Multidimensional Ito Lemma on 2 independent Brownian Motions

Question

I would appreciate any explanations on the following question:

For the function $F_t = f(t, X_1, X_2)$, $X_1, X_2 = W_t^1, W_t^2$ are independent Brownian motions respectively, then using Ito's multidimensional lemma, we obtain

\begin{align} dF_t &= \frac{\partial f}{\partial x}dW_t^1 + \frac{\partial f}{\partial y}dW_t^2 + \frac{1}{2}\frac{\partial^2 f}{\partial x^2}(dW_t^1)^2 + \frac{1}{2}\frac{\partial^2 f}{\partial y^2}(dW_t^2)^2 + \frac{\partial^2 f}{\partial x \partial y}dW_t^1dW_t^2 \\ &= \frac{\partial f}{\partial x}dW_t^1 + \frac{\partial f}{\partial y}dW_t^2 + \frac{1}{2}\left(\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial x^2}\right)dt \end{align}

Question: why is $\frac{\partial^2 f}{\partial x \partial y}dW_t^1dW_t^2 = 0$?
Specifically, why is $dW_t^1dW_t^2 = 0$? This is the bit I am unsure of.

Thanks in advance.

Answer 1

By subtracting the initial values, we may assume that $W_0^1=W_0^2=0$. By checking the finite-dimensional marginals, we then oberseve that the process $$X_t=\frac1{\sqrt 2}(W_t^1+W_t^2)$$ is also a Brownian motion. So $\langle X,X\rangle_t=t$, and using the bilinearity of the bracket, it follows that $\langle W^1,W^2\rangle_t=0$.

Here I use the notations in Le Gall’s book. $\langle M,N\rangle$ means the bracket of two continuous local martingales. Reference for notions: Le Gall’s Brownian Motion, Martingales, and Stochastic Calculus , section 4.4.