# Proof for non-positive semi-definite covariance matrix estimator

It is well known that the standard estimator of the covariance matrix can lose the property of being positive-semidefinite if the number of variables (e.g. number of stocks) exceeds the number of observations (e.g. trading days). I think the matrix can become singular. I have a clear idea why (inspired by the geometry of the problem) but does anybody have a short but rigorous proof for this fact?

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Me and a colleague tried to prove along the following lines:Let $X = (X_t^j)$ where $t = 1,\ldots,T$ and $j=1,\ldots,N$ and $T$ denotes the number of observations and $N$ the number of assets. Assuming that $T<N$ and that the mean returns are subtracted we have $$\hat{\Sigma} = \frac{1}{T-1} X^T X$$ and the claim follows as $$rank(\hat{\Sigma}) \le \min(rank(X),rank(X^T)) \le T$$ which implies $rank(\hat{\Sigma}) \le T < N$. Thus $\hat{\Sigma}$ is singular. Something more intuitive but still rigorous would be interesting. –  Richard Jun 29 '12 at 15:54
This isn't guaranteed to generate a positive semi-definite matrix –  pyCthon Jul 2 '12 at 2:17
This is what I am saying. But an easier proof for it, that's the question. In fact the matrix is singular for sure (as its rank is too little). –  Richard Jul 3 '12 at 20:09

The standard estimator of the covariance matrix is: $$\widehat{ \mathrm{cov}}(X) = \frac 1 {n-1} \sum_{i=1}^n (X_i-\bar X)(X_i-\bar X)^T,$$ where $X_i$ is the column vector containing the $i$th observation of all the observables. Each summand is an outer product of a vector with itself, i.e., a square matrix having rank at most one. Therefore $$\mathrm{rk\;}\widehat{\mathrm{cov}}(X) \le n$$ and the matrix not only can be but is always singular if $$n \lt \dim X,$$ i.e., if the number of observations is less than the number of variables.
Edit: regarding positive-semidefiniteness: $\widehat{ \mathrm{cov}}(X)$ is always positive-semidefinite because it is Gramian, even if its rank is not full. It loses the property of being positive-definite if and only if it is singular.