Suppose I know or have estimated the covariance matrix for one random variable (for example an asset) and have: $$ \begin{bmatrix} <\text{spot, spot}> & <\text{atmv, spot}> \\ <\text{spot, atmv}> & <\text{atmv, atmv}> \end{bmatrix} $$

where atmv is the at the money volatility (or can just be realized). Suppose then I know the beta or correlation of this asset A to asset B. How would you derive the covariance matrix B as a function of beta and covariance matrix for A?

  • $\begingroup$ you say the covariance matrix for one random variable but you need to view atmv as another random variable. So, you actually have three random variables. asset A, asset B and atmv. So, when you say covariance matrix of B, what is the random variable that B is correlated with ? $\endgroup$
    – mark leeds
    Commented Jul 14, 2023 at 18:41

1 Answer 1


$return(b)=cor_{A,B}*return(A)+noise_1$ so


so the correlation between b and atm vol depends on the correlation assumption between atmvol and noise_1. So there's no right answer if you have to go through A. Why don't you just directly correlate atmvol and asset B?

Note $cor_{A,B}$ is the correlation multiplied by the ratio of standard deviations of A and B (usual relationship between regression beta and correlation coefficient)

  • 2
    $\begingroup$ Arshdeep: I think you meant to write $\beta_{A,B}$ whereever you have written $cor_{A,B}$ above. Still, it's an insightful answer. Thanks. $\endgroup$
    – mark leeds
    Commented Jul 15, 2023 at 1:33
  • $\begingroup$ @markleeds yes I've clarified the notation in my comment, I was just being lazy and thought the point would be apparent either way. $\endgroup$
    – Arshdeep
    Commented Jul 15, 2023 at 1:43
  • 1
    $\begingroup$ No problem. But, in the first two formulae, it should be changed also. $\endgroup$
    – mark leeds
    Commented Jul 15, 2023 at 1:53
  • 1
    $\begingroup$ Even if the notation has been clarified, denoting beta as cor is unnecessarily confusing. $\endgroup$ Commented Jul 15, 2023 at 6:11
  • $\begingroup$ Whats the issue with simply assuming E(noise_1) = E(noise_2) = 0? $\endgroup$
    – roz
    Commented Jul 17, 2023 at 15:32

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