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

If such $\rho$ existed, then integrability and $H\cdot X\ge 0$ would giveBut $\rho >0$ almost surely and $H\cdot X$ is nonnegative and strictly positive with positive probability. Hence $\rho (H\cdot X)$ is nonnegative and nonzero with positive probability, so its expectation is strictly positive. This contradiction proves that no such positive state-price density exists.

Let $(h_k)$ be a bounded minimizing sequence. A subsequence converges to some $h_{\ast }$, and continuity gives $F(h_{\ast })=f$. Since $F$ is smooth and $h_{\ast }$ is an unconstrained minimizer,DefineThen $\rho >0$, $\mathbb{E}\rho =1$, andThus the normalized exponential tilt is the required state-price density.

We prove the contrapositive. LetThe function $F$ is constant along $N$, so minimize it on $N^{\perp }$. Suppose there is no $H$ with $H\cdot X\ge 0$ almost surely and strict inequality with positive probability. If a sequence $h_k\in N^{\perp }$ satisfies $\Vert h_k\Vert \to \infty$, pass to a subsequence withBecause $H\notin N$ and there is no arbitrage direction, $\mathbb{P}(H\cdot X<0)>0$. On that event, $e^{-h_k\cdot X}\zeta \to \infty$, and Fatou lemma gives ${\mathrm{liminf}}_{k}F(h_k)=\infty$. Hence every finite sublevel set of $F$ in $N^{\perp }$ is bounded. It is also closed, so $F$ attains its infimum there and has a bounded minimizing sequence. This contradicts the assumption. Therefore there is a unit vector $H$ satisfying

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.