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Why do you have to make a correlation matrix when calculating the parametric value at risk, if one of the assumptions for this method to work is that the assets of the portfolio must be independently distributed (i.e. their correlation must be equal 0)? Furthermore, $Var(X+Y) = Var(X) + Var(Y) + 2\cdot Cov(X,Y)$ is used when expanding the variance to get the portfolio variance, but this can only be used also if $X$ and $Y$ are not independent.

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Parametric simply means that a set of parameters govern the nature of the (joint) probability distribution of assets, some of those parameters being the correlations. It is not true in general to state that a parametric VaR model has cross-correlation of assets as zero.

I have never used a model that specifically precludes correlations. But if you defined one as such then your equations would be reduced as you state, but it is a very stringent assumption.

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  • $\begingroup$ As per Linda Allen's book "under the model's normality assumption, returns at any horizon are always independent and normally distributed" $\endgroup$ – matt_zarro Sep 12 '18 at 12:44
  • $\begingroup$ I searched her book for that quote. It occurs in the first few pages of the introduction in section 1.1.3. It does not refer to correlation with other assets but is actually referencing auto-correlation (the correlation of the asset movement with its own movement in a different time period), and claiming rightly the assumption of the model requires stationarity (i.e. zero correlation). However, it is phrased poorly and I can understand why it leads to confusion. $\endgroup$ – Attack68 Sep 12 '18 at 13:13
  • $\begingroup$ What about the var(x+y) formula part? Isn't that formula only applicable when the variables are not independent (and the assumptions claim that they must be independent)? $\endgroup$ – matt_zarro Sep 12 '18 at 13:29
  • $\begingroup$ Look at section 2.3, which discusses Returns Aggregations and specifically mentions the variance-covariance method and estimation of correlations. $\endgroup$ – Attack68 Sep 12 '18 at 13:37
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Under your assumptions the Cov(X,Y) expression is zero. The equation is still valid, just its result is determined by the Var(...) terms.

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