If such $\rho$ existed, then integrability and $H\cdot X\geq0$ would give
$$
0=H\cdot\mathbb E[\rho X]
=\mathbb E[\rho(H\cdot X)].
$$
But $\rho>0$ 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 \b[no such positive state-price density exists].


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-211/1/a/solution