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I have been using linear shrinkage to better estimate the covariance matrix when I do not have enough data. Let $\bf X$ be a $T\times n$ matrix representing the data (previously centered) and $${\bf S}={1\over{T}}{\bf X'X} $$ be the sample covariance of the data.

Now, let $\bf \hat\Sigma$ be the covariance estimated by linear shrinkage (proposed by Ledoit-Wolf). My question : is there a way to "retrieve" the implied data $\bf Y$ ( $T\times n$ ) such that its sample covariance will be equal to $\bf \hat\Sigma$, .i.e. such that $$ {\bf \hat\Sigma} ={1\over{T}}{\bf Y'Y} $$ Is there a guarantee that there exists such data matrix $\bf Y$ ? If so, how do we find the data that is the closest to $\bf X$ using some distance function ? Can somebody direct me to some papers addressing this question ? I asked specifically about the linear shrinkage estimator, but the question could be asked for any estimator of the covariance matrix. Thanks

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  • $\begingroup$ Some things about this question aren't quite right. First, $S$ is not the covariance matrix because the columns have not been centered, so are you actually asking about this sums-of-products matrix or do you truly intend to ask about the covariance matrix? Second, could you clearly indicate what "estimated by linear shrinkage" is? There may be several ways to interpret that. Third, could you articulate some sense of "closest"? How would you measure or quantify that? $\endgroup$
    – whuber ♦
    Commented Jan 13, 2021 at 15:47
  • $\begingroup$ Thanks for the comment, I will clarify the question $\endgroup$
    – quantguy
    Commented Jan 13, 2021 at 15:54
  • $\begingroup$ Thank you. Doesn't this question amount to asking "given a valid covariance matrix and a dataset size $T,$ do there exist data of length $T$ for which it is the (empirical) covariance matrix?" $\endgroup$
    – whuber ♦
    Commented Jan 13, 2021 at 17:02
  • $\begingroup$ Not exactly, we have some information on the process to obtain the "valid" covariance matrix : the data $\bf X$ used to obtain it. The question relates to the existence of $\bf Y$, but also how to obtain it so that it is as close as possible to $\bf X$. Do you have any ideas on how to do this ? $\endgroup$
    – quantguy
    Commented Jan 13, 2021 at 18:19
  • $\begingroup$ That sounds interesting, but first you need to explain what you mean by "as close as possible:" what metric do you have in mind? A potential problem is that typically the columns of a data matrix measure incommensurable things--height, weight, counts, etc.--but "close" means you have to condense all those differences into a single number. How? $\endgroup$
    – whuber ♦
    Commented Jan 13, 2021 at 19:13

1 Answer 1

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So there definitely is a solution

create y by rescaling X by $\sqrt \delta$ and then adding $\sqrt(1-\delta)$ independent noise with appropriate covariance structure

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