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I am trying to solve following expectation

$E \left [B^2_s \left( \int^t_s B_u dB_u \right)^2 \right]$

with $0 \leq s \leq t \leq T$ and $B_t$ a 1-dim. Brownian motion. Further using $E \left[ . \vert \mathcal{F}_s \right]$ is a hint.

I started by introducing the conditional expectation and seperating the integral

$E \left[ B^2_s E \left[ \left( \int^t_0 B_u dB_u - \int^s_0 B_u dB_u \right)^2 \vert \mathcal{F}_s \right] \right]$

I am not sure about the next step, but I think $\left( \int^t_0 B_u dB_u \right)_t$ are square integrable martingales and I can write

$E \left[ B^2_s E \left[ \left( \int^t_0 B_u dB_u \right)^2 - \left( \int^s_0 B_u dB_u \right)^2 \vert \mathcal{F}_s \right] \right]$

and with $\int^t_0 B_u dB_u = \dfrac{1}{2}(B^2_t - t)$

$E \left[ B^2_s E \left[ \left( \dfrac{1}{2}(B^2_t - t) \right)^2 - \left( \dfrac{1}{2}(B^2_s - s) \right)^2 \vert \mathcal{F}_s \right] \right]$

Any next step of making the equation easier did not have any real result.

Thank you for your help !

user113685
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1 Answers1

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Fix $s$ and condition by $\mathcal F_s$, then $M_t=\displaystyle\int_s^tB_sB_u\mathrm dB_u$ defines a martingale with quadratic variation process $\langle M,M\rangle_t=\displaystyle\int_s^t(B_sB_u)^2\mathrm du$ hence the expectation to be computed is $$ (\ast)=E(M_t^2)=E(E(M_t^2\mid\mathcal F_s))=E(\langle M,M\rangle_t)=\int_s^tE(B_s^2B_u^2)\mathrm du. $$ Now, for every $u\geqslant s$, $B_u^2=B_s^2+2B_sW+W^2$ where $W=B_u-B_s$ is independent of $B_s$ with $E(W)=0$ and $E(W^2)=u-s$, hence $$ (\ast)=\int_s^t(E(B_s^4)+(u-s)E(B_s^2))\mathrm du. $$ Plug in $E(B_s^2)=s$ and $E(B_s^4)=3s^2$ in the last integral to conclude that, unless I am mistaken, $$(\ast)=\tfrac12s(t+5s)(t-s).$$

Did
  • 279,727
  • I had problems with the step $E(E(M^2_t \vert \mathcal{F}_s)) = E(\langle M,M \rangle_t)$

    Could you maybe check my thoughts and see if they are correct?

    As you stated $M_t$ is a martingale with quadratic variation $\langle M, M \rangle_t$. Therefore we know that $M^2_t - \langle M,M \rangle_t$ is also a martingale.

     – user113685 Jan 17 '14 at 17:58
  • We use $E(M^2_t \vert \mathcal{F}_s) = E(M^2_t - \langle M,M \rangle_t + \langle M,M \rangle_t \vert \mathcal{F}_s) = E(M^2_t - \langle M,M \rangle_t \vert \mathcal{F}_s) + E( \langle M,M \rangle_t \vert \mathcal{F}_s) = E(M^2_s - \langle M,M \rangle_s) + E(\langle M,M \rangle_t \vert \mathcal{F}_s) = 0 + E(\langle M,M \rangle_t\vert \mathcal{F}_s)$

    And $E(E(M^2_t \vert \mathcal{F}_s)) = E(\langle M,M \rangle_t)$

    Thank you so much for your fast answer, it was of great help !

     – user113685 Jan 17 '14 at 17:59