We know that we have the following formula $d(XY)=XdY+YdX+d[X,Y]$, but how does this look when we have two actual processes?
Should it look something like this?
$d(XY)=XdY+YdX+d[X,Y]=X(\mu_{Y}dt+\sigma_{Y}dW)+Y(\mu_{X}dt+\sigma_{X}dW)+\sigma_{X}\sigma_{Y}dt=X \mu_{Y}+Y \mu_{X}+\sigma_{X}\sigma_{Y}dt +(\sigma_{Y}+\sigma_{X})dW=(\mu_{X}dt+\sigma_{X}dW )\mu_{Y}+(\mu_{Y}dt+\sigma_{Y}dW) \mu_{X}+\sigma_{X}\sigma_{Y}dt +(\sigma_{Y}+\sigma_{X})dW=$
$(\mu_{X}+\mu_{Y}+\sigma_{X}\sigma_{Y})dt+(\sigma_{X}\mu_{Y}+\sigma_{Y}\mu_{X}+\sigma_{Y}+\sigma_{X})dW$
where I used the definition of integration with respect to a semimartingale and the formula for the crossvariation of two Ito processes. Is the above equalites true for some hypothesis on the processes?
To be more general, let us suppose that we have two Ito process $X$ and $Y$ (respecting the usual conditions). Their dynamics is defined as follows: \begin{align} dX_t &= \mu_X(t,X_t)dt + \sigma_X(t,X_t)dW_t^X \\ dY_t &= \mu_Y(t,Y_t)dt + \sigma_Y(t,Y_t)dW_t^Y \end{align} where $(W^X,W^Y)$ are standard Brownians with correlation $d\langle{W_.^X,W_.^Y}\rangle_t=\rho_{XY}dt$. Applying the integration by parts, we have: \begin{align} d(X_tY_t) &= Y_tdX_t + X_tdY_t + d\langle{X_.,Y_.}\rangle_t \\ &= Y_t\left[\mu_X(t,X_t)dt + \sigma_X(t,X_t)dW_t^X\right] + X_t\left[\mu_Y(t,Y_t)dt + \sigma_Y(t,Y_t)dW_t^Y\right] + \sigma_X(t,X_t)\sigma_Y(t,Y_t)d\langle{W_.^X,W_.^Y}\rangle_t \\ &= Y_t\left[\mu_X(t,X_t)dt + \sigma_X(t,X_t)dW_t^X\right] + X_t\left[\mu_Y(t,Y_t)dt + \sigma_Y(t,Y_t)dW_t^Y\right] + \\&\sigma_X(t,X_t)\sigma_Y(t,Y_t)\rho_{XY}dt\\ \end{align} As far I know, that's all we can say.