Let us consider a simple equity portfolio that has exposures to only two factors: 0.5 exposure to value and 0.8 exposure to momentum. Let us assume that the volatilities of the two factors are 3% for value and 5% for momentum and the correlation between them is 0.2.

I do not understand this calculation, can anyone explain this formula: $$cov(r_{value},r_p) = cov(r_{value}, X_{value}r_{value}+X_{momentum}r_{momentum}=$$ $$X_{value}\sigma_{value}^2+X_{momentum}\sigma_{momentum}\sigma_{value}\rho=0.5\cdot0.03^2+0.8\cdot0.05\cdot0.03\cdot0.2=0.00069$$

I thought I would need the correlation between factor value and the portfolio in order to to find the covariance because $cor(\mathrm{value}, \mathrm{portfolio}) = \frac{cov( \mathrm{value}, \mathrm{portfolio})}{\sigma_{\mathrm{value}}\sigma_{\mathrm{portfolio}}}$?


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 $c=X_{momentum}$. This straightforward leads to your stated formula, after applying the general statement that $cov(X,Y)=\rho \sigma_X \sigma_Y$.

  • $\begingroup$ I'm still a bit lost. I understand that X is the variance of the value factor. I also understand that aY is the exposure to the value factor times the variance of the value factor. And that cZ is the exposure to the momentum factor times the variance of the momentum factor. Therefore the second term (after the comma) is the total variance of the portfolio. Therefore we assess the covariance of the value factor with the total portfolio variance (exactly what I needed). What I don't understand is how do you go from the left side of the "=" to the right side (extracting a and c)? $\endgroup$ – tweedi Nov 9 '18 at 17:05
  • $\begingroup$ This just follows from the definition of the covariance: $cov(aX,Y)=a \cdot cov(X,Y)$. $\endgroup$ – skoestlmeier Nov 9 '18 at 17:09

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