# Tag Info

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Multivariate volatility models for replacing the sample covariance matrix with in the mean-variance portfolio selection model: RiskMetrics 1996 EWMA (Exponentially weighted moving average) covariance matrix RiskMetrics 2006 EWMA covariance matrix Multivariate DCC-GARCH covariance matrix Jon Danielsson "Financial risk forecasting" has EWMA and GARCH for R ...

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You will have to add some constraints to get the weight vector of the eigen vector of the smallest eigen values, otherwise 0 is a trivial solution. Without going in the details of handling those extra constraints, the reason why the vector space associated with the smallest eigen value is relevant is because if you express variance of your portfolio in the ...

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If you take $A^T∗COV∗B$ then the result will be 1 x1 ( a scalar). (1xN * NxN * Nx1 = 1x1). I believe you forgot to take the transpose of A. The vector which pre-multiplies COV needs to be a row vector, because in your example it isn't you may be getting this weird result.

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Ledoit and Wolf have a new paper ( November 2018 ) called "Analytical Nonlinear Shrinkage of Large-Dimensional Covariance Matrices" which has MATLAB code for the procedure at the end of the paper. The paper can be downloaded at SSRN.

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As an addition to the already rich answers, I would suggest you to read the following paper by Marcos L. De Prado on the computation of Forward-Looking Correlation Matrices. Estimation of Theory-Implied Correlation Matrices https://papers.ssrn.com/sol3/papers.cfm?abstract_id=3484152

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The Ledoit-Wolf estimate cited by @develarist can be quite good, but as you say you already knew about "shrinking". It takes the population of correlations observed as an effective Bayesian prior for any given correlation, so it sort of inherently assumes that all pairs are similar an some sense. It would not work well, say, with known block sets of highly ...

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Quantile regression is considered a robust procedure but lacks the quality of being fully differentiable. There are also regularized regression models like ridge regression, lasso regression and elastic net regression that implicitly consider the covariance of the data like OLS, but additionally reduce volatility in estimates through the introduction of bias....

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A first hint: To convert Kendall's $\tau$ to the Pearson correlation coefficient $\rho$, one could use the relationship: $$\rho = \sin\Bigl(\frac{\pi}{2}\tau\Bigr)$$ But keep in mind that this only holds for the bivariate normal copula assumption, I don't know if this also holds to convert "plain vanilla" correlation matrices, see McNeil et al. (2005: ...

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I have actually considered the problem that you are working on, though configured somewhat differently. There isn't going to be a universal answer to your question. See, in particular, Holland, Paul W. Covariance Stabilizing Transformations. Ann. Statist. 1 (1973), no. 1, 84--92. Nonetheless, there are answers, some already mentioned. I would argue ...

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This is not a complete answer, more a different perspective to the answers already given. If you have some a-priori knowledge about the covariance structure and about the factors influencing it, you should try to reflect this in your statistical model. Three ideas: Divide your sample into subpopulations with identical factor values and estimate separately. ...

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In theory, the Ledoit and Wolf shrinkage estimator is supposed to guarantee a positive-definite matrix, given that it adds a positive-definite matrix (the target) to a semi-positive one (the sample covariance). I can see four reasons why you didn't get a positive-definite matrix: Your true covariance is effectively not full rank, i..e you have perfect ...

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You have to use a multivariate Garch indeed. Search for mGARCH versions like GARCH-BEKK or VECH GARCH or DCC.

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This is a common occurrence in monte carlo models. I suggest you look into Cholesky decomposition. The basic idea is that if you have a covariance matrix M that describes the relationship between your data, the cholesky decomposition will produce a lower triangular matrix L such that M = L * L' Now if you generate a vector X of random normals you can take L ...

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The covariance operator is linear and since $X$ and $Y$ are independent any covariance between them is zero. Therefore, the only part that matters is the X-part of the covariance: $Cov[X, Z] = \rho Cov[X, X] + Cov[..., ...Y] = \rho Var[X] = \rho$.

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as per definition $E[Y]=E[X]=0$, and also $\mathit{Var}(X)=\mathit{Var}(Y)=1$. \$\mathit{Var}(Z)=\mathit{Var}(\rho X+\sqrt{1-\rho^2}Y)=E\big[\big(\rho X+\sqrt{1-\rho^2}Y-\rho \langle X \rangle+\sqrt{1-\rho^2} \langle Y \rangle \big)^2\big]=E\big[\big(\rho X+\sqrt{1-\rho^2}Y \big)^2\big]=E\big[\rho^2 X^2+2\rho \sqrt{1-\rho^2}XY+(1-\rho^2)Y^2\big]=\rho^2E\big[...

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