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Let process $$I_t = \int_0^t f(s) W_s \,\mathrm d s $$
where $W_s$ is standard Brownian motion. My question are the following:

We know that $\mathbb{E} (I_{t})=0$ for all $t$ and $f$ a integrable function. Is there a general formula for the second-order moment i.e. $\mathbb{E}(I_{t}^2)$ ?

Thank you in advance for any comments, help, remarks or references related to this issue.

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    $\begingroup$ $f$ is stochastic? $\endgroup$
    – Canardini
    Dec 29, 2019 at 16:45
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    $\begingroup$ No, it's a deterministic function. $\endgroup$
    – KACEFMA.
    Dec 29, 2019 at 17:17

2 Answers 2

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Using Fubini's argument, assuming that $f$ is deterministic

$$E(I_t^2) = E\left(\int_0^t f(s) W_s ds\int_0^t f(u) W_u du\right)=\int_0^t\int_0^t{f(s)f(u)min(s,u)duds}$$

If $f$ is continuous(even piece wise) you can prove that $I_t$ is normally distributed.

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  • $\begingroup$ Thank you for your response. However, I still have a doubt since for $f(t)=1$ we find a variance equal to $\frac{t^3}{2}$ and not $\frac{t^3}{3}$. $\endgroup$
    – KACEFMA.
    Dec 29, 2019 at 19:23
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    $\begingroup$ I think you have to recalculate it, your expected result is correct $\endgroup$
    – Canardini
    Dec 29, 2019 at 20:18
  • $\begingroup$ Thank you for your precious help @Canardini $\endgroup$
    – KACEFMA.
    Dec 29, 2019 at 20:21
  • $\begingroup$ you are welcome, anytime $\endgroup$
    – Canardini
    Dec 29, 2019 at 20:24
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As @Canardini pointed out, \begin{align*} E\big(I_t^2\big) &= E\left(\int_0^t f(s) W_s ds\int_0^t f(u) W_u du\right)\\ &= \int_0^t\!\int_0^t f(s)f(u)\min(s,u)dsdu\\ &= \int_0^t\left(\int_0^u f(s)f(u) s ds + \int_u^t f(s)f(u) u ds \right)du\\ &= \int_0^t \int_0^u sf(s) f(u)ds du + \int_0^t \int_u^t uf(u)f(s) ds du\\ &=2\int_0^t uf(u) \int_u^t f(s) ds du\\ &=-u\left(\int_u^t f(s) ds\right)^2\Big|_0^t+\int_0^t\left(\int_u^t f(s) ds\right)^2 du\\ &=\int_0^t \left(\int_s^t f(u)du\right)^2 ds. \end{align*} Alternatively, note that \begin{align*} \int_0^t f(s) W_s ds &= W_t \int_0^t f(s) ds - \int_0^t \int_0^s f(u)du\, dW_s\\ &=\int_0^t f(s) ds\int_0^t dW_s - \int_0^t \int_0^s f(u)du\, dW_s\\ &=\int_0^t \int_s^t f(u)du\, dW_s. \end{align*} Then \begin{align*} E\big(I_t^2\big) &= \int_0^t \left(\int_s^t f(u)du\right)^2 ds. \end{align*}

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  • $\begingroup$ Thank you for your precious help @Gordon $\endgroup$
    – KACEFMA.
    Dec 29, 2019 at 20:24
  • $\begingroup$ You are welcome. $\endgroup$
    – Gordon
    Dec 29, 2019 at 20:34
  • $\begingroup$ @Gordon - can you please explain little further how the part after "Then" has actually come up? $\endgroup$
    – Bogaso
    Aug 2, 2020 at 9:47
  • $\begingroup$ By Ito's isometry $\endgroup$
    – Gordon
    Aug 2, 2020 at 12:32

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