Fubini's Theorem and Stochastic (Ito) Integrals

Question

Edit: in the comments, it is pointed out by @Mag that Ito Integral can - in fact - be defined pathwise, rather than an L2 limit. A bit of googling has led me to these Oxford university notes here for example. My question: would it be after all possible to apply Fubini's theorem to Ito Integral, defined pathwise?

Original question: I wonder whether Fubini's theorem is applicable to taking the expectation of Ito integrals. It is a well known fact that Ito integral posesses the "martingale property", i.e. for any deterministic or stochastic process $H(\omega,t)$, we have:

$$\mathbb{E}\left[\int_{h=0}^{h=t}H(\omega,h)dW_h\right]=0$$

From the above, it is obvious that for many types of integrands, Fubini's theorem "doesn't work", for example, we have that $\mathbb{E}[W_t^2]=t$, but:

$$\mathbb{E}\left[\int_{h=0}^{h=t}W_h^2dW_h\right]=0\neq \int_{h=0}^{h=t}\mathbb{E}\left[W_h^2\right]dW_h =\int_{h=0}^{h=t}hdW_h$$

Question: so what is it that makes Fubini's theorem not applicable above? The usual two conditions for being able to apply Fubini are:

• (i) $\mathbb{E}[|H(\omega,t)|]<\infty$
• (ii) $H(\omega,t)$ is measurable

In case of $H(\omega,t):=W_t^2$, both conditions are satisfied: (i) clearly, $\mathbb{E}[|W_t^2|]<\infty \forall t \in \mathbb{R}$ and (ii) $W_t$ is adapted and measurable (with respect to the filtration it generates and if the entire probability space is generated by $W_t$, then indeed $W_t$ is "measurable" with respect to $\left(\Omega,\mathbb{P},\mathcal{F}_t\right)$).

Answer 1

In Fubini's theorem we have two measures and two Lebesgue integrals. Here you have one Lebesgue integral (i.e the expectation) and an Ito integral. The Ito integral is not defined pathwise but as a $L^2$ limit. Also note that $dW_t$ is not a measure.

Also, it's not always true that $$\mathbb{E}\left[\int_0 ^t H_udW_u\right]=0. \tag{1}$$

The equality (1) is true when $\mathbb{E}\left[\int_0^t H^2_u du\right]<\infty$

Answer 2

I think the process $X_t:=\int_0^t H_s\mathbb dW_s$ is not a martingale in general. It is under very weak assumptions for $H$ a local martingale and if you have for example $\mathbb E(X_t^2)=\mathbb E(\int_0^t H_s^2\mathbb ds)<\infty$ for all $t\geq 0$ it is also a martingale. Another sufficent condition would be $\forall t\geq0: \mathbb E(\sup_{s\leq t}|X_s|)<\infty$.

The thing I mentioned in the comments to the other answer, that Itô integral is able to be defined pathwise, was just intended as an additional information, that the argument that it is defined as $L^2$-limit, is not a sufficent agrument against it (not saying anything about whether it actually possible or not). But being definied pathwise means still dependent on the path resp. $\omega$. Being definied pathwise or as $L^2$-limit is not the relevant part of your problem.

The main problem why there is definitly no Fubini-Theorem for Itô integrals combinied with integral with the measrue $P$ is, that $\mathbb E(\int_0^t W_s^2\mathbb dW_s)=\int_\Omega\int_0^t W_s^2\mathbb dW_s\mathbb dP$ is simply a real number, whereas $\int_0^t \mathbb E(W_s^2)\mathbb dW_s$ is stochastic process definitly depending on $\omega$.

But I remember an exercise regarding a Fubini-Theorem for Itô integrals with a Lebesgue integral, but it was a bit different to your application: Let $T>0$, $B\subset\mathbb R^n$ bounded and $$H:(\Omega,[0,T],B)\to\mathbb R: (\omega,t,x)\mapsto H(\omega,t,x)$$ a stochastic process not only depending on the time $t$ but also on $x$. Then under appropriate assumptions holds $$\int_0^T\int_BH(\omega,s,x)\mathbb dx\mathbb dW_s=\int_B\int_0^TH(\omega,s,x)\mathbb dW_s\mathbb dx. $$

If you are working with an Itô process $X$ there won't be much changes in the things above. For $\mathrm d X_t=b_t\mathrm dt +\sigma_t\mathrm dW_t$ you can simply replace the differentials as you are used and you are again in the situation above.

I hope this helps you a little bit for your problem.