How do they derive the last term

Question

Solution to question b) is :

What I don't understand is: How do they derive the last term? I know that $Cov(X_i,X_j)=σ_0^2$ for i different of j and $Cov(X_i,X_j)=σ_0^2+σ_1^2$ if i equals j. But then I'm not sure how they get the last term.. Any help that would let me understand it is appreciated

Answer 1

In terms of the Kronecker delta, $\operatorname{Cov}(X_i,\,X_j)=\sigma_0^2+\sigma_1^2\delta_{ij}$, so the total variance is$$\sigma_0^2\sum_{ij}w_iw_j+\sigma_1^2\sum_{ij}w_iw_j\delta_{ij}=\sigma_0^2\left(\sum_iw_i\right)^2+\sigma_1^2\sum_iw_i^2.$$This agrees with the claimed result because $\sum_iw_i=1$.