I am starting to study the multidimensional version of Ito's lemma . The book shows an exercise that I don't understand. The exercise is:
$(X_t^1,X_t^2)$ is a 2-dimensional Brownian motion and $F(t,x_1,x_2) = x_1x_2$.
Reviewing the theory well:
Two-dimensional case: $d(_t\int_{0}^{t}s \space dW_s) =$ ?
$X_t^1 = W_t$ and $X_t^2 = \int_{0}^{t}s \space dW_s\\$
I calculated the partial derivatives: $F(t,x_1,x_2) = x_1x_2$ ,
$\partial {x_1}F = x_2$
$\partial {x_2}F = x_1$
$\partial _{x_1x_1}F = \partial _{x_2x_2}F= 0$.
$\partial {x_1}{x_2}F = 1$
I pose: $dX_t^1 = dW_t\space \space$ and $dX_t^2 = t dW_t$
Apply Ito:
$dF(X_t^1,X^2_t)= X_t^1dX_t^2 + X_t^2dX_t^1 + \frac{1}{2}2d\langle X_t^1,X_t^2\rangle$
hence the result posted above $d(X_{t}^1X_{t}^2) = X_{t}^1 dX_{t}^2 + X_{t}^2 dX_{t}^1+ \beta dt$.
It's correct? Thanks.