Application of Ito's isometry in deduction of Wiener Ito Chaos expansion

Question

I am trying to learn about the Wiener Ito Chaos expansion and starting reading Oksendal's notes on Malliavin calculus where it is treated in Chapter 1. For a link to the notes, please see http://www.nhh.no/Files/Filer/institutter/for/dp/1996/wp0396.pdf . On page 1.2, in the derivation on the second half of the page, he says he is applying Ito's isometry. What I have a hard time understanding at the moment, is how he can actually do it because the integrands are different? Is this some kind of extension of the general Ito isometry or am I just missing something?

Thanks

Answer 1

Itô's formula states that

$$\mathbb{E} \left( \left| \int_0^T f(t_n) \, dW(t_n) \right|^2 \right) = \mathbb{E} \left( \int_0^T |f(t_n)|^2 \, dt_n \right)$$

for any "nice" function $f$. Applying this for

$$f(t_n) := \int_0^{t_n} \dots \int_0^{t_2} h(t_1,\ldots,t_n) \, dW_{t_1} \dots dW_{t_{n-1}}$$

yields

$$\mathbb{E} \left( \left| \int_0^T \int_0^{t_n} \dots \int_0^{t_2} h(t_1,\ldots,t_n) dW_{t_1} \ldots \, dW_{t_n} \right|^2 \right) = \mathbb{E} \left( \int_0^T \left| \int_0^{t_{n}} \dots \int_0^{t_2} h(t_1,\ldots,t_n) dW_{t_1} \ldots \, dW_{t_{n-1}} \right|^2 \, dt_{n} \right).$$

Using Tonelli's theorem, we can interchange the expectation $\mathbb{E}$ and the $t_{n}$-integration. Now the claim follows by iteration.