6
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ $f$ is stochastic? $\endgroup$ – Canardini Dec 29 '19 at 16:45
  • 1
    $\begingroup$ No, it's a deterministic function. $\endgroup$ – M. A. Kacef Dec 29 '19 at 17:17
6
$\begingroup$

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*}

$\endgroup$
  • $\begingroup$ Thank you for your precious help @Gordon $\endgroup$ – M. A. Kacef Dec 29 '19 at 20:24
  • $\begingroup$ You are welcome. $\endgroup$ – Gordon Dec 29 '19 at 20:34
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\frac{t^3}{3}$. $\endgroup$ – M. A. Kacef Dec 29 '19 at 19:23
  • 2
    $\begingroup$ I think you have to recalculate it, your expected result is correct $\endgroup$ – Canardini Dec 29 '19 at 20:18
  • $\begingroup$ Thank you for your precious help @Canardini $\endgroup$ – M. A. Kacef Dec 29 '19 at 20:21
  • $\begingroup$ you are welcome, anytime $\endgroup$ – Canardini Dec 29 '19 at 20:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.