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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.

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

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Community, SchrodingersCat, Claude Leibovici, Davide Giraudo
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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
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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*}

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    $\begingroup$ 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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