Partial derivative of function of correlated Brownian motions.

Question

I am facing the following problem. Provided with the following two processes existing under an arbitrary measure
$$dX_1(t) = \sigma(X_1(t))dW_1(t),$$
$$dX_2(t) = \sigma(X_2(t))dW_2(t),$$
with $\sigma$ some positive function and $W_1(t),W_2(t)$ two correlated Brownian motion such that $dW_1(t)dW_2(t)=\rho dt$. Now, imagine that we have the following function $f$, defined as
$$f(X_1(t),X_2(t))=aX_1(t)+bX_2(t), \quad a,b>0.$$
What holds then for
$$\frac{\partial f(X_1(t),X_2(t))}{\partial X_1} = \>?$$
I am curious if the correlation between the two processes has any impact on the outcome of taking this partial derivative or if the outcome is simply $a$. Any help is appreciated!

Answer 1

The correlation $\rho$ definitely has impact on the partial derivative.

Because $dW_1(t)dW_2(t)=\rho dt$, so there exists an brownian motion $Z(t)$ which is independant to $W_1(t)$ (and so, independant to $X_1(t)$) so that: $$dW_2(t)=\rho dW_1(t) +\sqrt{1-\rho^2}dZ(t) $$ Take for example $\sigma(x) = 1$, we have

\begin{align*} dX_2(t) &= dW_2(t)\\ &= \rho dW_1(t) +\sqrt{1-\rho^2}dZ(t) \\ &= \rho dX_1(t) +\sqrt{1-\rho^2}dZ(t) \end{align*} So, $$X_2(t) = \rho X_1(t) +\sqrt{1-\rho^2}Z(t) + const $$ We can now calculate the function $f$: \begin{align*} f(t) &= a X_1(t) + b X_2(t) \\ &= (a+ b\rho) X_1(t) +b\sqrt{1-\rho^2}Z(t) + const \end{align*}

So, because $Z(t)$ is independant to $X_1(t)$ \begin{align*} \frac{\partial f (X_1(t),X_2(t))}{\partial X_1(t)} &= (a+ b\rho) + b\sqrt{1-\rho^2} \frac{Z(t)}{\partial X_1(t)} \\ &= (a+ b\rho) \end{align*}