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Background

Consider the Wiener process: \begin{equation} W(t)=\int_0^t d W=\int_0^t \xi\left(t^{\prime}\right) d t^{\prime} \end{equation} where $\xi(t)$ is white Gaussian noise: $\langle \xi(t)\rangle = 0$ and $\langle \xi(t)\xi(t')\rangle = \delta(t-t')$.

Consider now the following stochastic integral:

$$ I=\int_0^T f\left(t^{\prime}\right) d W=\int_0^t f\left(t^{\prime}\right) \xi\left(t^{\prime}\right) d t^{\prime}$$

Where $f(t)$ is some function of $t$. If $f(t)$ is the Wiener process itself, we can show that $\mathbb{E}[ I]$ = 0 under the Ito interpretation, and $T/2$ in the Stratonovich interpretation.

Question

What is $\mathbb{E}[I]$ if I choose $f$ to be the integral of the Wiener process? I.e. let $$f(t)=\int_0^t W(t') d W$$

Edited Changed $\langle .\rangle$ to $\mathbb{E}[.]$.

Matt
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  • In stochastic calculus one should not use $\langle I\rangle$ to denote an expectation since this will be confused with quadratic variation. In short: $\mathbb E[\int_0^tW_s,dW_s]=0$ also. You can find that in every text book introducing stochastic integration. – Kurt G. Jan 30 '23 at 06:12
  • Careful, I am not asking for $\mathbb{E}\left[\int_0^T W_s d W_s\right]$, but I am asking for $\mathbb{E}\left[\int_0^T \left(\int_0^s W(r)dr\right) d W_s\right]$. This is not found in common textbooks. – Matt Jan 30 '23 at 06:17
  • I literally wrote that if $f(t)=W_t$ then $\mathbb{E}\left[\int_0^T f_t d W_s\right]=0$, you just repeated what I wrote with a different notation. I did not know about the use of the different notation, thanks. – Matt Jan 30 '23 at 06:21
  • Very confusing. When $f(t)=\int_0^t W(t'),dW$ you must be asking for $\mathbb E[\int_0^T(\int_0^s W_u,\color{red}{dW_u}),dW_s],.$ Anyway: in common text books you can find the general theorem that the expectation of pretty much every Ito integral is zero. – Kurt G. Jan 30 '23 at 06:23

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