3
$\begingroup$

I have a portfolio of 4 assets. I also have returns for 3 indices. I want to get the multivariate betas for these 4 assets-based on these assets. I only have the 7 x 7 covariance matrix estimated by a Exponential Weighted Moving Average Model (EWMA). How would I get these multivariate betas? I know from a univariate percpective I would use the covarience(Asset1Return,Index1Return)/Var(Index1) but this only gives the uni-variate beta.

$\endgroup$

2 Answers 2

1
$\begingroup$

With this solution you have to split your covariance matrix somewhat, but it should give you a vector with betas based on you conditional covariances.

Example with two indexes, $x1$ and $x2$, and one asset $y$.

$$[\sigma_{y,x1}, \sigma_{y, x2}]\begin{bmatrix} \sigma_{x1}^2 & \sigma_{x1,x2} \\ \sigma_{x1,x2} & \sigma_{x2}^2 \end{bmatrix}^{-1}$$

$\endgroup$
1
$\begingroup$

Say that you did the calculations in the classic regression way. If you stick the returns of your 4 asset returns in a $(T\times 4)$ matrix $Y$, and your 3 factor returns in a $(T\times 3)$ matrix $X$, then your betas would solve the multiple regressions, collected in a $(3\times 4)$ matrix $$Y = X\cdot \beta + \epsilon$$ You could also add a column of ones in $X$ if you want to also have a constant, but this does not matter. The OLS solution would be $$\beta = (X'X)^{-1}(X'Y)$$

You can see the analogy to the univariate cov/var formula that you describe. The $X'Y$ part corresponds to the $(3\times 4)$ asset-factor covariances, while the $X'X$ parts corresponds to the $(3\times 3)$ factor-factor covariances.

Therefore I would say that your large $(7\times 7)$ EWMA covariance matrix can be partitioned as $$\left( \begin{array}{cc} X'X & X'Y \\ Y'X & Y'Y \end{array} \right) $$ From which you can pick the right elements to calculate the betas.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.