For $c\in\{0,1\}$ set
$$
\zeta_c=\exp(cY-Y^2-\|X\|^2).
$$
The quadratic negative terms make
$$
F_c(h)=\mathbb E[e^{-h\cdot X}\zeta_c]
$$
everywhere 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) supplies
$$
\rho_c=\frac{
\exp(-h_c\cdot X+cY-Y^2-\|X\|^2)}{F_c(h_c)}
$$
with $\mathbb E\rho_c=1$ and $\mathbb E(\rho_cX)=0$. Uniqueness forces $\rho_0=\rho_1$. Taking logarithms and cancelling the common quadratic terms gives
$$
Y=(h_1-h_0)\cdot X+\log F_1(h_1)-\log F_0(h_0).
$$
Thus
$$
\boxed{Y=a+b\cdot X}
$$
with $b=h_1-h_0$ and $a=\log F_1(h_1)-\log F_0(h_0)$. Since $Y$ was arbitrary, the one-period market is complete.


Solution

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

 Solution

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