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Let $m(t,\omega)=\sum_{j \ge 0}B_{(j+1)2^{-n}}(\omega)I_{[j.2^{-n},(j+1)2^{-n})}(t)$ where $B(t)$ is the Brownian motion and $I_[.]$ is the standard indicator function, Can some body explain me why

$$ E[\int_0^T m(t,\omega)dB_t(\omega)]=\sum_{j \ge 0} E[(B_{t_{j+1}}-B_{t_j})^2] $$

This is a solved example from chapter 3 of stochastic differential equations, Oksandel.

Ron
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1 Answers1

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$$ E[\int_0^T m(t,\omega)dB_t(\omega)]=\sum_{j \ge 0} E[\int_{t_j}^{t_{j+1}} B_{(j+1)2^{-n}} dB_t] $$

$$ =\sum_{j \ge 0} E[B_{t_{j+1}} \left(B_{t_{j+1}} - B_{t_j} \right) ] = \frac12 \sum_{j \ge 0} E[\left(B_{t_{j+1}} - B_{t_j} \right)^2 ] $$

corindo
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  • @Cornido, the absence of 1/2 in the book crated a confusion. Thanks for the answer. – Ron Jun 17 '13 at 08:34