Let $B_t$ be standard Brownian Motion. Could someone please help me to show that $$E[(\int_{0}^{t}B_sdB_s)^2] = \int_{0}^{t}E[B_s^2]ds$$

I am sure that it has something to do with Ito's formula but I am pretty new to this stuff so some help would be greatly appreciated.


closed as off-topic by Did, user223391, SchrodingersCat, Claude Leibovici, Davide Giraudo Nov 11 '15 at 9:03

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  • $\begingroup$ The accepted solution computes each side, using various tricks, and sees they are equal. The exercise actually asks you to apply a very general formula, probably in your notes, which provides the result right away. This is one "advantage" of accepting answers on the spot. $\endgroup$ – Did Jun 11 '15 at 21:50

The RHS is direct to evaluate. The variance of $B_s$ is $s$ so: $$\int_0^t E[B_s^2] ds = \int_0^t s ds = \frac{t^2}{2}$$

You are right about use Ito's Lemma for the LHS. By Ito: $$B_t^2 = t + 2\int_0^t B_s dB_s$$ So \begin{eqnarray*} E[(\int_0^t B_s dB_s)^2] &=& E\left[\left(\frac{B_t^2 - t}{2}\right)^2 \right] \\ &=& \frac{1}{4}(E[B_t^4] - 2tE[B_t^2] + t^2)\\ &=& \frac{1}{4}(3t^2 - 2t\cdot t + t^2)\\ \end{eqnarray*}

  • 1
    $\begingroup$ Makes perfect sense, thanks for the clear answer. I was evaluating the LHS and trying to arrive at the RHS and I think that is why I was getting confused $\endgroup$ – EE_13 Jun 11 '15 at 16:23
  • $\begingroup$ Indeed, "evaluating the LHS and trying to arrive at the RHS" is what you are supposed to do, using a result in your notes. $\endgroup$ – Did Jun 12 '15 at 7:22

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