Context : we consider the 1-dimension processes $(X_t)_{t\geq0}$ and $(V_t)_{t\geq0}$ such that
- $dX_t=\mu dt+\sqrt{V_t}dW^X_t$
- $dV_t=\kappa(\theta-V_t)dt+\eta\sqrt{V_t}dW^V_t$.
Both brownians are correlated, and parameters are such that the system is stable (so that $\mathbb{E}[V_t]=\theta$).
Problem : We want to compute $\mathbb{E}[dX_{t+\tau}dX_t]$. By developping the expression, we obtain 4 distinct terms, among them $\mathbb{E}[\sqrt{V_{t+\tau}}dW^X_{t+\tau}\sqrt{V_t}dW^X_t]$, which is the problematic one.
My solution : we study \begin{align*} \mathbb{E}&\left[\int^{t+\tau}_0{\sqrt{V_s}dW^X_s}\int^t_0{\sqrt{V_s}dW^X_s}\right]. \end{align*} Through the law of iterated expectations, it equals \begin{align*} \mathbb{E}\left[\mathbb{E}\left[\int^{t+\tau}_0{\sqrt{V_s}dW^X_s}\;|\;\mathcal{F}_t\right]\int^t_0{\sqrt{V_s}dW^X_s}\right]. \end{align*} As $(\sqrt{V_t})_{t\geq0}$ is a $L^2$ process, $\left(\int^t_0{\sqrt{V_s}dW^X_s}\right)_{t\geq0}$ is a martingale, so the expression amounts to \begin{align*} \mathbb{E}\left[\left(\int^t_0{\sqrt{V_s}dW^X_s}\right)^2\right]. \end{align*} By Itô isometry, it is then equal to \begin{align*} \int^t_0{\mathbb{E}[V_s]ds}. \end{align*} Overall, this is equivalent to $\mathbb{E}[\sqrt{V_{t+\tau}}dW^X_{t+\tau}\sqrt{V_t}dW^X_t]=\mathbb{E}[V_t]dt$.
The problem is, I know the result to be false : it should amount to 0 (if the results of the article I am currently on are correct). I can even "intuitively" see why (independancy of the increments $dW_t$). So, two questions :
What makes the above reasonning false? Transition between simplified SDE and integral SDE forms? Incorrect manipulation of the $dt$ symbol at first? Any other thing?
How can we formally show the equality $\mathbb{E}[\sqrt{V_{t+\tau}}dW^X_{t+\tau}\sqrt{V_t}dW^X_t]=0$ ?
In advance, thank you (and sorry if I made any English mistakes).