Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 211 (Advanced Financial Models)

A portfolio $H\in {\mathbb{R}}^n$ is an arbitrage when $H\cdot P_0\le 0$, $H\cdot P_1\ge 0$ almost surely, and at least one inequality supplies a strict gain: either $H\cdot P_0<0$ or $\mathbb{P}(H\cdot P_1>0)>0$. It is a terminal-consumption arbitrage when $H\cdot P_0=0$, $H\cdot P_1\ge 0$ almost surely, and the terminal inequality is strict with positive probability.

A numéraire portfolio $\eta$ satisfies $\eta \cdot P_0>0$ and $\eta \cdot P_1>0$ almost surely. If an arbitrage $H$ already has zero initial cost, it is a terminal-consumption arbitrage. Otherwise $H\cdot P_0<0$; setThen $\tilde{H}\cdot P_0=0$, while its terminal payoff is the nonnegative payoff of $H$ plus a strictly positive multiple of $\eta \cdot P_1$. Hence it is strictly positive almost surely and is a terminal-consumption arbitrage.

If $A=\mathrm{Im}V={\mathbb{R}}^n$, the symmetric covariance matrix $V$ is positive definite. For every nonzero $H$, the scalar $H\cdot P_1$ is normal with variance $H^{T}VH>0$, so it has positive probability of being negative. It cannot be an arbitrage payoff. The zero portfolio provides no strict gain, proving absence of arbitrage.

For a numéraire portfolio $\eta$, the normal random variable $\eta \cdot P_1$ is strictly positive almost surely. A nondegenerate normal variable has support on all of $\mathbb{R}$, so it must be degenerate: ${\eta }^{T}V\eta =0$, equivalently $V\eta =0$. Thus $\eta \cdot P_1=\eta \cdot \mu >0$ deterministically. The scaled portfoliohas terminal value $B\cdot P_1=1$ almost surely and therefore replicates a risk-free bond.

Let $d=\mu -(1+r)P_0$. Since $V$ is symmetric, $A=\mathrm{Im}V=(\ker V{)}^{\perp }$. If $d\in A$, every $H\in \ker V$ satisfiesAny $H\notin \ker V$ has a nondegenerate normal terminal value and cannot be nonnegative almost surely. Any $H\in \ker V$ has the displayed deterministic relation, which excludes an arbitrage because $1+r=(\eta \cdot \mu )/(\eta \cdot P_0)>0$.

Conversely, if $d\notin A$, choose $h\in \ker V$ with $h\cdot d>0$. The zero-cost portfoliohas deterministic terminal payoffIt is a terminal-consumption arbitrage. Thus no arbitrage is equivalent to $d\in A$, the Arbitrage in a one-period Gaussian market criterion.

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:

The bond pays one at maturity. If $\tau =\inf \{t<T:P_t^T\le 0\}$ had positive probability of being finite, buy one bond at $\tau$. A negative price supplies immediate consumption and a positive terminal payoff; a zero price supplies a free positive terminal payoff. Trading only on the stopping event gives an arbitrage. Therefore $P_t^T>0$ almost surely for every $t<T$.

For $K_1<K_2$, the lower-strike payoff dominates:If $C_t^{T,K_1}<C_t^{T,K_2}$, buy the cheaper lower-strike call and sell the higher-strike call. This gives positive initial consumption and a nonnegative terminal payoff, an arbitrage. Hence the monotonicity of a European call price in strike gives $C_t^{T,K_1}\ge C_t^{T,K_2}$.

First, no arbitrage implies the lower boundotherwise buy the call and $K$ maturity-$(T+1)$ bonds and short one non-dividend-paying stock; the initial receipt is positive and the terminal payoff is nonnegative. At time $T$, the assumption $P_T^{T+1}\le 1$ therefore gives $C_T^{T+1,K}\ge (S_T-K{)}^{+}$.

If $C_t^{T,K}>C_t^{T+1,K}$, sell the shorter call and buy the longer one. At time $T$, the longer call's no-arbitrage value covers the shorter call's payoff, with a strictly positive initial receipt. This is impossible, so $T\mapsto C_t^{T,K}$ is nondecreasing.

Order the support as $K_1<\cdots <K_N$ and putOn the finite support,This follows by telescoping: at $S_T=K_j$, only terms through $j$ survive and reconstruct successive increments of $g$. The static replication on a finite terminal support therefore has no-arbitrage price

Let $B_t=e^{rt}$ and let ${\psi }_t$ be the bank-account holding. Then $X_t={\psi }_t{B}_t+{\theta }_t{S}_t$. The self-financing portfolio condition gives

SetBy the Girsanov theorem, $W_t^Q=W_t+\lambda t$ is Brownian motion under $Q$. Consequentlyso discounted stock price is a martingale and $Q$ is the Risk-neutral measure for the Black-Scholes model.

Define the risk-neutral claim valueThen $V(0,S_0)=x$, $V(T,s)=g(s)$, and the Black-Scholes equation holds. Differentiation under the integral gives the deltaA Gaussian shift $Z\mapsto Z+\sigma \sqrt{\tau }$ rewrites this aswhich is exactly the stated ${\theta }_t$ at $s=S_t$.

Apply Itô formula to $V(t,S_t)$. The PDE givesThis is the same wealth equation as part a, with the same initial value $x$. Uniqueness therefore gives $X_t^{x,\theta }=V(t,S_t)$ and hence