Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 211 / 1 / d / Solution

For $c\in \{0,1\}$ setThe quadratic negative terms makeeverywhere finite and smooth. Existence of the assumed $\rho$ rules out the arbitrage direction in part (c) by part (a). Hence each $F_c$ has a bounded minimizing sequence, and part (b) supplieswith $\mathbb{E}{\rho }_c=1$ and $\mathbb{E}({\rho }_{c}X)=0$. Uniqueness forces ${\rho }_0={\rho }_1$. Taking logarithms and cancelling the common quadratic terms givesThuswith $b=h_1-h_0$ and $a=\log F_1(h_1)-\log F_0(h_0)$. Since $Y$ was arbitrary, the one-period market is complete.