Suppose I have two processes.
$A_t = A_0 \exp((a-\frac{1}{2}\sigma_A^2)t+\sigma_A W_t^A$
$B_t = B_0 \exp((b-\frac{1}{2}\sigma_B^2)t+\sigma_B W_t^B$
I would like to calculate $E[A_s B_t]$ where s < t.
Attempt:
I can rewrite $B_t$ as $B_t = B_s \exp((b-\frac{1}{2}\sigma_B^2)(t-s) + \sigma_B W_{t-s}^B)$. Then, $A_sB_t = A_sB_s\exp((b-\frac{1}{2}\sigma_B^2)(t-s) + \sigma_B W_{t-s}^B)$.
Thus, $E(A_s B_t) = E(A_s B_s)E(\exp((b-\frac{1}{2}\sigma_B^2)(t-s) + \sigma_B W_{t-s}^B))$
$= A_0 B_0 \exp((a+b+\rho\sigma_A \sigma_B)s)+\sigma(b(t-s))$.
Could anyone confirm whether this approach is correct. How would I then be able do compute $E(A_s B_t B_r)$ where $ s < t < r$. Would I simply do the same thing but use
$B_r = B_s\exp((b-\frac{1}{2}\sigma_B^2)(r-t)+\sigma_BW_{r-t}^B)\exp((b-\frac{1}{2}\sigma_B^2)(t-s)+\sigma_BW_{t-s}^B)\exp((b-\frac{1}{2}\sigma_B^2)s+\sigma_BW_{s}^B)$
My second part of this question relates to the forward. Assume $s<t<r$ and $F_{t,r}^A = E(A_r|A_t)$. How would I be able to compute $E(B_r F_{t,r}^A )$? Am I allowed to do $E(B_r)E( F_{t,r}^A )$ even though the time periods overlap because I am working with a forward?