Let $x\_1,\dots,x\_n$ be i.i.d. $N(0,I\_{p\times p})$, with $n>p$. Let $\hat S=\frac1n\sum\_{i=1}^n x\_i x\_i^T$ be the sample covariance.


Assume the asymptotic setting where $\frac pn\to \alpha<1$.


Is there a result about the concentration of $\mathbb{E}\left[\operatorname{tr}\left(\hat{S}^{-1}\right)\right]$ in the asymptotic setting? By using Lemma 3.2 in [this paper](http://arxiv.org/pdf/1401.3424v4.pdf) coupled with the Sherman-Morrison formula I believe that $\mathbb{E}\left[\operatorname{tr}\left(\hat{S}^{-1}\right)\right]=\frac{1}{1-\alpha}$, but I haven't been able to make it rigorous.


Are there any existing results about this?



Trace of the inverse sample covariance as the number of samples and dimension scale to infinity
Writing $\hat{S}= \frac1n XX^T$ where $X$ has columns $x\_1,\dots,x\_n$, we can express 
$$
\text{tr}(\hat{S}^{-1}) = \sum\_{i=1}^p \lambda\_i(\hat{S})^{-1} = n\sum\_{i=1}^p \sigma\_i(X)^{-2}
$$
where $\sigma\_1(X), \dots, \sigma\_p(X)$ are the nonzero singular values of $X$. Then we can use the inverse second moment identity (observed by Tao and Vu here <http://arxiv.org/pdf/0807.4898v5.pdf>; the rectangular case is given as Lemma 4.14 here: <https://arxiv.org/pdf/1109.3343v4.pdf>) to express 
$$\sum\_{i=1}^p \sigma\_i(X)^{-2}=\sum\_{i=1}^p \text{dist}(R\_i,R\_{-i})^{-2}$$
where $\text{dist}(R\_i,R\_{-i})$ is the distance from the $i$th row of $X$ to the span of the remaining rows. Since $X$ has normally distributed entries the variables $\text{dist}(R\_i,R\_{-i})^{2}$ have chi-squared distribution, so the expectation of this sum can be computed explicitly. Unless I have made some mistake it comes out to $\mathbb{E}\frac1p\text{tr}(\hat{S}^{-1}) = \frac{n}{n-p-1} = \frac{1}{1-\alpha}+o(1)$ (note we need the normalization $1/p$).


