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Hot answers tagged random-variables

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The original mean-variance model was static and assumed that the mean vector $\mu$ and covariance matrix $\Sigma$ are known. These determine the optimal portfolio weights that in this case are deterministic as well. However, in practice people do two types of modifications. First because these means and covariances are generally time-varying, we instead use ...

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They are independent. The point is that $y$ is derived from your easily sampled distribution $g$ randomly. Now you have a random test (via $v$) that decides whether to accept $y$ or not as part of the random sample of the harder to sample $f$. The procedure uses $M$ in the accept-reject method and whilst you can derive conservative estimates with $M$ quite ...

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@Wombat: I think it's best to think of it this way. Suppose you have a time series and the joint distribution of $n$ elements ( in a certain, specific order) of the series is normal with mean zero and some non-diagonal covariance matrix $\Sigma$. Then, in this case, exchangeability means that you could take the $n$ elements in any order and the joint ...

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let's take X1=1 and X2=2: The covariance matrix is zero (X1 and X2 are independents) and X1, X2 have different distributions!!!

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In the original Markowitz papers, no. In the so called 'resampled efficiency' or 'resampling frontier' method by Michaud, the weights are recalculated over and over from perturbed versions of the covariance matrix, to account for the fact that the covariance matrix is not known exactly (estimation error). In this case yes, the weights are random variables.

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There are ways that you might think of portfolio weights as estimates and thus random variables. If you are working with the optimizer, you may be able to get the inverse Hessian out of it. If so, that can be used to get you estimates of the standard errors of your portfolio weights. One big caveat though: any weight with a binding constraint will likely ...

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