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Let $x_1,\ldots,x_n$ be drawn iid from such "nice" distribution on $\mathbb R^p$ (but possibly very general!), and let $X$ be the $n$-by-$p$ matrix formed by vertically stacking the $x_i$'s.

Question. What is the limiting distribution of the "scatter matrix" $\frac{1}{n}XX^T:=\frac{1}{n}\sum_{i=1}^nx_ix_i^T$ as $n \rightarrow \infty$ ?

Observations

  • If the $x_i$'s are from a centered multivariate Gaussian, then $XX^T$ has a Wishart distribution.
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$\newcommand{\tr}{\operatorname{tr}}$ Let $x=[x^1,\dots,x^p]^T:=x_1$, $y:=xx^T$, $\mu:=Ey$, $w:=y-\mu$, and $s:=\sum_1^n x_ix_i^T$. Then, by the appropriate laws of large numbers, $s/n\to Ey$ almost surely and hence in probability and in distribution, provided that $Ey$ exists in $\mathbb R^{p\times p}$. Everywhere here, the convergence is for $n\to\infty$.

Assume now that $E|x|^4<\infty$, where $|x|$ is the Euclidean norm of $x$. Note that $y$ is a random matrix in the vector space $\mathbb R^{p\times p}$, which is naturally endowed by the inner product $A\cdot B:=\tr AB^T$ for $A$ and $B$ in $\mathbb R^{p\times p}$. With respect to this inner product, the covariance operator (say $R$) of the random matrix $y$ is given by the formula $$(Rf)_{k,l}=\sum_{i,j=1}^pf_{i,j}\operatorname{Cov}(x^ix^j,x^kx^l) $$ for $f=(f_{i,j})\in\mathbb R^{p\times p}$ and $k,l=1,\dots,p$. So, by the multivariate central limit theorem, $(s-n\mu)/\sqrt n$ converges in distribution to a zero-mean Gaussian random matrix in $\mathbb R^{p\times p}$ with covariance operator $R$.

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  • $\begingroup$ Thanks for the response. A (perhaps) naive question though, in case the $x_i$'s are draw from a centered multivariate Gaussian, we know that $s$ has a Wishart distribution. Reasoning along your lines, is it ok that the translate $(s-n\mu)/\sqrt{n}$ converges to a Gaussian, as this would immediately imply $s/\sqrt{n}$ also converges to a Gaussian ? This sounds weird. $\endgroup$
    – dohmatob
    Commented Jul 2, 2019 at 12:11
  • $\begingroup$ If the $x_i$'s are non-degenerate Gaussian, then $\mu$ is a nonzero (positive-semidefinte) matrix, and so, the convergence of $(s-n\mu)/\sqrt n$ implies that $s/\sqrt n$ does not converge. $\endgroup$
    – Iosif Pinelis
    Commented Jul 2, 2019 at 12:20
  • $\begingroup$ Oops, that was an oversight on my part. Indeed, the difference is $\sqrt{n}\mu$ which diverges (except when $\mu=0$). Thanks! $\endgroup$
    – dohmatob
    Commented Jul 2, 2019 at 12:35
  • $\begingroup$ Any obvious conditions which would imply the convergence is to a Ginibre (i.e matrices with iid Gaussian entries) ? $\endgroup$
    – dohmatob
    Commented Jul 2, 2019 at 12:38
  • $\begingroup$ @dohmatob : Concerning your latest comment: That will never be the case if $Var(x^i x^j)\ne0$ for some $i,j$ such that $i\ne j$. Indeed, then $(i,j)\ne(j,i)$, whereas $Cov(x^ix^j,x^jx^i)=Var(x^i x^j)\ne0$, and so, the covariance operator $R$ is not a multiple of the identity operator and hence cannot be the covariance operator of a Gaussian matrix with iid entries. $\endgroup$
    – Iosif Pinelis
    Commented Jul 2, 2019 at 12:52

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