Let $\xi \sim \mathcal{N}(0, \Sigma)$ be a Gaussian random vector in $\mathbb{R}^n$. I would like to calculate the covariance matrix of the normalized vector $\frac{\xi}{\lVert \xi \rVert}$, i.e.:
$$ C = \mathbb{E}\left[\frac{\xi \otimes \xi}{\lVert \xi \rVert^2} \right] $$
where $\lVert \cdot \rVert$ denotes the Euclidean $l_2$-norm.
What we know:
- If $\Sigma = I$, then $C = n^{-1} I$, where $I$ is the identity matrix, according to this post, but it is not clear how to generalise.
- Using the spectral decomposition $\Sigma = V\Lambda V^\mathtt{T}$, it is enough to estimate $C$ for $\xi \sim \mathcal{N}(0, \Lambda)$, where $\Lambda$ is a diagonal matrix, since $\mathbb{E}\left[\frac{(V \xi) \otimes (V \xi)}{\lVert V\xi \rVert^2}\right] = VCV^\mathtt{T}$.
Any help would be appreciated. In particular, I am interested in cases where $n\to\infty$ and $\xi$ lies in a (separable) Hilbert space, which is well defined when $\Sigma$ is trace class and positive definite.
Best Answer
Assume that $A=\mathrm{diag}(a^2_1,\ldots,a^2_n)$ and write $\xi_i=a_iZ_i$ where the $Z_1,\ldots, Z_n$ are independent and $N(0,1).$ Write also $Y_i=Z_i^2/2$ which has density $e^{-y}y^{-1/2}/\sqrt{\pi}.$ The trick is to use $\frac{1}{U}=\int_0^{\infty}e^{-s U}ds$ applied to $U=a_1^2Y_1+\cdots+a_n^2Y_n$ for computing $A=E( Y_1/U)$ and $B=E(\sqrt{Y_1Y_2}/U).$ We get
$$A=\int_0^{\infty}E(Y_1e^{-sa_1^2Y_1})\prod_{i=2}^nE(e^{-sa_i^2Y_i})ds=\frac{1}{2}\int_0^{\infty}\frac{1}{1+a_1^2 s}\times \prod_{i=1}^n\frac{1}{(1+a_i^2 s)^{1/2}}ds$$
$$B=\int_0^{\infty}E(\sqrt{Y_1}e^{-sa_1^2Y_1})E(\sqrt{Y_2}e^{-sa_2^2Y_2})\prod_{i=3}^nE(e^{-sa_i^2Y_i})ds$$$$=\frac{1}{\pi}\int_0^{\infty}\frac{1}{(1+a_1^2s)^{1/2} }\frac{1}{(1+a_2^2s)^{1/2} }\times \prod_{i=1}^n\frac{1}{(1+a_i^2 s)^{1/2}}ds$$ In other terms if you consider the matrices $M=(m_{ij})$ with $m_{ii}=1/2$ and $m_{ij}=1/\pi$ for $i\neq j$ and $$D(s)=\mathrm{diag}\left(\frac{a_1}{(1+sa_1^2)^{1/2}},\ldots ,\frac{a_n}{(1+sa_n^2)^{1/2}}\right)$$ you have $$E\left(\frac{\xi\otimes \xi}{\|\xi^2\|}\right)=\int_0^{\infty}D(s)MD(s)\times \prod_{i=1}^n\frac{1}{(1+a_i^2 s)^{1/2}}ds$$