# Integrability in Ito isometry

Itō isometry from Wikipedia:

Let $W : [0, T] \times \Omega \to \mathbb{R}$ denote the canonical real-valued Wiener process defined up to time $T > 0$, and let $X : [0, T] \times \Omega \to \mathbb{R}$ be a stochastic process that is adapted to the natural filtration $\mathcal{F}_{*}^{W}$ of the Wiener process. Then $$\mathbb{E} \left[ \left( \int_{0}^{T} X_{t} \, \mathrm{d} W_{t} \right)^{2} \right] = \mathbb{E} \left[ \int_{0}^{T} X_{t}^{2} \, \mathrm{d} t \right],$$

In other words, the Itō stochastic integral, as a function, is an isometry of normed vector spaces with respect to the norms induced by the inner products $$( X, Y )_{L^{2} (W)} := \mathbb{E} \left( \int_{0}^{T} X_{t} \, \mathrm{d} W_{t} \int_{0}^{T} Y_{t} \, \mathrm{d} W_{t} \right) = \int_{\Omega} \left( \int_{0}^{T} X_{t} \, \mathrm{d} W_{t} \int_{0}^{T} Y_{t} \, \mathrm{d} W_{t} \right) \mathrm{d} \mathrm P (\omega)$$ and $$( A, B )_{L^{2} (\Omega)} := \mathbb{E} ( A B ) = \int_{\Omega} A(\omega) B(\omega) \, \mathrm{d} \mathrm P (\omega).$$

Since it suggests an isometry between two normed spaces, I was wondering

• On the LHS, why is the ito integral $\int_{0}^{T} X_{t} \, \mathrm{d} W_{t}$ a $L^2(\Omega)$ function, so that we can talk about its $L^2$ norm?
• On the RHS, does $\mathrm E (\int_0^T X_t^2 dt)$ equal $\int_{\Omega \times [0,T]} X^2 d(\mathrm{P} \times \lambda)$, and $X \in L^2(\Omega \times [0,T])$, so that we can talk about the $L^2(\Omega \times [0,T])$ norm of $X$? Here $\mathrm{P}$ is the probability measure on $\Omega$, $\lambda$ is the restrction of the Lebesgue measure on $[0,T]$, and $\mathrm{P} \times \lambda$ means their product measure. I can't apply Fubini't theorem here, because its conditions seem not apply here.

Note that in Wikipedia version of Ito isometry, the process $X$ is only required to be adapted to the filtration of the Wiener process. In Shreve's Stochastic Calculus for Finance, the Ito isometry is under the asssumption that $\mathbb{E} [\int_0^T |X(t)|^2 dt] < \infty$. Under both cases, I am not able to figure out the answers to my questions above.

Thanks and regards!

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You seem to mix $\gamma$ and $P$. I am not aware of Ito isometry results for any adapted processes, it seems to be necessary to satisfy some integrability conditions. – Ilya Feb 21 '13 at 10:25
@Ilya: $\gamma$ is from the Wikipedia article, representing the wiener measure when $\Omega := \mathbb R^{[0,T]}$. I use $\mathrm P$ in place of $\gamma$, to allow a more general $\Omega$ which may not be $\mathbb R^{[0,T]}$ or part of it. Could you point out the verion of Ito isometry you have seen, which may need some integrability conditions? – Tim Feb 21 '13 at 13:22
Oksendal, Corollary 3.1.7 - the same assumption on integrability as in Shreve. – Ilya Feb 21 '13 at 13:46
@Ilya: Thanks! So is the assumption $\mathbb{E} [\int_0^T |X(t)|^2 dt] < \infty$? Then my post was asking: does the assumption imply $\int_0^T |X(t)|^2 dt$ is in $L^2(\Omega)$, and $X$ in $L^2(\Omega \times [0,T])$? – Tim Feb 21 '13 at 13:58
No, it does not imply that $\int_0^T |X(t)^2|dt$ is in $L^2(\Omega)$, but I don't see why would that be needed. For the second part, what about Fubini? – Ilya Feb 21 '13 at 14:07