I have to show that, $$E \left[\frac{\Delta S_i}{S_i}\frac{\Delta S_j}{S_j} \right]=\left[\beta_i\beta_j+\sigma_i^2\delta_{ij}\right]\Delta t+\left[\mu_i\mu_j+\frac{1}{2} \beta_i^2\beta_j^2+\beta_i\beta_j(\mu_i+\mu_j)+\left(\frac{1}{2}(\tilde{\sigma}_i^4-\beta_i^4)+2\mu_i\sigma_i^2 \right)\delta_{ij} \right]\Delta t$$
I know that I have to use Ito's product rule and I think it is better to determine the product $\frac{d S_i}{S_i}\frac{d S_j}{S_j}$ first, then take the expectation and then take the discrete form.
For help the definitions:
$\frac{d S_i}{S_i}=\mu_idt+\beta_idz_0+\sigma_idz_i$ which implies $S_i=S_0\exp((\mu_i-\frac{1}{2}\tilde{\sigma}_i^2)t+\beta_idz_0+\sigma_idz_i)$ and
$\frac{\Delta S_i}{S_i}=\exp\big((\beta_i z_0+\sigma_i z_i)\Delta t^{1/2}+(\mu_i-\frac{1}{2}\tilde{\sigma}_i^2)\Delta t\big)-1$
another useful identity is $\sigma^2+\beta^2=\tilde{\sigma}^2$.
$z=(z_0,...,z_N)$ is an ($N$+1) dimensional standard brownian motion with $z(t)=\mathcal{N}(0,tI)$.
It would be very nice if someone could help me with the product rule at this special case.