Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 211 / 1 / h / Solution

Let $Y$ and $\tilde{Y}$ be normalized martingale deflators. For any date $T$ and bounded ${\mathcal{F}}_T$-measurable $X_T$, condition g supplies a strategy whose only prescribed cash flow is $X_T$ at $T$. Applying the martingale identity from part b givesThus $\mathbb{E}[(Y_T-{\tilde{Y}}_T)X_T]=0$ for every bounded ${\mathcal{F}}_T$-measurable $X_T$. Taking indicators, or the sign of the difference, shows $Y_T={\tilde{Y}}_T$ almost surely. Since $T$ was arbitrary, the normalized martingale deflator is unique.