Expectation of $\frac{\mathbf{x} \mathbf{x}^{\rm{T}}}{\| \mathbf{x} \|^{2}}$ for $\mathbf{x} \sim \mathcal{N}(\boldsymbol{\mu},\boldsymbol{\Sigma})$

Question

Let $\mathbf{x} \sim \mathcal{N}(\boldsymbol{\mu},\boldsymbol{\Sigma})$. I would like to compute the following expectation:

$$\mathbb{E} \bigg[ \frac{\mathbf{x} \mathbf{x}^{\mathrm{T}}}{\| \mathbf{x} \|^{2}} \bigg] = \mathbb{E} \bigg[ \frac{\mathbf{X}}{\mathrm{tr}(\mathbf{X})} \bigg]$$

where I have defined $\mathbf{X} = \mathbf{x} \mathbf{x}^{\mathrm{T}}$. It is know that $\mathbb{E}[\mathbf{x} \mathbf{x}^{\mathrm{T}}] = \mathbb{E}[\mathbf{X}] = \boldsymbol{\mu} \boldsymbol{\mu}^{\mathrm{T}} + \boldsymbol{\Sigma}$, but this does not help much since numerator and denominator cannot be considered separately.

I would expect that many people have encountered this problem before, but its solution seems far from straightforward.

Answer 1

$||x||^2=x^tx=x_1^2+x_2^2+\cdots+x_n^2$. Note that the $i^{th}$ diagonal element of $xx^t$ is b$x_i^2$.