Proof of Itô Isometry

We prove $\mathbb{E}[(\int_0^T H_t \, dB_t)^2] = \mathbb{E}[\int_0^T H_t^2 \, dt]$ for simple processes, then extend by density.

Step 1: For simple $H_t = \sum_{i=0}^{n-1} H_i \mathbf{1}_{[t_i, t_{i+1})}$, expand the square:

$\left(\int_0^T H_t \, dB_t\right)^2 = \left(\sum_i H_i \Delta B_i\right)^2 = \sum_i H_i^2 (\Delta B_i)^2 + 2\sum_{i<j} H_i H_j \Delta B_i \Delta B_j.$

Step 2: Take expectations. The cross terms vanish by independence: $\mathbb{E}[H_i H_j \Delta B_i \Delta B_j] = \mathbb{E}[H_i H_j] \mathbb{E}[\Delta B_i] \mathbb{E}[\Delta B_j] = 0$ for $i \neq j$.

Step 3: For the diagonal terms: $\mathbb{E}[H_i^2 (\Delta B_i)^2] = \mathbb{E}[H_i^2] (t_{i+1} - t_i)$.

Summing over $i$ gives the isometry. Extension to general $L^2$ processes follows by approximation and continuity of the $L^2$ norm.