Variance of a time integral with respect to a Brownian Motion function

Question

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.

Answer 1

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.

Answer 2

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*}