= Solution
For a bounded <simple predictable process> $H$ and a <continuous martingale> $M\in\mathcal H^{2,c}$, the <Itô isometry> states
$$
\boxed{\mathbb E\left|\int_0^tH_s\,dM_s\right|^2
=\mathbb E\int_0^tH_s^2\,d\langle M\rangle_s.}
$$
Here $\langle M\rangle$ is the <predictable quadratic variation>, equal to $[M]$ for a <continuous martingale>. On the whole half-line, the terminal form is
$$
\mathbb E|(H\mathbin\cdot M)_\infty|^2
=\mathbb E\int_0^\infty H_s^2\,d\langle M\rangle_s.
$$
The <Itô integral> starts at zero even when $M_0\ne0$. For <Brownian motion>, $d\langle B\rangle_s=ds$.
For completeness, cross terms of the finite increment sum have <expectation> zero by conditioning on the earlier endpoint of the later increment. Also $M^2-\langle M\rangle$ is a <martingale>, so the remaining terms satisfy
$$
\mathbb E[\xi_j^2(M_v-M_u)^2]
=\mathbb E[\xi_j^2(\langle M\rangle_v-\langle M\rangle_u)]
\quad(u=t_j<t,\ v=t_{j+1}\wedge t).
$$
This proves the displayed identity without a change-of-variables formula. Completing bounded <simple predictable processes> in the norm $(\mathbb E\int H^2d\langle M\rangle)^{1/2}$ defines the <Itô integral> of every <square-integrable stochastic integrand>. This completion will justify the unbounded Brownian integrands below.
Back to article page
