Okay, So this is the whole question. It's a bit far. But i just have one question about it:

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So in question c) we have the equation for the sharpe ratio as: $$\frac{E(w_1R_{1A} + w_2R_{2A})}{\sigma_c}$$


$$\sigma_C = w_1^2 var(\beta_1 R_{1t}) + w_2^2 var(\beta_2 R_{2t}) + 2w_1 w_2 Cov(R_1R_2)$$

We have the $2cov$ here, but in e) question. The right answer is that they not have the $2cov$ here. Otherwise it is the same equation.




1 Answer 1


Because as your covariance-variance matrix shows the covariance is 0 between them.

The question e asks you to use the “true” covariance matrix from d presumably.

It’s not that you exclude it, that’s sort of misinterpretation, it is just zero ($cov(R_1,R_2)=0 \implies 2w_1 w_2 Cov(R_1R_2) =0$) and per convention zero is usually dropped from equations.

  • $\begingroup$ @1mflon1 thank you :* But how did you see that $cov(R1, R2) =0$? $\endgroup$
    – soetirl13
    Commented Nov 20, 2018 at 11:40
  • $\begingroup$ @soetirl13 from that table in Qd each entry in the matrix gives you covariance. The first entry between r1 and r1 itself second between r1 and r2 and so on. You can see that between r1 and r2 it’s 0. Also note that “covariance” of variable with itself is just variance hence the name variance covariance matrix. $\endgroup$
    – 1muflon1
    Commented Nov 20, 2018 at 13:32

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