[隨機分析] Ito Isometry Property in H^2 Space

已知 $f \in \cal{H}^2[0,t]$，我們有 Ito Isometry 如下：
$$E\left[ \left( \int_0^t f(s) dB_s \right)^2 \right] = E \left [ \int_0^t f(s)^2 ds \right]$$
現在我們看看 cross term 會怎麼樣?

考慮 $f,g \in \cal{H}^2[0,t]$
$$E\left[ {\int_0^t f (s)d{B_s} \cdot \int_0^t g (s)d{B_s}} \right] = E\left[ {\int_0^t f (s)g\left( s \right)ds} \right]$$
Proof:
觀察下式：
$$\begin{array}{l} E\left[ {{{\left( {\int_0^t {f(s)} d{B_s} + \int_0^t {g\left( s \right)} d{B_s}} \right)}^2}} \right]\\ \begin{array}{*{20}{c}} {}&{} \end{array} = E\left[ {{{\left( {\int_0^t {f(s)} d{B_s}} \right)}^2} + 2\int_0^t {f(s)} d{B_s}\int_0^t {g\left( s \right)} d{B_s} + {{\left( {\int_0^t {g(s)} d{B_s}} \right)}^2}} \right]\\ \begin{array}{*{20}{c}} {}&{} \end{array} = E\left[ {{{\left( {\int_0^t {f(s)} d{B_s}} \right)}^2}} \right] + E\left[ {2\int_0^t {f(s)} d{B_s}\int_0^t {g\left( s \right)} d{B_s}} \right]\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{}&{} \end{array} + E\left[ {{{\left( {\int_0^t {g(s)} d{B_s}} \right)}^2}} \right]\\ \begin{array}{*{20}{c}} {}&{} \end{array} = E\left[ {\int_0^t {{f^2}(s)} ds} \right] + E\left[ {2\int_0^t {f(s)} d{B_s}\int_0^t {g\left( s \right)} d{B_s}} \right]\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{}&{}&{} \end{array} + E\left[ {\int_0^t {{g^2}(s)} ds} \right] \ \ \ \  (*) \end{array}$$ 但注意到我們所觀察的式子亦可寫成
$$\begin{array}{l} E\left[ {{{\left( {\int_0^t {f(s)} d{B_s} + \int_0^t {g\left( s \right)} d{B_s}} \right)}^2}} \right] = E\left[ {{{\left( {\int_0^t {\left( {f(s) + g\left( s \right)} \right)} d{B_s}} \right)}^2}} \right]\\ \begin{array}{*{20}{c}} {}&{} \end{array} = E\left[ {\int_0^t {{{\left( {f(s) + g\left( s \right)} \right)}^2}} ds} \right]\\ \begin{array}{*{20}{c}} {}&{} \end{array} = E\left[ {\int_0^t {{f^2}(s)} ds} \right] + 2E\left[ {\int_0^t {f(s)g\left( s \right)} ds} \right] + E\left[ {\int_0^t {{g^2}\left( s \right)} ds} \right] \ \ \ \ \ (\star) \end{array}$$ 比較 $(*)$ 與 $(\star)$，可得
$$E\left[ {\int_0^t {f(s)} d{B_s}\int_0^t {g\left( s \right)} d{B_s}} \right] = E\left[ {\int_0^t {f(s)g\left( s \right)} ds} \right]$$