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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$\begingroup$ 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. $\endgroup$– Richi WaJun 29, 2012 at 15:54
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$\begingroup$ This isn't guaranteed to generate a positive semi-definite matrix $\endgroup$– pyCthonJul 2, 2012 at 2:17
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$\begingroup$ 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). $\endgroup$– Richi WaJul 3, 2012 at 20:09
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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.
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$\begingroup$ I appreciate your answer - but isn't it quite similar to the calculation in the question. Of course you added valuable comments and details. What I am looking for is a more financial/applied answer. $\endgroup$– Richi WaJul 27, 2012 at 6:19
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$\begingroup$ Yes, it is the same as the calculation in your comment, just summed out entrywise for each observation, although I did arrive at the rank inequality slightly differently. So you are looking for a financial/applied sort of intuitive way to see this both rigorously and obviously.... $\endgroup$– JL344Jul 27, 2012 at 7:41
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$\begingroup$ Yes, this is what I mean - sorry that this didn't come out clearly. Something obvious and intuitive .. thank you! $\endgroup$– Richi WaJul 27, 2012 at 8:25
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$\begingroup$ A comment on your edit: this is important: I can have portfolios of zero variance but never ever negative variance ... that's the meaning in financial terms - right? $\endgroup$– Richi WaJul 28, 2012 at 12:06
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$\begingroup$ I think you're starting to get at something: in financial terms, if the number of observations is less than the number of securities whose returns are observed each time, then it is always possible (in retrospect, of course) to construct a portfolio that would have remained constant in value during that time, but, as you say, it can never have a negative variance, unless you can trade imaginary shares on the stock market! $\endgroup$– JL344Jul 29, 2012 at 0:39