Solution: Optional variation process of a transformed martingale

We use a similar approach as the proof of \((2.14)\) in the book. From \((2.13)\) we have \((H\bullet M)_n = \sum_{i=1}^nH_i\Delta M_i\) and thus \(\Delta (H \bullet M)_i = H_i \Delta M_i\). From \((2.8)\) we have \([M]_n = \sum_{i=1}^n(\Delta M_i)^2\) and so \(\Delta [M]_i = (\Delta M_i)^2\).

Using \((2.8)\) for the process \((H\bullet M)\) we thus get \begin{align*}[H\bullet M]_n &= \sum_{i=1}^n(\Delta (H \bullet M)_i)^2 \\ &= \sum_{i=1}^n(H_i \Delta M_i)^2 \\ &= \sum_{i=1}^nH_i^2 (\Delta M_i)^2 \\ &= \sum_{i=1}^nH_i^2 \Delta [M]_i,\end{align*} which proves the statement.