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I am trying to compute the standard error of the sample spectral risk measure, which is used as a metric for portfolio risk. Briefly, a sample spectral risk measure is defined as $q = \sum_i w_i x_{(i)}$, where $x_{(i)}$ are the sample order statistics, and $w_i$ is a sequence of monotonically non-increasing non-negative weights that sum to $1$. I would like to compute the standard error of $q$ (preferrably not via bootstrap). I don't know much about L-estimators, but it looks to me like $q$ is a kind of L-estimator (but with extra restrictions imposed on the weights $w_i$), so this should probably be an easily solved problem.

edit: per @srikant's question, I should note that the weights $w_i$ are chosen a priori by the user, and should be considered independent from the samples $x$.

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  • $\begingroup$ What is the distribution of $x$? I suppose we can assume that the weights are known. $\endgroup$ – user28 Aug 27 '10 at 18:34
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    $\begingroup$ right, sorry, the weights are independent from $x$. The restrictions on the weights characterize, evidently, the class of 'coherent' risk measures. I have not seen any suggestions about how to set them, though, and this appears to be a black art. However, perhaps one would choose them to get the best discriminatory power (i.e. smallest standard error)... $\endgroup$ – shabbychef Aug 27 '10 at 18:56
  • $\begingroup$ @Glen_b: I am looking for a non-parametric estimator. I just found 2 papers by Stigler which, though very dense, appear to give a form of the asymptotic variance for the parametric case. I am now hunting down papers by Downton, Stillito, as referred to in Elamir & Seheult's papers on L-moment standard errors. $\endgroup$ – shabbychef Aug 27 '10 at 23:35
  • $\begingroup$ Oh, okay, you mean asymptotically. But you will still need certain assumptions about the distribution for the results to apply, and the results will depend on characteristics of the distribution that are unknown (such as the variance). $\endgroup$ – Glen_b Aug 28 '10 at 3:24
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As you know, from

$$Var[q] = Var[\sum_i w_i x_{(i)}] = \sum_i\sum_j w_i w_j Cov[x_{(i)}, x_{(j)}]$$

it follows you need only compute the variances and covariances of the order statistics. To do this, diagonalize the covariance matrix! Although this cannot be done in general, M. A. Stephens has obtained (heuristically) an asymptotic diagonalization. (The eigenvectors are Hermite polynomials.) In the spirit of PCA, limiting your calculations to the largest few eigenvalues can greatly reduce the computational effort and might produce a reasonable approximation, depending on the structure of the $w_i$. In fact, if you adjust that weight vector to be a linear combination of a small number of the eigenvectors, that will assure a simple calculation of $Var[q]$ and perhaps not cost you much in terms of the accuracy of $q$ itself. At worst, a preliminary eigendecomposition of $\vec{w}$ will then require only $O(N)$ calculations of the variance rather than $O(N^2)$.

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    $\begingroup$ as it turns out, for most weighting schemes, only the first few $w_i$ are non-zero, which should also simplify the computation. $\endgroup$ – shabbychef Nov 6 '10 at 19:54
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It looks like I am probably stuck with a bootstrap. One interesting possibility here is to compute the 'exact bootstrap covariance', as outlined by Hutson & Ernst. Presumably the bootstrap covariance gives a good estimate of the standard error, asymptotically. However, the approach of Hutson & Ernst requires computation of the covariance of each pair of order statistics, and so this method is quadratic in the number of samples. Maybe I should just stick with the bootstrap!

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