In this question solutions the following calculation is used. Var(R)=0.8*{E^2(R1)+Var(R1)}+0.2[E^2[R2]+Var[R2]}-E^2[R] Why are we able to do this? Is it just because the variance of R is the variance of R1 80% of the time and it is the Variance of R2 20% of the time and we are not summing the 2 distributions R1 and R2 at the same time? Why can't we use 0.8*Var[R1]+0.2*Var[R2] in the above situation? If R=0.8R1+0.2R2 as in both R1 and R2 would be incorporated in every instance of R then the above would not work I don't think. We would have to use Var[R]=0.8^2Var[R1]+0.2^2Var[R2]+2*0.2*0.8*cov[R1,R2]? Thank you
R is not a weighted sum of the returns R1 and R2, so Var(R) =/= Var(0.8*R1 + 0.2*R2). We can capture that probabilistic behaviour of R when looking at its expectations E[R] and E[R^2]. It's from these that the variance is derived.
