Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 211 / 2

A one-period martingale deflator is a pair $Y_0>0$, $Y_1>0$ such thatIf $Y^0,Y^1$ are deflators and ${\epsilon }_0,{\epsilon }_1>0$, their positive linear combination is strictly positive andIt is therefore another martingale deflator.

The one-period fundamental theorem of asset pricing, applied as a separating-hyperplane theorem to the cone of attainable payoffs, gives the superhedging dualitywhere the supremum is over martingale deflators. The assumed strict inequalities make the right side strictly below ${\xi }_0$. Hence some $H$ satisfies $H\cdot P_0\le {\xi }_0$ and $H\cdot P_1\ge {\xi }_1$ almost surely.

Apply part b with initial capital ${\xi }_0+\epsilon$ and let $\epsilon \downarrow 0$. The finite-dimensional attainable space is closed, so a limit portfolio $H$ satisfies $H\cdot P_0\le {\xi }_0$ and $H\cdot P_1\ge {\xi }_1$. Let $\bar{Y}$ be a deflator for which equality holds. ThenStrict positivity of ${\bar{Y}}_1$ forces $H\cdot P_1={\xi }_1$ almost surely, and the middle identity then gives $H\cdot P_0={\xi }_0$.

By definition of the concave conjugate, $u(x)\le \hat{u}(y)+xy$ for all $x,y>0$. Put $x=H\cdot P_1$ and $y=Y_1$, take expectations, and use the deflator identity:The maximizing condition in the definition of $\hat{u}$ is $u^{\prime }(x)=y$, so equality holds when $u^{\prime }(H\cdot P_1)=Y_1$ almost surely. This is utility duality with martingale deflators.

For every deflator $Y$ and $\epsilon \ge 0$, $Y^{\ast }+\epsilon Y$ is a deflator. Minimality and right differentiation at zero giveTaking $Y=Y^{\ast }$ and varying the positive scalar multiple $(1+\epsilon )Y^{\ast }$ in both directions around one gives equality.

Set ${\xi }_1=-{\hat{u}}^{\prime }(Y_1^{\ast })$ and ${\xi }_0=X_0$. The preceding inequality saysfor every deflator, with equality at $Y^{\ast }$. Part c produces $H^{\ast }$ with $H^{\ast }\cdot P_0=X_0$ and $H^{\ast }\cdot P_1={\xi }_1$. The inverse relation between conjugate derivatives gives $u^{\prime }({\xi }_1)=Y_1^{\ast }$, so equality holds in part d: