Suppose a numéraire strategy $\eta$ existed and write $N_t=\eta_{t+1}\cdot P_t=\eta_t\cdot P_t>0$, using zero consumption. Since $\eta_{t+1}$ is $\mathcal F_t$-measurable and $M_t=(-1)^tZ_tP_t$ is a martingale,
$$
\mathbb E[\eta_{t+1}\cdot M_{t+1}\mid\mathcal F_t]
=\eta_{t+1}\cdot M_t.
$$
Thus
$$
-(-1)^t\mathbb E[Z_{t+1}N_{t+1}\mid\mathcal F_t]
=(-1)^tZ_tN_t,
$$
or
$$
\mathbb E[Z_{t+1}N_{t+1}\mid\mathcal F_t]
=-Z_tN_t.
$$
The left side is nonnegative and the right side nonpositive, so $Z_tN_t=0$ almost surely. Strict positivity of $N_t$ implies $Z_t=0$ almost surely for every $t$, contradicting
$$
\mathbb P(Z_t=0\text{ for all }t)=0.
$$
Therefore \b[the market has no numéraire strategy].


Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 211 / 4 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-211/4/c/solution