= Solution
Put $t_j=j2^{-n}$ and $\Delta_j=X_{t_{j+1}}-X_{t_j}$. Telescoping $X_{t_{j+1}}^2-X_{t_j}^2=2X_{t_j}\Delta_j+\Delta_j^2$ gives
$$
M_1^{(n)}=\sum_{j=0}^{2^n-1}X_{t_j}\Delta_j.
$$
The <martingale transform> summands are orthogonal in $L^2$: for an earlier summand, conditioning on the sigma-algebra at the start of the later increment makes the cross expectation zero. Hence
$$
\mathbb E(M_1^{(n)})^2=\sum_j\mathbb E[X_{t_j}^2\Delta_j^2]\le C^2\sum_j\mathbb E\Delta_j^2.
$$
The <martingale> increments themselves are also orthogonal. Since $X_0=0$, their variance sum equals $\mathbb E X_1^2\le C^2$. Therefore
$$
\boxed{\mathbb E(M_1^{(n)})^2\le C^4.}
$$
Only discrete martingale orthogonality is used here; no pre-existing <quadratic variation> calculation is needed.
Back to article page
