I was wondering how to prove/compute the differential of the product of two Brownian motions. I know how to do it in case they are independent as follows:
Suppose $dX_t= \mu_t dt +\sigma_t dW_t$ and $dY_t = v_t dt + \rho_t d \bar{W}_t$, where $W_t$ and $\bar{W}_t$ are independent, i.e. $(dW_t d\bar{W}_t= 0)$
Then I can prove $d(X_t Y_t)$ with the following trick:
I start with decomposition: $(X_t+Y_t)^2=X_t^2+Y_t^2+ 2X_t Y_t$,
Which leads by differentiation to $d(X_t Y_t)= \frac{1}{2}[d(\{X_t +Y_t\}^2) - d(X_t^2) - d(Y_t^2)]$
Then I apply Ito-lemma to all three parts separately as follows: $d(X_t^2)=(2\mu_tXt+\sigma_t^2)dt+ 2\sigma_t X_tdW_t= 2X_t dX_t + \sigma_t^2dt$ $d(Y_t^2)=(2v_t Y_t+\rho_t^2)dt+ 2\rho_t Y_t d\bar{W}_t= 2Y_t dY_t+\rho_t^2 dt$ $d((X_t+Y_t)^2) = 2(X_t+Y_t)dX_t+ 2(X_t+Y_t)dY_t + (\sigma_t^2+\rho_t^2)dt$
Combined I eventually find: $d(X_t Y_t) = Y_t dX_t + X_t dY_t$.
Now I was wondering how to do the prove in a similar fashion in the case of correlated brownian motions. Any help would be appreciated.
Well it's simpler to tackle it multidimensionnal Itô with $f(x,y)=x.y$, this gives you : $$d(XY)_t=f_{x}(X_t,Y_t)dX_t +f_{y}(X_t,Y_t)dY_t + \frac{1}{2} f_{xx}(X_t,Y_t)d<X>_t+\frac{1}{2}f_{yy}(X_t,Y_t)d<Y>_t +f_{xy}(X_t,Y_t)d<X,Y>_t$$
$$d(XY)_t=Y_t.dX_t +X_t.dY_t + \frac{1}{2}.0.d<X>_t+\frac{1}{2}.0.d<Y>_t +1.d<X,Y>_t$$ $$d(XY)_t=Y_t.dX_t +X_t.dY_t + d<X,Y>_t$$ Now $d<X,Y>_t=\sigma_t.\rho_td<B,\tilde{B}>_t=c.\sigma_t.\rho_t.dt$ where $d<B,\tilde{B}>=c.dt$ so that : $$d(XY)_t=Y_t.dX_t +X_t.dY_t + c.\sigma_t.\rho_t.dt$$