I cannot make sense of this:
Suppose X is a RV with mean $ \mu_i $ and variance $ \sigma_i^2$. Consider a sequence of $n$ random variables X; the variance of the sequence is:
$ Var(S) = \sum_i^{n} \sum_j^{n} \sigma_i\sigma_j E[z_iz_j] $
where $ z_i = \frac{X_i - \mu_i}{\sigma_i}$.
I do not understand why $ \sum_i^{n} \sum_j^{n} \sigma_i\sigma_j $ is multiplied by the second term $E[z_iz_j]$. The term $ \sum_i^{n} \sum_j^{n} \sigma_i\sigma_j $ already includes the sum of the variances and all the combination of the covariances. I undestand that for $i=j$ we have $E[z_iz_j]=1$, but for $i \neq j $ don't we have the additional extra term $E[z_iz_j]$?
\begin{eqnarray} \operatorname{Var}(S)&=&E\left[\left(\sum_iX_i\right)^2\right]-E\left[\sum_iX_i\right]^2 \\ &=& \sum_{i,j}\left(E[X_iX_j]-E[X_i]E[X_j]\right) \\ &=& \sum_{i,j}\left(E[(\sigma_iz_i+\mu_i)(\sigma_jz_j+\mu_j)]-E[\sigma_iz_i+\mu_i]E[\sigma_jz_j+\mu_j]\right) \\ &=& \sum_{i,j}\sigma_i\sigma_jE[z_iz_j] \;, \end{eqnarray}
since everything else cancels.
It's not true that $\sum_{i,j}\sigma_i\sigma_j$ already includes the covariances. The $\sigma_i^2$ are only the variances of the individual variables; the factor $E[z_iz_j]$ takes into account the covariance.