Fubini's Theorem on Ito's integral?

Question

It is well known that Fubini's Theorem allows us to swap two interated integrals. In stochastic calculus, Fubini's Theorem (under certain conditions) ensures that $$\mathbb{E}\int_0^t g(W_s) ds = \int_0^t \mathbb{E}g(W_s) ds$$ for some function $g$ and $W_s$ is a Brownian motion.

Is it possible to apply Fubini's Theorem to Ito's integral? More precisely,

Do we have $$\mathbb{E}\int_0^t g(W_s) dW_s = \int_0^t\mathbb{E} g(W_s) dW_s?$$

Answer 1

I will mention a Fubini-type theorem for double integrals involving a stochastic integral and a Lebesgue integral. But keep in mind that there are results for double stochastic integrals as well.

Let $g: X\times[0,\infty)\times\Omega \mapsto \mathbb{R}$ be a parametrized stochastic process with the parameter taking values in $X$ and $\mu$ be a $\sigma$-finite measure on a $\sigma$-algebra defined on this space. $g$ satisfies certain measurability properties on the triple product $\sigma$-algebra, which I won't get into as this is a very technical area.

Suppose that $$\int_X\left(\int_0^t \lvert g(x, s, \omega)\rvert^2ds\right)^{\frac{1}{2}}d\mu(x) < \infty$$ Then, $$\int_X\int_0^t g(x, s)dW_sd\mu(x) = \int_0^t \int_Xg(x, s)d\mu(x)dW_s$$

The example you gave does not fit into this framework as the outer integral on the left hand side is on $\Omega$, not a separate parameter space. You cannot treat it as a parameter space as it is coupled to the inner integral. I don't see how you can exchange two integrals that take place on the same domain in different ways. Well, technically the stochastic integral is on $[0,t]$ but it is defined using convergence in $\Omega$.