Ito Isometry - Definitions

Question

Three different lecturers have provided three different definitions of Ito Isometry. These are:

Lecturer A \begin{align*} \mathbb{E}\left[ \left(\int_{0}^{\infty} h_{s}\,dW_{s}\right)^{2} \right] = \mathbb{E}\left[\int_{0}^{\infty}|h_{s}|^{2}\,ds\right] < \infty \end{align*} Lecturer B \begin{align*} \mathbb{E}\left[\left(\int_{0}^{T}K_{s}\,dW_{s}\right)^{2}\right] &= \mathbb{E}\left[\int_{0}^{T} K_{s}^{2}\,ds\right]\\ &= \int_{0}^{T}\mathbb{E}\left[K_{s}^{2}\right]\,ds \end{align*} Lecturer C \begin{align*} \mathbb{E} \left[\left|\int_{0}^{t} \Phi(s)\,dW(s)\right|^{2} \right] &= \int_{0}^{t} \mathbb{E}\left[\left|\Phi(s)\right|^{2}\right]\,ds \end{align*}

The books by Klebaner (Introduction to stochastic calculus with applications) and Milkosch (Elementary stochastic calculus) both agree with lecturer B. Lecturer B is my old lecturer. Lecturers A and C are my current lecturers.

Any insight into the following questions would be appreciated.

1 - Why is lecturer A using infinity as a limit? Does it matter?

2 - Why are lecturers A and C using the modulus sign?

Many thanks,

John

Answer 1

1. It simply depends whether one wants to consider integrals with infinite time-horizon or not. Usually, stochastic integrals of the form $$\int_0^T K_s \, dW_s \tag{1}$$ are considered where $T<\infty$. In this case, $$\mathbb{E} \left[ \left( \int_0^T K_s \, dW_s \right)^2 \right] = \mathbb{E}\left( \int_0^T K_s^2 \, ds \right) \tag{2}$$ is called Itô's isometry. Now if we want to allow $T=\infty$ in $(1)$, then, accordingly, we have to replace $T$ by $\infty$ in $(2)$. Note however, that $$\mathbb{E}\left[ \left( \int_0^{\infty} K_s \, dW_s \right)^2 \right] < \infty$$ implies in particular $$\mathbb{E} \left[\left( \int_0^T K_s \, dW_s \right)^2\right]< \infty$$ for any $T>0$.
2. It doesn't matter since $x^2=|x|^2$ holds for any real number $x \in \mathbb{R}$.