Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 211 (Advanced Financial Models)

If such $\rho$ existed, then integrability and $H\cdot X\ge 0$ would giveBut $\rho >0$ 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 no such positive state-price density exists.

Let $(h_k)$ be a bounded minimizing sequence. A subsequence converges to some $h_{\ast }$, and continuity gives $F(h_{\ast })=f$. Since $F$ is smooth and $h_{\ast }$ is an unconstrained minimizer,DefineThen $\rho >0$, $\mathbb{E}\rho =1$, andThus the normalized exponential tilt is the required state-price density.

We prove the contrapositive. LetThe function $F$ is constant along $N$, so minimize it on $N^{\perp }$. Suppose there is no $H$ with $H\cdot X\ge 0$ almost surely and strict inequality with positive probability. If a sequence $h_k\in N^{\perp }$ satisfies $\Vert h_k\Vert \to \infty$, pass to a subsequence withBecause $H\notin N$ and there is no arbitrage direction, $\mathbb{P}(H\cdot X<0)>0$. On that event, $e^{-h_k\cdot X}\zeta \to \infty$, and Fatou lemma gives ${\mathrm{liminf}}_{k}F(h_k)=\infty$. Hence every finite sublevel set of $F$ in $N^{\perp }$ is bounded. It is also closed, so $F$ attains its infimum there and has a bounded minimizing sequence. This contradicts the assumption. Therefore there is a unit vector $H$ satisfying

For $c\in \{0,1\}$ setThe quadratic negative terms makeeverywhere 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) supplieswith $\mathbb{E}{\rho }_c=1$ and $\mathbb{E}({\rho }_{c}X)=0$. Uniqueness forces ${\rho }_0={\rho }_1$. Taking logarithms and cancelling the common quadratic terms givesThuswith $b=h_1-h_0$ and $a=\log F_1(h_1)-\log F_0(h_0)$. Since $Y$ was arbitrary, the one-period market is complete.

For $\theta =p+iq$ with $0\le p\le 1$,The elementary inequality $x^p\le 1+x$ for $x>0$ givesTherefore the Mellin transform $M(\theta )=\mathbb{E}[S^{\theta }]$ is absolutely well-defined throughout the strip and

HereThe Laplace transform of an exponential distribution is $\mathbb{E}[e^{-sX}]=(1+s{)}^{-1}$ for $\mathrm{Re}s>-1$. Since $\mathrm{Re}(t\theta )\ge 0$,

By the bound in part (a), Fubini's theorem applies. Conditional on $S$,where the Characteristic function of the Cauchy distribution was used. Since $\mathbb{E}S=1$,The formula expresses a European call option value through complex moments of $S$. In an affine stochastic-volatility model such as the Heston model, those moments are available from an explicit transform, so call prices reduce to a one-dimensional Fourier expectation or integral.

Condition on $S$ and use the characteristic function of a standard normal distribution:The integrand has modulus one, so Fubini's theorem is immediate. Taking expectation over $S$ yields

A local martingale deflator makes both $YB$ and $YS$ local martingales. Since the filtration is generated by $W$, the martingale representation theorem and the finite-variation drift forced by $YB$ givefor a continuous adapted $\lambda$, where $\eta =-Y\lambda$. Applying the Itô product rule to $YS$ gives driftIt vanishes exactly when

SetThe boundedness of ${\xi }_T$ and positivity of the deflator make $M$ a nonnegative true martingale. By the Brownian martingale representation theorem, $d{M}_t={\eta }_{t}d{W}_t$. A self-financing wealth process with stock holding ${\pi }_t$ satisfiesThe product $YV$ then has diffusion coefficientChooseThen $Y_t{V}_t=M_t$, so $V_T={\xi }_T$ and the strategy replicates the claim. It is admissible because $V=M/Y$ is nonnegative.

