I am considering the following SDEs:
$$dX_1=-\theta(X_1-a_1)dt+\sqrt{X_1}(1-X_1)dW_1-X_1\sqrt{X_2}dW_2$$ $$dX_2=-\theta(X_2-a_2)dt-X_2\sqrt{X_1}dW_1+\sqrt{X_2}(1-X_2)dW_2$$
Here $W_1$ and $W_2$are both Brownian motions and they are independent.
Is there anyway that I can calculate the following expectation? $$E\left(\int_s^t\sqrt{1-X_1}dW_1 \cdot \int_s^t\sqrt{\frac{X_1X_2}{1-X_1}}dW_2\right)$$
I know if the integrand is adapted to the filtration, we can use Ito isometry, but here I am not sure if I can still use that.
Thank you in advance.
Edit: Also, how can I calculate $E(\int_s^t X_1 dW_1)$, I know if the integrand is adapted to $\mathcal{F}_t$ then the expectation is zero, but here $X_1$ depends on both $W_1$ and $W_2$, seems it is not adapted to the filtration of only $W_1$ or $W_2$..
This expectation is zero, if $dW_1(s)$ and $dW_2(t)$ are independent of each other $\forall s, t$. Let $\omega$ stand for an event. We have
$$E\Big[\int_s^t f(\omega,x)dW_1(x) \int_s^t g(\omega,y)dW_2(y)\Big] = E\Big[\lim\limits_{\delta\rightarrow 0}\sum_{x,y} f(\omega,x)g(\omega,x) E\Big[\big(W_1(x+\delta x)-W_1(x)\big)\big(W_2(x+\delta x)-W_2(x)\big)\big|\mathcal F_x \Big] \Big] = 0,$$ by Riemann sum (definition of Ito integral) and the law of iterated expectation and the fact that. $$E\Big[f(\omega,x)\big(W_1(x+\delta x)-W_1(x)\big)g(\omega,x)\big(W_2(x+\delta x)-W_2(x)\big)\big|\mathcal F_x \Big] = f(\omega,x)g(\omega,x)E\Big[\big(W_1(x+\delta x)-W_1(x)\big)\big(W_2(x+\delta x)-W_2(x)\big)\big|\mathcal F_x \Big] =f(\omega,x)g(\omega,x)E\Big[\big(W_1(x+\delta x)-W_1(x)\big)\big|\mathcal F_x\Big]E\Big[\big(W_2(x+\delta x)-W_2(x)\big)\big|\mathcal F_x\Big] = 0,$$ due to the martingale property of the Brownian motion.