Deriving Beta From Covariance

Question

I am having trouble understanding an equation in a note I saw. The note has a covariance matrix and it uses this matrix to derive betas.

I have a covar matrix:

   a  b  c
a  1  2  3
b  2  3  5
c  3  5  6

And lets hypothetically assume the variances are 1,3,6 for a,b,c respectively. Now standard beta of two variables is beta(r,m) = cov(r,m)/var(m). So in the above, the beta(a,b) = 2 / 3.

My question is how do you calculate beta(a,a-b)?

From my notes, it says that the beta is calculated as beta(a,a-b) = [ Cov(a,a) - Cov(a,b) ] / Var(a-b)

Is the above correct? If so is there a proof or a website for more information on this? I'm a noob in statistics.

thanks

Answer 1

$$\beta(a, a-b) = \dfrac{\sigma_{aa}-\sigma_{ab}}{\sqrt{\sigma_{aa}(\sigma_{aa}+\sigma_{bb}-2\sigma_{ab})}}$$ where $\sigma$ refers to the covariances.

Answer 2

When you find the $\beta$, you usually find the beta with reference to some benchmark. You wouldn't find $a$ relative to $(b-a)$ It would make more sense to find $(b-a)$ relative to $a.$ And then you would want to find the beta of everything relative to $a.$

Suppose: $A = \sigma_a Z_1$, where $Z_1$ is a standard normal variable.

$B = m Z_1 + n Z_2$, with $Z_2$ as an independent standard normal variable.

$\sigma_b^2 = m^2 + n^2$

$Cov(a,b) = \sigma_a m\\ \rho_{a,b} = \frac {\sigma_a p}{\sigma_a \sigma_b} = \frac {m}{\sigma_b}\\ \beta{b,a} = \frac {\sigma_b}{\sigma_a} \rho_{a,b} = \frac {m}{\sigma_a}\\ B = \beta_{b,a} A + n Z_2 $

$A-B = (1 - \beta_{b,a}) A - qZ_2$, Since $Z_2$ is an independent random variable we can just as easily say:

$A-B = (1 - \beta_{b,a}) A + qZ_2$

$\beta {a-b,a} = 1-\beta_{b,a}$