For any other admissible replicating wealth $\tilde{V}$, the nonnegative local martingale $Y\tilde{V}$ is a supermartingale. HenceThe constructed strategy hasso this is the minimal replication cost.

For constant coefficients, the density process $Y_t{B}_t/B_0$ is a true exponential martingale and defines the risk-neutral measure $Q$. Under $Q$,The minimal value process is thereforeThe Markov property and the lognormal transition law make this a deterministic function of $(t,s)$, and

Apply Itô formula to $V(t,S_t)$. Its Brownian coefficient isThe self-financing portfolio's Brownian coefficient is ${\pi }_t\sigma S_t$. Since $S_t\sigma >0$, equality of the two value processes forcesThus the stock holding is the claim's option delta.

For initial capital zero and a predictable strategy $H$, let $C_t^{0,H}$ denote consumption after the time-$t$ portfolio payoff and before choosing the next holdings. An investment-consumption arbitrage haswith strictly positive consumption at some date with positive probability. A terminal-consumption arbitrage is a finite-horizon such strategy whose consumption is zero before its terminal date $T$, while $C_T^{0,H}\ge 0$ almost surely and $\mathbb{P}(C_T^{0,H}>0)>0$.

A numéraire strategy $\eta$ has zero consumption and strictly positive wealthat every date. Given an investment-consumption arbitrage $H$, retain its holdings and invest each nonnegative consumption $C_s^{0,H}$ in the numéraire. Withthe self-financing identity for $\eta$ gives zero intermediate consumption for $K$. At a deterministic $T$ after a date at which positive consumption occurs with positive probability, liquidating givesIt is strictly positive with positive probability. Hence $K$ is a terminal-consumption arbitrage.

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 $M_t=(-1{)}^t{Z}_t{P}_t$ is a martingale,ThusorThe left side is nonnegative and the right side nonpositive, so $Z_t{N}_t=0$ almost surely. Strict positivity of $N_t$ implies $Z_t=0$ almost surely for every $t$, contradictingTherefore the market has no numéraire strategy.

At maturity, $P_T^T=1$. If $P_t^T\le 0$ on an event $A\in {\mathcal{F}}_t$ of positive probability, buying one bond on $A$ has nonpositive cost and certain payoff $1$ on $A$ at $T$; any negative purchase cost can also be consumed or retained. This is an arbitrage. Therefore absence of arbitrage implies

The one-period spot interest rate is defined byand the bank account byA probability measure $Q$ equivalent to the physical measure is a risk-neutral measure when every discounted zero-coupon bond priceis a $Q$-martingale. Equivalently,

If $T\mapsto P_t^T$ is nonincreasing, thenso $1+r_t=(P_t^{t+1}{)}^{-1}\ge 1$ and $r_t\ge 0$.

Conversely, if every spot rate is nonnegative, then $B_{T+1}=B_T(1+r_T)\ge B_T$. Under a risk-neutral measure,Hence

Risk-neutral valuation givesFor $s\ge t$,ThereforeThe future ${\zeta }_j$ are independent and identically distributed under the stated model, sowith an empty product equal to one.

Apply the multidimensional Itô formula to ${\xi }_t=F(t,S_t,v_t)$. The stated PDE cancels its drift to $rFdt$, leavingConsequently $e^{-rt}{S}_t$ and $e^{-rt}{\xi }_t$ are local martingales under the physical measure $P$. Thus $P$ itself is an equivalent local martingale measure for the augmented market relative to the bank account. The continuous-time fundamental theorem of asset pricing says that existence of such a measure for locally bounded prices implies no free lunch with vanishing risk, and hence no arbitrage. The terminal condition also gives ${\xi }_T=\sqrt{S_T}$ as required.

Forone hasSubstitution into the PDE and collection of the coefficient of $v$ give the Riccati differential equationThe terminal condition also requires $B(T)=0$.

The terms independent of $v$ in the substituted PDE satisfysoIntegrating backward from $T$ yieldsThus the requested constant is