 Solution

Write $\Delta M_n=M_n-M_{n-1}$. Conditional on ${\mathcal{F}}_{n-1}$, the variables $X_{n-1}$, $M_{n-1}$, and $A_n$ are known, while $\mathbb{E}[\Delta M_n\mid {\mathcal{F}}_{n-1}]=0$. HenceThe denominator is positive almost surely, so the recovery of a martingale-transform integrand by conditional covariance gives