Solution
= Solution
Predictability gives $E[H_k(X_k-X_{k-1})]=E[H_kE(X_k-X_{k-1}\mid\mathcal F_{k-1})]\le0$. Sum over $k\le n$.
= Solution
Predictability gives $E[H_k(X_k-X_{k-1})]=E[H_kE(X_k-X_{k-1}\mid\mathcal F_{k-1})]\le0$. Sum over $k\le n$.