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.


1 Answer 1


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.

  • $\begingroup$ 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, 2018 at 15:23
  • 2
    $\begingroup$ The rest is just calculus. $\endgroup$
    – Gordon
    Dec 9, 2018 at 15:27
  • $\begingroup$ @Gordon I guess you need to justify the first equality by Fubini. $\endgroup$
    – FunnyBuzer
    Jan 16, 2019 at 10:47
  • $\begingroup$ @FunnyBuzer, yes, it is indeed based on Fubini. $\endgroup$
    – Gordon
    Jan 16, 2019 at 13:46

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