Let ($B_1$, $B_2$) be a bi-dimensional correlated Brownian motions Calculate the stochastic differential equation of the process $\cos(B_{1,t}B_{2,t})$.
Attempt:
Let $X_t$ be the stochastic process defined by $X_t = \cos(B_{1,t}B_{2,t})$.
Call $Y_t:= (B_{1,t}B_{2,t})$, so the process becomes $X_t := \cos(Y_t)$
We calculate the stochastic differential of $X_t$: $$dX_t = – \sin (Y_t)dY_t – \frac{1}{2} \cos (Y_t) d\langle Y_t \rangle$$
The stochastic differential for $Y_t$ is: $dY_t = B_{1,t} \space dB_{2,t} + B_{1,t} \space dB_{1,t} + t \space dt$
I don't get out of this exercise at all. Thanks for the tips.
Best Answer
I would advise that you apply the multidimensional version of Itô's formula directly. Set $f(x,y) = \cos (xy)$, so that the derivatives of interest are: $$\begin{align*} f_x(x,y) &= -y\sin (xy) \\ f_{xx}(x,y) &= - y^2 \cos(xy) \\ f_y(x,y) &= -x \sin (xy) \\ f_{yy}(x,y) &= -x^2 \cos (xy) \\ f_{xy}(x,y) &= -xy \cos(xy) \end{align*}$$ Itô's formula then gives us for $Z_t = \cos (B_t^{(1)} B_t^{(2)})$ that: $$ \begin{align*} dZ_t &= f_x(B_t^{(1)}, B_t^{(2)})dB_t^{(1)} + f_y(B_t^{(1)}, B_t^{(2)})dB_t^{(2)} + \frac{1}{2}\left( f_{xx}(B_t^{(1)}, B_t^{(2)}) + f_{yy}(B_t^{(1)}, B_t^{(2)}) \right)dt + f_{xy}(B_t^{(1)}, B_t^{(2)})d\langle B^{(1)}, B^{(2)}\rangle \end{align*}$$
where I have substituted $dt = d\langle B^{(1)}\rangle_t = d\langle B^{(2)} \rangle_t$. You may also substitute $d\langle B^{(1)}, B^{(2)}\rangle = \rho dt$, where $\rho \in [-1,1]$ is the correlation coefficient.
Can you take it from here?