4
$\begingroup$

I want to calculate $$\operatorname{E} \left[ \int_0^1{W(t)dt \cdot \int_0^1{t^2W(t)dt}} \right].$$

I discovered that the first integral is $\operatorname{N}(0, \frac{1}{3})$ but I don't know how to get the other one and the full answer of their multiplied expectation.

Improve this question
$\endgroup$
0
7
$\begingroup$

Note that \begin{align*} E\left(\int_0^1 W_t\, dt \int_0^1 t^2W_t\, dt \right) &= E\left(\int_0^1\!\!\!\int_0^1 s^2 W_s W_t\, dsdt \right)\\ &=\int_0^1\!\!\!\int_0^1 s^2 E(W_s W_t)\, dsdt\\ &=\int_0^1\!\!\!\int_0^1 s^2 (s\wedge t)\, dsdt\\ &=\int_0^1 s^2\,ds \int_0^1 s\wedge t\, dt\\ &=\int_0^1 s^2\,ds \left(\int_0^s t\,dt + \int_s^1 s\,dt\right). \end{align*} The remaining is now straightforward.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thank you very much. But the rest can be solved after that? Or just calculate 1/3(s^2/2+s-s^2) $\endgroup$ – Hobong Dec 9 '18 at 15:23
  • 2
    $\begingroup$ The rest is just calculus. $\endgroup$ – Gordon Dec 9 '18 at 15:27
  • $\begingroup$ @Gordon I guess you need to justify the first equality by Fubini. $\endgroup$ – FunnyBuzer Jan 16 '19 at 10:47
  • $\begingroup$ @FunnyBuzer, yes, it is indeed based on Fubini. $\endgroup$ – Gordon Jan 16 '19 at 13:46

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.