To do this, we can consider the stochastic integral as a limit of Riemann sums. Let $t_n = \frac{T}{2^n}$ for $n \geq 0$. Then we can write $\xi(t)$ as:
$$ \xi(t) = \lim_{n \to \infty} \sum_{k=1}^{2^n} W(t_k)(W(t_k) - W(t_{k-1})) $$
Since $W(t)$ is adapted to
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