Itô isometry

Remarks

For a simple process $H = \sum_i \xi_i \mathbf{1}_{(t_i, t_{i+1}]}$, expand the square of $\sum_i \xi_i \Delta B_i$. A cross term $E[\xi_i \xi_j \Delta B_i \Delta B_j]$ with $i < j$ vanishes, because $\Delta B_j$ has mean zero and is independent of $\mathcal{F}_{t_j}$. A diagonal term equals $E[\xi_i^2] (t_{i+1} - t_i)$, because $\Delta B_i$ is independent of $\mathcal{F}_{t_i}$ with variance $t_{i+1} - t_i$. The general case follows by approximation, and this is how the integral is defined. Example: for $H_t = B_t$ with $B_0 = 0$, the right side is $\int_0^T t \, dt = T^2/2$, and indeed $\int_0^T B_t \, dB_t = \tfrac12 (B_T^2 - T)$ has variance $\tfrac14 (3T^2 - 2T^2 + T^2) = T^2/2$.

Going deeper

For a student

The isometry is the key step in the construction of the Itô integral: it extends the integral from simple processes to all square-integrable adapted ones. Its polarised form gives $E\left[\int H \, dB \int K \, dB\right] = E\left[\int H K \, dt\right]$. For a general continuous martingale $M$ in place of $B$, the same identity holds with $dt$ replaced by $d\langle M \rangle_t$.