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"Like" Ito: $$d (B^2) = B dB + B dB + dB dB$$ That is $$B dB = \frac{1}{2} d (B^2) - \frac{1}{2} dB dB$$ Integrate. Last term is 1/2 the quadratic variation. I understand the questions as follows: In iii) one has to define what $dB dB$ stands for and one has to "proof" the first line in my answer. In ii) one may use Ito to "know" that $dB dB = dt$.

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For Itô Processes $dX(t) = \mu(t) \mathrm{d}t + \sigma(t) \mathrm{d}W(t)$ you have the result that (under appropriate assumptions which ensure that the local martingale is a martingale, e.g. $E( (\int \sigma(t)^2 \mathrm{d}t )^{1/2} ) < \infty$, etc.): $X$ is a martingale $\Leftrightarrow$ $\mu(t) = 0$. So in order to check if a process $X$ is a ...

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i picked this off from Shreve. Start with the definition of sampled quadratic variation: (1) $\frac{1}{2}Q_\pi = \frac{1}{2}\sum\nolimits_{j=0}^{n-1} (W_{j+1}) - W_j)) ^2$ where $\pi$ = {0,1,2...,n} is a partition of $[0,T]$ (Note we took $\frac{1}{2}$ of both sides for reasons that will be clear in the next line.) Now we know (1) is equal to ...

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