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Let $(W_{t})_{t \geq 0}$ ba a $d$-dimensional Brownian motion, $k \in {\mathbb N}$ and $i_{1},i_{2},\ldots,i_{k} \in \{1,2,\ldots,d\}$. Then can we calculate the following multiple stochastic integral?

$$ {\mathbb E}\left[\prod_{j=1}^{k}\int_{0}^{t}f_{i_{j}}(s){\rm d}W_{s}^{i_{j}}\right]=??? $$

For example, for $k=1$ and $k=2$, $$ {\mathbb E}\left[\int_{0}^{t}f_{i_{1}}(s){\rm d}W_{s}^{i_{1}}\right]=0, \quad {\mathbb E}\left[\prod_{j=1}^{2}\int_{0}^{t}f_{i_{j}}(s){\rm d}W_{s}^{i_{j}}\right]={\mathbb E}\left[\int_{0}^{t}f_{i_{1}}(s)f_{i_{2}}(s){\rm d}s\right]\delta_{i_{1},i_{2}}. $$

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For $k=3$, using the It\^o formula, we obtain \begin{align*} \prod_{j=1}^{3}\int_{0}^{t}f_{i_{j}}(s){\rm d}W_{s}^{i_{j}} &=\int_{0}^{t}\left(\int_{0}^{s}f_{i_{2}}(u){\rm d}W_{u}^{i_{2}}\right)\left(\int_{0}^{s}f_{i_{3}}(u){\rm d}W_{u}^{i_{3}}\right)f_{i_{1}}(s){\rm d}W_{s}^{i_{1}} \\ &\hspace{0.5cm}+\int_{0}^{t}\left(\int_{0}^{s}f_{i_{1}}(u){\rm d}W_{u}^{i_{1}}\right)\left(\int_{0}^{s}f_{i_{3}}(u){\rm d}W_{u}^{i_{3}}\right)f_{i_{2}}(s){\rm d}W_{s}^{i_{2}} \\ &\hspace{0.5cm}+\int_{0}^{t}\left(\int_{0}^{s}f_{i_{1}}(u){\rm d}W_{u}^{i_{1}}\right)\left(\int_{0}^{s}f_{i_{2}}(u){\rm d}W_{u}^{i_{2}}\right)f_{i_{3}}(s){\rm d}W_{s}^{i_{3}} \\ &\hspace{0.5cm}+\frac{1}{2}\int_{0}^{t}\int_{0}^{s}f_{i_{3}}(u){\rm d}W_{u}^{i_{3}}\delta_{i_{1},i_{2}}{\rm d}s \\ &\hspace{0.5cm}+\frac{1}{2}\int_{0}^{t}\int_{0}^{s}f_{i_{2}}(u){\rm d}W_{u}^{i_{2}}\delta_{i_{1},i_{3}}{\rm d}s \\ &\hspace{0.5cm}+\frac{1}{2}\int_{0}^{t}\int_{0}^{s}f_{i_{1}}(u){\rm d}W_{u}^{i_{1}}\delta_{i_{2},i_{3}}{\rm d}s. \end{align*} Thus by Fubini's theorem, we have \begin{align*} {\mathbb E}\left[\prod_{j=1}^{3}\int_{0}^{t}f_{i_{j}}(s){\rm d}W_{s}^{i_{j}}\right] &=\frac{1}{2}{\mathbb E}\left[\int_{0}^{t}\int_{0}^{s}f_{i_{3}}(u){\rm d}W_{u}^{i_{3}}{\rm d}s\right]\delta_{i_{1},i_{2}} \\ &\hspace{0.5cm}+\frac{1}{2}{\mathbb E}\left[\int_{0}^{t}\int_{0}^{s}f_{i_{2}}(u){\rm d}W_{u}^{i_{2}}{\rm d}s\right]\delta_{i_{1},i_{3}} \\ &\hspace{0.5cm}+\frac{1}{2}{\mathbb E}\left[\int_{0}^{t}\int_{0}^{s}f_{i_{1}}(u){\rm d}W_{u}^{i_{1}}{\rm d}s\right]\delta_{i_{2},i_{3}} \\ &=\frac{1}{2}\int_{0}^{t}{\mathbb E}\left[\int_{0}^{s}f_{i_{3}}(u){\rm d}W_{u}^{i_{3}}\right]{\rm d}s\delta_{i_{1},i_{2}} \\ &\hspace{0.5cm}+\frac{1}{2}\int_{0}^{t}{\mathbb E}\left[\int_{0}^{s}f_{i_{2}}(u){\rm d}W_{u}^{i_{2}}\right]{\rm d}s\delta_{i_{1},i_{3}} \\ &\hspace{0.5cm}+\frac{1}{2}\int_{0}^{t}{\mathbb E}\left[\int_{0}^{s}f_{i_{1}}(u){\rm d}W_{u}^{i_{1}}\right]{\rm d}s\delta_{i_{2},i_{3}} \\ &=0. \end{align*} The general case follows similarly.

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