation, but asks specifically for the expectation formula, while I'm asking if there's some reasonable approximation.

Edit: Cross posted in CV here

Answer 1

If we let the $x\in \mathbb R^n$ be distributed as $X\sim\mathcal N(\mu,\sigma^2 I)$, then it is clear that by symmetry the expectation value $\mathbb E\left( \frac{X}{|X|^k}\right)$ is parallel to $\mu$. The value can be computed exactly as $$ \mathbb E\left( \frac{X}{|X|^k}\right) = \frac{\Gamma\left(\frac{n}{2}+1-\frac{k}{2}\right)}{(2\sigma^2)^{k/2}\Gamma\left(\frac{n}{2}+1\right)}{}_1F_1\left(\frac{k}{2};\frac{n+2}{2};-\frac{|\mu|^2}{2\sigma^2}\right)\,\mu,\tag{$\ast$} $$ where ${}_1F_1(a;b;z)$ is Kummer's confluent hypergeometric function.

For certain values of $n$ and $k$ this can written in terms of more elementary functions e.g for $n=2$ $k=2$ $$ \mathbb E_{\mathbb R^2}\left( \frac{X}{|X|^2}\right) = \frac{1-e^{-\frac{\mu^2}{2\sigma^2}}}{|\mu|^2}\, \mu. $$

The $n$-dimensional integral for $(\ast)$ can be solved by utilizing the symmetry around $\mu$. Choose $r$ to be a radial coordinate and $\theta$ the angle to $\mu$, then the projection of the expectation value onto $\mu$ can be computed as $$ \mathbb E\left( \frac{\mu\cdot X}{|\mu||X|^k}\right) =(2\pi\sigma^2)^{-n/2}e^{-\frac{\mu^2}{2\sigma^2}}\int_0^\infty dr\int_0^\pi d\theta \,r\,A_{n-2}(r\sin\theta)e^{-\frac{r^2}{2\sigma^2}+\frac{\mu r}{\sigma^2}\cos\theta}\frac{\cos\theta}{r^k}, $$ where $A_{n-1}(r)=\frac{2\pi^{n/2}}{\Gamma(n/2)}r^{n-1}$ is the area of a $(n-1)$-sphere of radius $r$.

Answer 2

Let $X\sim\mathcal N(\mu,\sigma^2 I)$. Let $k \in \mathbb{R}_{> 0}$.

Using the identity ($q > 0, m > 0$) $$ \frac{1}{q^m} = \frac{1}{\Gamma(m)}\int_0^\infty \mathrm{e}^{-qt}t^{m - 1}\,\mathrm{d} t, $$ we have \begin{align*} \mathbb{E}\left(\frac{X}{\|X\|^k}\right) &= \mathbb{E}\left(\frac{X}{(\|X\|^2)^{k/2}}\right) \\[6pt] &= \mathbb{E}\left(X \cdot \frac{1}{\Gamma(k/2)}\int_0^\infty \mathrm{e}^{-t\|X\|^2}t^{k/2 - 1}\,\mathrm{d} t \right) \\[6pt] &= \frac{1}{\Gamma(k/2)}\int_0^\infty \mathbb{E}\left(X\mathrm{e}^{-t\|X\|^2}\right)t^{k/2 - 1}\,\mathrm{d} t \\ &= \mu \cdot \frac{1}{\Gamma(k/2)}\int_0^\infty \frac{1}{(2\sigma^2 t + 1)^{n/2 + 1}}\,\mathrm{exp}\left({-\frac{\|\mu\|^2 t}{2\sigma^2 t + 1}}\right) \cdot t^{k/2 - 1}\,\mathrm{d} t \end{align*} where we use $$\mathbb{E}\left(X\mathrm{e}^{-t\|X\|^2}\right) = \mu \cdot \frac{1}{(2\sigma^2 t + 1)^{n/2 + 1}}\,\mathrm{exp}\left({-\frac{\|\mu\|^2 t}{2\sigma^2 t + 1}}\right).$$

For example, when $k=2, n=2$, we have $$\mathbb{E}\left(\frac{X}{\|X\|^2}\right) = \mu \cdot \int_0^\infty \frac{1}{(2\sigma^2 t + 1)^{2}}\,\mathrm{exp}\left({-\frac{\|\mu\|^2 t}{2\sigma^2 t + 1}}\right) \,\mathrm{d} t = \mu \cdot \frac{1 - \mathrm{e}^{-\frac{\|\mu\|^2}{2\sigma^2}}}{\|\mu\|^2}$$ which is the same as @Ben's result.