5

By: bilinearity of covariance, independence of Brownian increments, and Itô's isometry, we obtain: $$\begin{align} & \text{Cov}\left(\int^{t_1}_0\sigma(t)dW_t,\int^{t_2}_0\sigma(t)dW_t\right) \\[6pt] & \qquad = \text{Cov}\left(\int^{t_1\wedge t_2}_0\sigma(t)dW_t,\int^{t_1\wedge t_2}_0\sigma(t)dW_t +\int^{t_1\vee t_2}_{t_1\wedge t_2}\sigma(t)...


3

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.


2

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 ...


2

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.


2

If $X$, $Y$, and $Z$ are real-valued random variables and $a$, $b$, $c$, $d$ are constant (i.e. non-random), then the following fact is a consequence of the definition of the covariance: $$cov\left(X, (aY+b)+(cZ+d)\right)=a\cdot cov\left(X,Y\right)+c\cdot cov\left(X,Z\right)$$ For your formula, set $b=d=0$, $X=Y=r_{value}$, $a=X_{value}$ and $c=X_{momentum}$...


1

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


1

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 ...


1

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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