I was studying the expected value of the outer product of a normalized non-centered Gaussian vector and I found this very interesting and solved question and I am looking to generalize to a Student t vector.
The set-up is the following: let $X\sim t_\nu (\mu, \sigma^2 I_d)$. I want to study an high-dimensionality approximation of $$\mathbb{E}\left(\frac{XX^\top}{X^\top X}\right).$$
The Gaussian case becomes $$\mathbb{E}\left(\frac{XX^\top}{X^\top X}\right) = \frac{1}{d}\left(I_d + \frac{\mu\mu^\top}{\sigma^2}\right)+O\left(\frac{1}{d^2}\right).$$
This is my approach to the Student t case:
If $X\sim t_\nu (\mu, \sigma^2 I_d)$, we can write it as $$ X = {\mu} + \frac{\sigma Z}{\sqrt{S / \nu}}, \quad \text{with } Z \sim \mathcal{N}(0, I_d),\; S \sim \chi^2_\nu \text{ independent}. $$ Let $ \tau := \frac{\sigma}{\sqrt{S/\nu}} $, so $ X = {\mu} + \tau Z $, and: $$ \frac{X X^\top}{X^\top X} = \frac{({\mu} + \tau Z)({\mu} + \tau Z)^\top}{\|{\mu} + \tau Z\|^2}. $$ Conditioning on $ S $, we have $ X \mid S \sim \mathcal{N}({\mu}, \tau^2 I_d) $. From the Gaussian case, for large $ d$: $$ \mathbb{E}\left[ \frac{X X^\top}{X^\top X} \, \middle| \, S \right] = \frac{1}{d} \left( I + \frac{{\mu} {\mu}^\top}{\tau^2} \right) + \mathcal{O}\left(\frac{1}{d^2}\right). $$ Since $ \tau^2 = \frac{\sigma^2 \nu}{S} $, we write: $$ \mathbb{E}\left[ \frac{X X^\top}{X^\top X} \right] = \mathbb{E}_S\left[ \frac{1}{d} \left( I + \frac{S}{\sigma^2 \nu} {\mu} {\mu}^\top \right) \right] + \mathcal{O}\left(\frac{1}{d^2}\right). $$ Using $ \mathbb{E}[S] = \nu $ for $ S \sim \chi^2_\nu $, we get: $$ \mathbb{E}\left[ \frac{X X^\top}{X^\top X} \right] = \frac{1}{d} \left( I + \frac{1}{\sigma^2} {\mu} {\mu}^\top \right) + \mathcal{O}\left(\frac{1}{d^2}\right). $$
Which seems a bit odd, especially since it does not depend on $\nu$. Do you see any flaws with this proof?
