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For functions $f$ satisfying appropriate (good) conditions, the following property is called to be isometric.

$$ E\left[\left|\int_{a}^{b}f(t,\omega)dB_{t}(\omega)\right|^{2}\right]=E\left[\int_{a}^{b}|f(t,\omega)|^{2}dt\right] $$

Here $B_{t}(\omega)$ is the Brownian motion.

Why is this called to be isometric?

If one defines the $L^{2}$-norm by $$ \|f\|_{L^{2}(P)}=(E[|f|^{2}])^{1/2}, $$ then the above equality is rewritten as $$ \left\|\int_{a}^{b}f_{t} dB_{t}\right\|_{L^{2}(P)}^{2}=\int_{a}^{b}\|f\|_{L^{2}(P)}^{2}(t)dt $$ provided that Fubini theorem is valid. However I don't look isometry as in functional analysis in Hilbert space, i.e., $\|Tf\|_{H_{1}}=\|f\|_{H_{2}}$ for some operator $T$. What I don't connect between Ito integral and Riemann integral is also a reason that I don't know why it's called to be isometric.

I'm glad if you tell me a reason.

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1 Answers1

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Consider a function $f: [0,T] \times \Omega \to \mathbb{R}$ which is also in your "good conditions" space, denoted $\mathcal{V}$. Consider the operator $I : \mathcal{V} \to L^2(\Omega, \mathcal{F}_T, P)$ given by $$ I_T(f) := \int_0^T f(s,\omega) \, dB_t(\omega). $$ Ito's isometry tells us $$ E\left( \left( \int_0^T f(s) \, dB_t \right)^2 \right) = \int_0^T E(f^2(s)) ds. $$ Now, note \begin{align} ||f||_{L^2([0,T] \times \Omega)}^2 & = \int_{[0,T] \times \Omega} f^2(s,\omega) \, d(Leb \otimes P)(s,\omega) \\ & = \int_0^T \int_{\Omega} f^2(s,\omega) \,dP(\omega) \, ds\\ & = \int_0^T E (f^2(s)) ds \\ & = \int_{\Omega} \left( \int_0^T f(s,\omega)\, dB_t(\omega)\right)^2 \, dP(\omega) \\ & = \int_{\Omega} \left(I_T(f)(\omega)\right)^2 \, dP(\omega) \\ & = ||I_T(f)||_{L^2(\Omega)}^2. \end{align} The second equality is Tonelli's theorem, valid since $f^2 \geq 0$ and both measures are $\sigma$-finite and the fourth equality is Ito's isometry.

Writing out the entire measure spaces, the opreator $I_T$ is an isometry from $$ L^2([0,T] \times \Omega, \mathcal{B} \otimes \mathcal{F}_T, Leb \otimes P) \to L^2(\Omega, \mathcal{F}_T, P), $$ so long as your function $f \in \mathcal{V}$.

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